Chapter 05 Review Problems

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Stoichiometry

Balance the following equation.

\[ \begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{NH_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Cl_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NCl_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{HCl} \end{align*} \]

Additionally, what mass (in g) of HCl is produced if 1.29 g of NH₃ reacts with 4.50 g of Cl₂?

Solution

Balance the following equation.

\[ \begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_3H_6} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NH_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_3H_3N} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} \end{align*} \]

Additionally, determine the following:

What is the limiting reactant if 4.20 g of C₃H₆ reacts with 3.10 g of NH₃ and 6.10 g of O₂?
What mass (in g) of C₃NH₃ can theoretically be produced given the information in (a)?
What mass (in g) of C₃H₃N, NH₃, and O₂ would theoretically be leftover given the information in (a)?
If 1.90 g of C₃H₃N was produced in an experiment, what is the percent yield?

Solution

Answer:

The balanced chemical reaction is

\[ \begin{align*} \mathrm{2~C_3H_6} + \mathrm{2~NH_3} + \mathrm{3~O_2} \longrightarrow \mathrm{2~C_3H_3N} + \mathrm{6~H_2O} \end{align*} \]

C₃H₆
5.29 g C₃NH₃
0 g C₃H₆; 1.40 g NH₃; 1.31 g O₂
35.9 %

Concept: balancing equations; limiting reactant; percent yield; stoichiometry

a. Limiting Reactant

First, determine the limiting reactant by calculating the moles of the product, C₃H₃N, that can be formed from the given mass of each reactant.

\[ \begin{align*} n(\mathrm{C_3H_3N}) &= n(\mathrm{C_3H_6}) ~ r(\mathrm{C_3H_3N,C_3H_6}) \\[1.5ex] &= m(\mathrm{C_3H_6}) ~ M(\mathrm{C_3H_6})^{-1} ~ r(\mathrm{C_3H_3N,C_3H_6}) ~ \\[1.5ex] &= 4.2\bar{0}~\mathrm{g~C_3H_6} ~ \left ( \dfrac{\mathrm{mol~C_3H_6}}{\mathrm{42.0\bar{9}~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3H_3N}}{2~\mathrm{mol~C_3H_6}} \right )\\[1.5ex] &= 0.099\bar{7}86~\mathrm{mol} \\[1.5ex] n(\mathrm{C_3H_3N}) &= n(\mathrm{NH_3}) ~ r(\mathrm{C_3H_3N,NH_3}) \\[1.5ex] &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{C_3H_3N,NH_3}) ~ \\[1.5ex] &= 3.1\bar{0}~\mathrm{g~NH_3} ~ \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.0\bar{4}~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3H_3N}}{2~\mathrm{mol~NH_3}} \right )\\[1.5ex] &= 0.18\bar{1}92~\mathrm{mol} \\[3ex] n(\mathrm{C_3H_3N}) &= n(\mathrm{O_2}) ~ r(\mathrm{C_3H_3N,O_2}) \\[1.5ex] &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{C_3H_3N,O_2}) ~ \\[1.5ex] &= 6.1\bar{0}~\mathrm{g~O_2} ~ \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.0\bar{0}~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3H_3N}}{3~\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 0.12\bar{7}08~\mathrm{mol} \end{align*} \]

Since C₃H₆ produces the fewest moles of product, C₃H₆ is the limiting reactant.

b. Theoretical Yield of C₃H₃N

The theoretical yield is the mass of product formed from the limiting reactant.

\[ \begin{align*} m(\mathrm{C_3H_3N}) &= 0.099\bar{7}86~\mathrm{mol} \left ( \dfrac{53.0\bar{6}~\mathrm{g}}{\mathrm{mol~C_3H_3N}} \right ) \\[1.5ex] &= 5.2\bar{9}46~\mathrm{g} \\[1.5ex] &= 5.29~\mathrm{g} \end{align*} \]

5.29 g of C₃H₃N can be produced.

c. Mass of Leftover Reactants

The amount of each excess reactant consumed is determined by the amount of product formed.

\[ \begin{align*} m(\mathrm{C_3H_6})_{\mathrm{leftover}} &= m(\mathrm{C_3H_6})_{\mathrm{initial}} - m(\mathrm{C_3H_6})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{C_3H_6})_{\mathrm{initial}} - \left [ n(\mathrm{C_3H_3N}) ~ r(\mathrm{C_3H_6,C_3H_3N}) ~ M(\mathrm{C_3H_6}) \right ] \\[1.5ex] &= 4.2\bar{0}~\mathrm{g~C_3H_6} - \left [ 0.099\bar{7}86~\mathrm{mol~C_3H_3N} \left ( \dfrac{\mathrm{2~mol~C_3H_6}}{2~\mathrm{mol~C_3H_3N}} \right ) \left ( \dfrac{42.0\bar{9}~\mathrm{g}}{\mathrm{mol~C_3H_6}} \right ) \right ] \\[1.5ex] &= 4.2\bar{0}~\mathrm{g} - 4.1\bar{9}99~\mathrm{g} \\[1.5ex] &= 0.0\bar{0}00~\mathrm{g} \\[1.5ex] &= 0~\mathrm{g} \\[3ex] m(\mathrm{NH_3})_{\mathrm{leftover}} &= m(\mathrm{NH_3})_{\mathrm{initial}} - m(\mathrm{NH_3})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{NH_3})_{\mathrm{initial}} - \left [ n(\mathrm{C_3H_3N}) ~ r(\mathrm{NH_3,C_3H_3N}) ~ M(\mathrm{NH_3}) \right ] \\[1.5ex] &= 3.1\bar{0}~\mathrm{g~NH_3} - \left [ 0.099\bar{7}86~\mathrm{mol~C_3H_3N} \left ( \dfrac{\mathrm{2~mol~NH_3}}{2~\mathrm{mol~C_3H_3N}} \right ) \left ( \dfrac{17.0\bar{4}~\mathrm{g}}{\mathrm{mol~NH_3}} \right ) \right ] \\[1.5ex] &= 3.1\bar{0}~\mathrm{g} - 1.7\bar{0}03~\mathrm{g} \\[1.5ex] &= 1.3\bar{9}97~\mathrm{g} \\[1.5ex] &= 1.40~\mathrm{g}\\[3ex] m(\mathrm{O_2})_{\mathrm{leftover}} &= m(\mathrm{O_2})_{\mathrm{initial}} - m(\mathrm{O_2})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{O_2})_{\mathrm{initial}} - \left [ n(\mathrm{C_3H_3N}) ~ r(\mathrm{O_2,C_3H_3N}) ~ M(\mathrm{O_2}) \right ] \\[1.5ex] &= 6.1\bar{0}~\mathrm{g~O_2} - \left [ 0.099\bar{7}86~\mathrm{mol~C_3H_3N} \left ( \dfrac{\mathrm{3~mol~O_2}}{2~\mathrm{mol~C_3H_3N}} \right ) \left ( \dfrac{32.0\bar{0}~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \right ] \\[1.5ex] &= 6.1\bar{0}~\mathrm{g} - 4.7\bar{8}97~\mathrm{g}\\[1.5ex] &= 1.3\bar{1}03~\mathrm{g} \\[1.5ex] &= 1.31~\mathrm{g} \end{align*} \]

d. Percent Yield

\[ \begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m(\mathrm{actual})}{m(\mathrm{theoretical})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{1.9\bar{0}~\mathrm{g}}{5.2\bar{9}46~\mathrm{g}}\right ) \times 100~\% \\[1.5ex] &= 35.\bar{8}85~\% \\[1.5ex] &= 35.9~\% \end{align*} \]

Balance the following equation.

\[ \begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_5NO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_{14}O_4} ~\overset{\mathrm{Zn}}{ \underset{\mathrm{KOH}}{\longrightarrow}}~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{(C_6H_5N)_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_{12}O_4} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} \end{align*} \]

Additionally, determine the following:

What volume (in mL) of C₆H₅NO₂ (ρ = 1.20 g mL^–1) must be allowed to react with an excess of C₆H₁₄O₄ to produce 6.25 g of (C₆H₅N)₂ if the percent yield is 83.7 %?
If 0.16 L of C₆H₅NO₂ (ρ = 1.20 g mL^–1) and 0.51 L C₆H₁₄O₄ (ρ = 1.12 g mL^–1) react to yield 64.3 g of (C₆H₅N)₂, what is the limiting reactant and what is the percent yield of the reaction?

Solution

Answer:

\[ \begin{align*} \mathrm{2~C_6H_5NO_2} + \mathrm{4~C_6H_{14}O_4} \longrightarrow \mathrm{(C_6H_5N)_2} + \mathrm{4~C_6H_{12}O_4} + \mathrm{4~H_2O} \end{align*} \]

8.41 mL C₆H₅NO₂
C₆H₅NO₂ is limiting; 45 %

Concept: balancing equations; limiting reactant; percent yield; stoichiometry

First, determine the theoretical mass of (C₆H₅N)₂ needed to produce an actual yield of 6.25 g.

\[ \begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{actual}}}{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{theoretical}}} \right ) \times 100~\% \longrightarrow \\[1.5ex] m[\mathrm{(C_6H_5N)_2}]_{\mathrm{theoretical}} &= \left ( \dfrac{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{actual}}}{\%~\mathrm{yield}} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{6.2\bar{5}~\mathrm{g}}{83.\bar{7}~\%} \right ) \times 100~\% \\[1.5ex] &= 7.4\bar{6}71~\mathrm{g} \end{align*} \]

Now, calculate the volume of C₆H₅NO₂ required to produce this theoretical mass.

\[ \begin{align*} V(\mathrm{C_6H_5NO_2}) &= m(\mathrm{C_6H_5NO_2}) ~ \rho (\mathrm{C_6H_5NO_2})^{-1} \\[1.5ex] &= n[\mathrm{C_6H_5NO_2}] ~ M(\mathrm{C_6H_5NO_2}) ~ \rho (\mathrm{C_6H_5NO_2})^{-1} \\[1.5ex] &= n[\mathrm{(C_6H_5N)_2}] ~ r[\mathrm{C_6H_5NO_2,(C_6H_5N)_2}] ~ M(\mathrm{C_6H_5NO_2}) ~ \rho (\mathrm{C_6H_5NO_2})^{-1} \\[1.5ex] &= m[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]^{-1} ~ r[\mathrm{C_6H_5NO_2,(C_6H_5N)_2}] ~ M(\mathrm{C_6H_5NO_2}) ~ \rho(\mathrm{C_6H_5NO_2})^{-1}\\[1.5ex] &= 7.4\bar{6}71~\mathrm{g~(C_6H_5N)_2} ~ \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{182.2\bar{4}~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_6H_5NO_2}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \\ &\phantom{=}~ \left ( \dfrac{123.1\bar{2}~\mathrm{g}}{\mathrm{mol~C_6H_5NO_2}} \right ) \left ( \dfrac{\mathrm{mL~}}{1.2\bar{0}~\mathrm{g~C_6H_5NO_2}} \right ) \\[1.5ex] &= 8.4\bar{0}78~\mathrm{mL} \\[1.5ex] &= 8.41~\mathrm{mL} \end{align*} \]

First, find the limiting reactant by calculating the theoretical mass of (C₆H₅N)₂ produced from each reactant.

From C₆H₅NO₂:

\[ \begin{align*} m[\mathrm{(C_6H_5N)_2}] &= n[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]\\[1.5ex] &= n[\mathrm{C_6H_5NO_2}] ~ r[\mathrm{(C_6H_5N)_2,C_6H_5NO_2}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= m[\mathrm{C_6H_5NO_2}] ~ M(\mathrm{C_6H_5NO_2})^{-1}~ r[\mathrm{(C_6H_5N)_2,C_6H_5NO_2}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= V(\mathrm{C_6H_5NO_2}) ~ \rho(\mathrm{C_6H_5NO_2}) ~ M(\mathrm{C_6H_5NO_2})^{-1} ~ r[\mathrm{(C_6H_5N)_2,C_6H_5NO_2}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= 0.1\bar{6}~\mathrm{L~C_6H_5NO_2} ~ \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.2\bar{0}~\mathrm{g~C_6H_5NO_2}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~C_6H_5NO_2}}{123.1\bar{2}~\mathrm{g}} \right ) \\ &\phantom{=}~ \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{2~\mathrm{mol~C_6H_5NO_2}} \right ) \left ( \dfrac{182.2\bar{4}~\mathrm{g}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \\[1.5ex] &= 1\bar{4}2.0~\mathrm{g} \end{align*} \]

From C₆H₁₄O₄:

\[ \begin{align*} m[\mathrm{(C_6H_5N)_2}] &= n[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]\\[1.5ex] &= n[\mathrm{C_6H_{14}O_4}] ~ r[\mathrm{(C_6H_5N)_2,C_6H_{14}O_4}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= m[\mathrm{C_6H_{14}O_4}] ~ M[\mathrm{C_6H_{14}O_4}]^{-1} ~ r[\mathrm{(C_6H_5N)_2,C_6H_{14}O_4}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= V(\mathrm{C_6H_{14}O_4}) ~ \rho(\mathrm{C_6H_{14}O_4}) ~ M(\mathrm{C_6H_{14}O_4})^{-1} ~ r[\mathrm{(C_6H_5N)_2,C_6H_{14}O_4}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= 0.5\bar{1}~\mathrm{L~C_6H_{14}O_4} ~ \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.1\bar{2}~\mathrm{g~C_6H_{14}O_4}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~C_6H_{14}O_4}}{150.2\bar{0}~\mathrm{g}} \right ) \\ &\phantom{=}~ \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{4~\mathrm{mol~C_6H_{14}O_4}} \right ) \left ( \dfrac{182.2\bar{4}~\mathrm{g}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \\[1.5ex] &= 1\bar{7}3.26~\mathrm{g} \end{align*} \]

Since C₆H₅NO₂ produces the smaller amount of product, it is the limiting reactant.

Next, calculate the percent yield using the actual yield of 64.3 g.

\[ \begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{actual}}}{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{theoretical}}} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{64.\bar{3}~\mathrm{g}}{1\bar{4}2.0~\mathrm{g}}\right ) \times 100~\% \\[1.5ex] &= 4\bar{5}.28~\% \\[1.5ex] &= 45~\% \end{align*} \]

Balance the following equation.

\[ \begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Fe} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Br_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Na_2CO_3} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NaBr} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{CO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Fe_3O_4} \end{align*} \]

Additionally, an alloy contains 81.2 % Fe and 18.8 % Ni (by mass). A 6.40 g sample of the alloy reacts with 8.50 g Na₂CO₃ with excess Br₂.

What mass (in g) of Fe₃O₄ can be produced?
What mass (in g) of Fe₃O₄ is produced if the percent yield of the reaction is 94.2 %?

Solution

Answer:

\[ \begin{align*} \mathrm{3~Fe} + \mathrm{4~Br_2} + 4~\mathrm{Na_2CO_3} \longrightarrow \mathrm{8~NaBr} + \mathrm{4~CO_2} + \mathrm{Fe_3O_4} \end{align*} \]

4.64 g Fe₃O₄
4.37 g Fe₃O₄

Concept: balancing equations; limiting reactant; percent yield; stoichiometry

a. Theoretical Yield

First, determine the limiting reactant by calculating the theoretical mass of Fe₃O₄ that can be produced from each reactant.

From the iron in the alloy:

\[ \begin{align*} m(\mathrm{Fe_3O_4}) &= n(\mathrm{Fe_3O_4}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= n(\mathrm{Fe}) ~ r(\mathrm{Fe_3O_4,Fe}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= m(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} ~ r(\mathrm{Fe_3O_4,Fe}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= m(\mathrm{sample}) ~ w(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} ~ r(\mathrm{Fe_3O_4,Fe}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= 6.4\bar{0}~\mathrm{g~sample}~ \left ( \dfrac{81.\bar{2}~\%~\mathrm{Fe}}{100~\%~\mathrm{sample}} \right ) \left ( \dfrac{\mathrm{mol~Fe}}{55.8\bar{5}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Fe_3O_4}}{\mathrm{3~mol~Fe}} \right ) \left ( \dfrac{231.5\bar{5}~\mathrm{g}}{\mathrm{mol~Fe_3O_4}} \right ) \\[1.5ex] &= 7.1\bar{8}18~\mathrm{g} \\[1.5ex] &= 7.18~\mathrm{g} \end{align*} \]

From Na₂CO₃:

\[ \begin{align*} m(\mathrm{Fe_3O_4}) &= n(\mathrm{Fe_3O_4}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= n(\mathrm{Na_2CO_3}) ~ r(\mathrm{Fe_3O_4,Na_2CO_3}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= m(\mathrm{Na_2CO_3}) ~ M(\mathrm{Na_2CO_3})^{-1} ~ r(\mathrm{Fe_3O_4,Na_2CO_3}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= 8.5\bar{0}~\mathrm{g~Na_2CO_3}~ \left ( \dfrac{\mathrm{mol~Na_2CO_3}}{105.9\bar{9}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Fe_3O_4}}{\mathrm{4~mol~Na_2CO_3}} \right ) \left ( \dfrac{231.5\bar{5}~\mathrm{g}}{\mathrm{mol~Fe_3O_4}} \right ) \\[1.5ex] &= 4.6\bar{4}23~\mathrm{g} \\[1.5ex] &= 4.64~\mathrm{g} \end{align*} \]

Since Na₂CO₃ produces the lesser amount of product, it is the limiting reactant. The theoretical yield is 4.65 g.

b. Actual Yield

Calculate the actual yield with an efficiency of 94.2 %.

\[ \begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m(\mathrm{Fe_3O_4})_{\mathrm{actual}}}{m(\mathrm{Fe_3O_4})_{\mathrm{theoretical}}} \right ) \times 100~\% \longrightarrow \\[1.5ex] m(\mathrm{Fe_3O_4})_{\mathrm{actual}} &= \left ( \dfrac{\mathrm{\%~yield}}{100~\%} \right )~ m(\mathrm{Fe_3O_4})_{\mathrm{theoretical}}\\[1.5ex] &= \left ( \dfrac{94.2~\%}{100~\%} \right ) \times 4.6\bar{4}23~\mathrm{g}~\mathrm{Fe_3O_4} \\[1.5ex] &= 4.3\bar{7}30~\mathrm{g} \\[1.5ex] &= 4.37~\mathrm{g} \end{align*} \]

A mixture contained no fluorine compound except methyl fluoroacetate, FCH₂COOCH₃ (M(FCH₂COOCH₃) = 92.08 g mol^–1). When chemically treated, all the fluorine was converted to CaF₂ (M(CaF₂) = 78.08 g mol^–1). The mass of CaF₂ obtained was 120.1 g. Find the mass (in g) of methyl fluoroacetate in the original mixture.

Solution

A 1.600 g sample of a metal chloride, MCl₂, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.412 g. Calculate the molar mass (in g mol^–1) of the metal, M²⁺, and identify the metal.

Solution

Consider four individual samples of phosphine (PH₃), water, hydrogen sulfide, and hydrogen fluoride, each with a mass of 221 g. Rank the compounds from the least to the greatest number of hydrogen atoms contained in each sample.

Solution

Answer: HF < H₂S < PH₃ < H₂O

Concept: moles; number of particles

PH₃

\[ \begin{align*} N(\mathrm{H}) &= n(\mathrm{H}) ~ N_{\mathrm{A}}\\[1.5ex] &= n(\mathrm{PH_3}) ~ r(\mathrm{H,PH_3}) ~ N_{\mathrm{A}}\\[1.5ex] &= m(\mathrm{PH_3}) ~ M(\mathrm{PH_3})^{-1} ~ r(\mathrm{H,PH_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 22\bar{1}~\mathrm{g~PH_3} ~ \left ( \dfrac{\mathrm{mol~PH_3}}{34.0\bar{0}~\mathrm{g}} \right ) \left ( \dfrac{3~\mathrm{mol~H}}{\mathrm{mol~PH_3}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 1.1\bar{7}42\times 10^{25}~\mathrm{H} \\[1.5ex] &= 1.17\times 10^{25}~\mathrm{H} \end{align*} \]

H₂O

\[ \begin{align*} N(\mathrm{H}) &= n(\mathrm{H}) ~ N_{\mathrm{A}}\\[1.5ex] &= n(\mathrm{H_2O}) ~ r(\mathrm{H,H_2O}) ~ N_{\mathrm{A}}\\[1.5ex] &= m(\mathrm{H_2O}) ~ M(\mathrm{H_2O})^{-1} ~ r(\mathrm{H,H_2O}) ~ N_{\mathrm{A}} \\[1.5ex] &= 22\bar{1}~\mathrm{g~H_2O} ~ \left ( \dfrac{\mathrm{mol~H_2O}}{18.0\bar{2}~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H}}{\mathrm{mol~H_2O}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 1.4\bar{7}70\times 10^{25}~\mathrm{H} \\[1.5ex] &= 1.48\times 10^{25}~\mathrm{H} \end{align*} \]

H₂S

\[ \begin{align*} N(\mathrm{H}) &= n(\mathrm{H}) ~ N_{\mathrm{A}}\\[1.5ex] &= n(\mathrm{H_2S}) ~ r(\mathrm{H,H_2S}) ~ N_{\mathrm{A}}\\[1.5ex] &= m(\mathrm{H_2S}) ~ M(\mathrm{H_2S})^{-1} ~ r(\mathrm{H,H_2S}) ~ N_{\mathrm{A}} \\[1.5ex] &= 22\bar{1}~\mathrm{g~H_2S} ~ \left ( \dfrac{\mathrm{mol~H_2S}}{34.0\bar{8}~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H}}{\mathrm{mol~H_2S}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 7.8\bar{1}02\times 10^{24}~\mathrm{H} \\[1.5ex] &= 7.81\times 10^{24}~\mathrm{H} \end{align*} \]

\[ \begin{align*} N(\mathrm{H}) &= n(\mathrm{H}) ~ N_{\mathrm{A}}\\[1.5ex] &= n(\mathrm{HF}) ~ r(\mathrm{H,HF}) ~ N_{\mathrm{A}}\\[1.5ex] &= m(\mathrm{HF}) ~ M(\mathrm{HF})^{-1} ~ r(\mathrm{H,HF}) ~ N_{\mathrm{A}} \\[1.5ex] &= 22\bar{1}~\mathrm{g~HF} ~ \left ( \dfrac{\mathrm{mol~HF}}{20.0\bar{1}~\mathrm{g}} \right ) \left ( \dfrac{1~\mathrm{mol~H}}{\mathrm{mol~HF}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 6.6\bar{5}09\times 10^{24}~\mathrm{H} \\[1.5ex] &= 6.65\times 10^{24}~\mathrm{H} \end{align*} \]

A mixture of BaCl₂ and NaCl is analyzed by precipitating all of the barium as BaSO₄. After the addition of excess Na₂SO₄ to a 4.988 g sample of the mixture, the mass of precipitate collected is 2.123 g. What is the mass percentage of barium chloride in the mixture?

Solution

A 15.00 g sample of an alloy containing only Pb and Sn was dissolved in nitric acid. Sulfuric acid was added to this solution, which precipitated 3.90 g of PbSO₄. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (M(PbSO₄) = 303.26 g mol^–1)

Solution

Hydrogen cyanide is produced industrially from a reaction between gaseous ammonia, oxygen, and methane via the Andrussow process.

\[ \begin{align*} \mathrm{2~NH_3(g)} + \mathrm{3~O_2(g)} + \mathrm{2~CH_4(g)} \longrightarrow \mathrm{2~HCN(g)} + \mathrm{6~H_2O(g)} \end{align*} \]

If 6.00 × 10³ kg of each reactant react, what mass (in kg) of each product would be produced (assuming a 100 % yield)?

Solution

Answer:

3.38 × 10³ kg HCN
6.76 × 10³ kg H₂O

Concept: stoichiometry; limiting reactant

First, determine the limiting reactant by calculating the moles of a product (we will use HCN) that can be formed from the given mass of each reactant.

\[ \begin{align*} n(\mathrm{HCN}) &= n(\mathrm{NH_3}) ~ r(\mathrm{HCN,NH_3}) \\[1.5ex] &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{HCN,NH_3}) \\[1.5ex] &= 6.0\bar{0} \times 10^{3}~\mathrm{kg~NH_3} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.0\bar{4}~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{2~mol~NH_3}} \right )\\[1.5ex] &= 3.5\bar{2}11\times 10^{5}~\mathrm{mol} \end{align*} \]

\[ \begin{align*} n(\mathrm{HCN}) &= n(\mathrm{O_2}) ~ r(\mathrm{HCN,O_2}) \\[1.5ex] &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{HCN,O_2}) \\[1.5ex] &= 6.0\bar{0} \times 10^{3}~\mathrm{kg~O_2} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.0\bar{0}~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{3~mol~O_2}} \right )\\[1.5ex] &= 1.2\bar{5}00\times 10^{5}~\mathrm{mol} \end{align*} \]

\[ \begin{align*} n(\mathrm{HCN}) &= n(\mathrm{CH_4}) ~ r(\mathrm{HCN,CH_4}) \\[1.5ex] &= m(\mathrm{CH_4}) ~ M(\mathrm{CH_4})^{-1} ~ r(\mathrm{HCN,CH_4}) \\[1.5ex] &= 6.0\bar{0} \times 10^{3}~\mathrm{kg~CH_4} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~CH_4}}{\mathrm{16.0\bar{5}~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{2~mol~CH_4}} \right )\\[1.5ex] &= 3.7\bar{3}83\times 10^{5}~\mathrm{mol} \end{align*} \]

Since O₂ produces the fewest moles of product, O₂ is the limiting reactant. All subsequent calculations must be based on the amount of product formed from O₂.

Now, calculate the mass of each product formed in kilograms.

\[ \begin{align*} m(\mathrm{HCN}) &= n(\mathrm{HCN}) ~ M(\mathrm{HCN}) \\[1.5ex] &= 1.2\bar{5}00\times 10^{5}~\mathrm{mol~HCN} \left ( \dfrac{27.0\bar{3}~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 3.3\bar{7}87\times 10^{3}~\mathrm{kg} \\[1.5ex] &= 3.38\times 10^{3}~\mathrm{kg}\\[3ex] m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= n(\mathrm{HCN}) ~ r(\mathrm{H_2O,HCN}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 1.2\bar{5}00\times 10^{5}~\mathrm{mol~HCN} \left ( \dfrac{6~\mathrm{mol~H_2O}}{2~\mathrm{mol~HCN}} \right ) \left ( \dfrac{18.0\bar{2}~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 6.7\bar{5}75\times 10^{3}~\mathrm{kg} \\[1.5ex] &= 6.76\times 10^{3}~\mathrm{kg} \end{align*} \]

Chemist’s Note: The Andrussow Process

The reaction described in this problem is known as the Andrussow process, developed by German chemist Leonid Andrussow in the late 1920s. It remains one of the primary industrial methods for producing hydrogen cyanide (HCN).

This is a high-stakes, gas-phase reaction that is typically carried out at extremely high temperatures (over 1000 °C) and in the presence of a platinum catalyst. The process is remarkably efficient, but it must be carefully controlled. The reactants—ammonia, oxygen, and methane—can form an explosive mixture. Hydrogen cyanide itself is a volatile and highly toxic liquid, famously used as a chemical weapon in the past, but it is also a critically important precursor for manufacturing a vast range of materials.

Today, the HCN produced via the Andrussow process is primarily used to make adiponitrile (a precursor to Nylon 66), methyl methacrylate (for Plexiglas^®), and sodium cyanide (used in gold mining).

Acrylonitrile (C₃H₃N) is the starting material for many synthetic carpets and fabrics and is produced by the following reaction.

\[ \begin{align*} \mathrm{2~C_3H_6(g)} + \mathrm{2~NH_3(g)} + \mathrm{3~O_2(g)} \longrightarrow \mathrm{2~C_3H_3N(g)} + \mathrm{6~H_2O(g)} \end{align*} \]

If 25.00 g C₃H₆, 10.00 g NH₃, and 10.00 g O₂ react, what mass (in g) of acrylonitrile can be theoretically produced?

Solution

Balance the following equation.

\[ \begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag_2O(s)} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag(s)} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2(g)} \end{align*} \]

Additionally, consider a 4.260 × 10³ mg sample of impure silver oxide that, when completely decomposes, yields 283 mg of O₂(g). Assuming that the silver oxide is the only source of oxygen, what is the mass percent of silver oxide in the sample?

Solution

A 2.000 g sample of a pure, unknown metal (M) containing compound (M₂SO₄) was dissolved in water and treated with an excess of aqueous calcium chloride. All the sulfate ions precipitated as calcium sulfate which was collected, dried, and found to be 1.916 g. What is the relative atomic weight and the identity of the unknown metal?

Solution

Answer: 23.02; Na

Concept: stoichiometry; precipitation; relative atomic weight

The balanced chemical equation for the precipitation reaction is:

\[ \begin{align*} \mathrm{M_2SO_4(aq)} + \mathrm{CaCl_2(aq)} \longrightarrow \mathrm{CaSO_4(s)} + \mathrm{2~MCl(aq)} \end{align*} \]

A Note on the Calculation Strategy: The goal is to find the relative atomic weight of the unknown metal, M. This is a unitless quantity (A_r that is numerically equal to its molar mass (M). The most direct way to solve this is to construct a single equation for A_r(M) based on its fundamental definition: the mass of the metal divided by the moles of the metal. We can determine the mass of M by subtracting the mass of the sulfate from the total sample mass. We can determine the moles of M by relating it stoichiometrically to the moles of the CaSO₄ precipitate. The following symbolic derivation builds this complete equation step-by-step.

\[ \begin{align*} A_{\mathrm{r}}(\mathrm{M^+}) &= m(\mathrm{M^+}) ~ n(\mathrm{M^+})^{-1} ~ M_{\mathrm{u}}{^{-1}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - m(\mathrm{SO_4{^{2-}}}) \biggl\} ~ \biggl\{ n(\mathrm{M_2SO_4}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~ M_{\mathrm{u}}{^{-1}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - m(\mathrm{SO_4{^{2-}}}) \biggl\} ~ \biggl\{ n(\mathrm{SO_4{^{2-}}}) ~ r(\mathrm{M_2SO_4,SO_4{^{2-}}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~ M_{\mathrm{u}}{^{-1}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - m(\mathrm{SO_4{^{2-}}}) \biggl\} ~ \biggl\{ n(\mathrm{CaSO_4}) ~ r(\mathrm{SO_4{^{2-}},CaSO_4}) ~ r(\mathrm{M_2SO_4,SO_4{^{2-}}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~ M_{\mathrm{u}}{^{-1}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) ~-~ \biggr[ m(\mathrm{CaSO_4}) ~ M(\mathrm{CaSO_4})^{-1} ~ r(\mathrm{SO_4{^{2-}},CaSO_4}) \biggr] \biggl\} ~ \\[1.5ex] &\phantom{=~~~} \biggl\{ m(\mathrm{CaSO_4}) ~ M(\mathrm{CaSO_4})^{-1} ~ r(\mathrm{M_2SO_4,CaSO_4}) ~ r(\mathrm{M_2SO_4,SO_4{^{2-}}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~ M_{\mathrm{u}}{^{-1}} \\[1.5ex] &= \Biggl\{ 2.00\bar{0}~\mathrm{g~M_2SO_4} ~-~ \Biggr[ 1.91\bar{6}~\mathrm{g~CaSO_4} \left ( \dfrac{\mathrm{mol~CaSO_4}}{136.1\bar{4}~\mathrm{g}} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~SO_4{^{2-}}}}{\mathrm{mol~CaSO_4}} \right ) \left ( \dfrac{96.06~\mathrm{g}}{\displaystyle \mathrm{mol~SO_4{^{2-}}}} \right ) \Biggr] \Biggl\} \\[1.5ex] &\phantom{=~~~} \Biggl\{ 1.91\bar{6}~\mathrm{g~CaSO_4} \left ( \dfrac{\mathrm{mol~CaSO_4}}{136.1\bar{4}~\mathrm{g}} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~SO_4{^{2-}}}}{\mathrm{mol~CaSO_4}} \right ) \left ( \dfrac{\mathrm{mol~M_2SO_4}}{\mathrm{\displaystyle mol~SO_4{^{2-}}}} \right ) \left ( \dfrac{\mathrm{2~mol~M^+}}{\mathrm{mol~M_2SO_4}} \right ) \Biggl\}^{-1} \left ( \dfrac{\mathrm{mol}}{\mathrm{g}} \right ) \\[1.5ex] &= \biggl\{ 2.00\bar{0}~\mathrm{g~M_2SO_4} - 1.35\bar{1}92~\mathrm{g~SO_4{^{2-}}} \biggl\} ~ \biggl\{ 0.0281\bar{4}74~\mathrm{mol~M^{+}} \biggl\}^{-1} \left ( \dfrac{\mathrm{mol}}{\mathrm{g}} \right ) \\[1.5ex] &= \biggl\{ 0.648\bar{0}80~\mathrm{g~M^+} \biggl\} ~ \biggl\{ 0.0281\bar{4}74~\mathrm{mol~M^{+}} \biggl\}^{-1} \left ( \dfrac{\mathrm{mol}}{\mathrm{g}} \right ) \\[1.5ex] &= \left ( \dfrac{0.648\bar{0}80~\mathrm{g~M^+}}{0.0281\bar{4}74~\mathrm{mol~M^+}} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{g}}\right ) \\[1.5ex] &= 23.0\bar{2}45 \\[1.5ex] &= 23.02 \end{align*} \]

The calculated relative atomic weight of 23.02 closely corresponds to Sodium (Na).

Chemist’s Note: Gravimetric Analysis

The procedure described in this problem is a classic example of gravimetric analysis, one of the most accurate and precise methods for quantitative chemical analysis. The name comes from “gravity,” as the core of the technique is the precise measurement of mass (weight).

The strategy is always the same: a component of interest in a sample, which may be difficult to weigh directly, is converted into a new compound (a precipitate) with a well-known, stable formula. For this technique to work, the precipitate must be highly insoluble. Calcium sulfate (CaSO₄) is an excellent choice here because it precipitates almost completely from the solution, ensuring that nearly every sulfate ion from the original sample is captured and weighed.

By using the mass of the final precipitate, chemists can work backward stoichiometrically to determine the mass or percentage of a specific element or compound in the original, unknown sample with very high accuracy.

6.00 g of barium chloride was added to 225 mL of a 1.40 M solution of sodium sulfide.

Write the balanced molecular equation and include phase labels.
Write the full ionic equation and include phase labels.
Write the net ionic equation and include phase labels. If there is no net ionic equation, write “no net ionic equation.”
Indicate the precipitate (if any).
Is there a limiting reactant for this process?

Solution

Concentration and Dilution

Which solution has the greatest molar concentration of SO₄²⁻?

0.060 M H₂SO₄
0.27 M MgSO₄
0.17 M Na₂SO₄
0.098 M Al₂(SO₄)₃
0.22 M CuSO₄

Solution

A solution is prepared by dissolving 4.20 g NaCl, 0.170 g KCl, and 0.180 g CaCl₂ in water. The volume of the solution is 500.0 mL. What is the molar concentration (in mol L^–1) of Cl⁻ in the solution?

Solution

In the following reaction (Hint: is it balanced?), 55.0 mL of potassium sulfate solution was added to excess lead acetate. What is the molar concentration (in g mol⁻¹) of K⁺ in the potassium sulfate solution if 1.20 g of PbSO₄ was produced?

\[ \begin{align*} \mathrm{K_2SO_4(aq)} + \mathrm{Pb(C_2H_3O_2)_2(aq)} \longrightarrow \mathrm{KC_2H_3O_2(aq)} + \mathrm{PbSO_4(s)} \end{align*} \]

Solution

If 13.0 g of AgNO₃ is available, what volume (in L) of 0.25 M AgNO₃ can be prepared?

Solution

A solution of ethanol (C₂H₆O) in water is prepared by dissolving 70.0 mL of ethanol (ρ = 0.79 g cm⁻³) in enough water to make a 250.0 mL solution. What is the molar concentration (in mol L⁻¹) of the ethanol in this solution?

Solution

If all of the chloride in a 4.105 g sample of an unknown metal chloride is precipitated as AgCl with 70.80 mL of 0.2000 M AgNO₃, what is the percentage (by mass) of chloride in the sample?

Solution

A 0.685 g sample of an unknown diprotic acid requires a 42.57 mL 0.5188 M aqueous NaOH solution to be completely neutralized. What is the molar mass (in g mol^–1) of the acid?

Solution

What mass (in g) of Na₂CrO₄ is required to precipitate all of the silver ions from a 75.0 mL 0.150 M aqueous solution of AgNO₃?

Solution

Which of the following aqueous solutions contains the largest number of ions?

100.0 mL of 0.200 M NaOH
50.0 mL of 0.300 M BaCl₂
75.0 mL of 0.175 M Na₃PO₄

Solution

A 45.0 mL sample of HCl(aq) requires 24.16 mL of 0.106 M NaOH for complete neutralization. What is the molar concentration (in mol L^–1) of the original HCl(aq) solution?

Solution

Combine a 55.0 mL 1.10 M aqueous silver nitrate solution with a 25.0 mL 0.65 M sodium chloride solution. What mass (in g) of silver chloride is produced?

Solution

What mass (in g) of NaOH is required to completely react with a 35.0 mL 2.2 M aqueous H₂SO₄ solution?

Solution

You have 75.0 mL of a 1.50 M aqueous solution of Na₂CrO₄ and 125.0 mL of a 1.15 M aqueous solution of AgNO₃. Calculate the molar concentration (in mol L⁻¹) of CrO₄²⁻ after the two solutions are mixed together.

Solution

Answer: [CrO₄²⁻] = 0.20 mol L⁻¹

Concept: balancing equations; stoichiometry; molarity; solubility rules; limiting reactant

\[ \begin{align*} \mathrm{Na_2CrO_4(aq)} + \mathrm{2~AgNO_3(aq)} \longrightarrow \mathrm{Ag_2CrO_4(s)} + \mathrm{2~NaNO_3(aq)} \end{align*} \]

Find the limiting reactant.

\[ \begin{align*} n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= c(\mathrm{Na_2CrO_4}) ~ V(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= \left ( \dfrac{1.5\bar{0}~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \left ( \dfrac{75.\bar{0}~\mathrm{mL~Na_2CrO_4}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{mol~Na_2CrO_4}} \right )\\[1.5ex] &= 0.12\bar{2}50~\mathrm{mol} \\[3.0ex] n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= c(\mathrm{AgNO_3}) ~ V(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= \left ( \dfrac{1.1\bar{5}~\mathrm{mol~AgNO_3}}{\mathrm{L}} \right ) \left ( \dfrac{125.\bar{0}~\mathrm{mL~AgNO_3}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{2~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.071\bar{8}75~\mathrm{mol} \end{align*} \]

AgNO₃ is limiting.

\[ \begin{align*} c(\mathrm{CrO_4{^{2-}}}) &= n(\mathrm{CrO_4{^{2-}}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Na_2CrO_4})_{\mathrm{leftover}} ~ r(\mathrm{CrO_4{^{2-}},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Na_2CrO_4})_{\mathrm{initial}} - n(\mathrm{Na_2CrO_4})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{CrO_4{^{2-}},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ c(\mathrm{Na_2CrO_4}) ~ V(\mathrm{Na_2CrO_4}) \biggr] ~ - ~ \biggr[ n(\mathrm{Ag_2CrO_4})_{\mathrm{formed}} ~ r(\mathrm{Na_2CrO_4,Ag_2CrO_4}) \biggr] \Biggl\} ~ r(\mathrm{CrO_4{^{2-}},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \left ( \dfrac{1.5\bar{0}~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \left ( \dfrac{75.\bar{0}~\mathrm{mL~Na_2CrO_4}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \biggr] ~-~ \biggr[ 0.071\bar{8}75~\mathrm{mol~Ag_2CrO_4} \left ( \dfrac{\mathrm{mol~Na_2CrO_4}}{\mathrm{mol~Ag_2CrO_4}} \right ) \biggr] \biggl\} \left ( \dfrac{\mathrm{mol~CrO_4{^{2-}}}}{\mathrm{mol~Na_2CrO_4}} \right ) \left ( \dfrac{1}{(75.\bar{0} + 125.\bar{0})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ \left ( 0.11\bar{2}5 - 0.071\bar{8}75 \right )~\mathrm{mol~Na_2CrO_4} \biggl\} \left ( \dfrac{\mathrm{mol~CrO_4{^{2-}}}}{\mathrm{mol~Na_2CrO_4}} \right ) \left ( \dfrac{1}{(75.\bar{0} + 125.\bar{0})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \left ( 0.04\bar{0}62~\mathrm{mol~Na_2CrO_4} \right ) \left ( \dfrac{\mathrm{mol~CrO_4{^{2-}}}}{\mathrm{mol~Na_2CrO_4}} \right ) \left ( \dfrac{1}{200.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.2\bar{0}31~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.20~\mathrm{mol~L^{-1}} \end{align*} \]

Sulfamic acid (HSO₃NH₂) is a strong monoprotic acid that can be used to standardize a strong base. A 1.790 g sample of HSO₃NH₂ is required to completely neutralize a 190.4 mL aqueous KOH solution. What is the molar concentration (in mol L^–1) of the KOH solution?

Solution

You mix 275.0 mL of 1.22 M aqueous lead(II) nitrate with 310.0 mL of 1.56 M aqueous potassium iodide. Determine the following.

the balanced molecular equation for this reaction
the limiting reactant
the final molar concentration (in mol L^–1) of Pb²⁺
the mass (in g) of lead(II) iodide formed
the final molar concentration (in mol L^–1) of K⁺
the final molar concentration (in mol L^–1) of NO₃^–

Solution

Answer:

Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)
KI
[Pb²⁺] = 0.160 mol L^–1
111 g PbI₂(s)
[K⁺] = 0.827 mol L^–1
[NO₃^–] = 1.15 mol L^–1

Concept: solubility rules; stoichiometry; molarity; limiting reactant

a. Balanced Molecular Equation

\[ \begin{align*} \mathrm{Pb(NO_3)_2(aq)} + \mathrm{2~KI(aq)} \longrightarrow \mathrm{PbI_2(s)} + \mathrm{2~KNO_3(aq)} \end{align*} \]

b. Limiting Reactant

To find the limiting reactant, calculate the moles of PbI₂ that can be formed from each reactant.

\[ \begin{align*} n(\mathrm{PbI_2}) &= n(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{PbI_2,Pb(NO_3)_2}) \\[1.5ex] &= V(\mathrm{Pb(NO_3)_2}) ~ c(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{PbI_2,Pb(NO_3)_2}) \\[1.5ex] &= 275.\bar{0}~\mathrm{mL~Pb(NO_3)_2} ~ \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{1.2\bar{2}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~PbI_2}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \\[1.5ex] &= 0.33\bar{5}50~\mathrm{mol} \\[3ex] n(\mathrm{PbI_2}) &= n(\mathrm{KI}) ~ r(\mathrm{PbI_2,KI}) \\[1.5ex] &= V(\mathrm{KI}) ~ c(\mathrm{KI}) ~ r(\mathrm{PbI_2,KI}) \\[1.5ex] &= 310.\bar{0}~\mathrm{mL~KI} ~ \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{1.5\bar{6}~\mathrm{mol~KI}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~PbI_2}}{2~\mathrm{mol~KI}} \right ) \\[1.5ex] &= 0.24\bar{1}80~\mathrm{mol} \end{align*} \]

Since KI produces fewer moles of product, KI is the limiting reactant.

c. Final Molar Concentration of Pb²⁺

Since the product, PbI₂, is not soluble in water, the only dissolved Pb²⁺ ions come from the unreacted Pb(NO₃)₂ reactant.

\[ \begin{align*} c(\mathrm{Pb^{2+}})_{\mathrm{final}} &= n(\mathrm{Pb^{2+}})_{\mathrm{remaining}} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Pb(NO_3)_2})_{\mathrm{remaining}} ~ r(\mathrm{Pb^{2+},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} - n(\mathrm{Pb(NO_3)_2})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Pb^{2+},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) \biggr] ~-~ \biggr[ n(\mathrm{PbI_2})_{\mathrm{formed}} ~~ r(\mathrm{Pb(NO_3)_2, PbI_2}) ~ \biggr] \Biggl\} ~ r(\mathrm{Pb^{2+},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{\biggr[ \left ( \dfrac{1.2\bar{2}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{275.\bar{0}~\mathrm{mL~Pb(NO_3)_2}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] ~-~ \biggr[ \left ( 0.24\bar{1}80~\mathrm{mol~PbI_2} \right ) \left ( \dfrac{\mathrm{mol~Pb(NO_3)_2}}{\mathrm{mol~PbI_2}} \right ) \biggr]\Biggl\}\\ &\phantom{= \Biggl\{} \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{(275.\bar{0} + 310.\bar{0}) ~\mathrm{mL~solution}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggr[ \left ( 0.33\bar{5}0 - 0.24\bar{1}80 \right )~\mathrm{mol~Pb(NO_3)_2} \biggr] \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{585.\bar{0}0 ~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \left ( 0.09\bar{3}70~\mathrm{mol~Pb(NO_3)_2} \right ) \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{585.\bar{0}0 ~\mathrm{mL~solution}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.16\bar{0}17~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.160~\mathrm{mol~L^{-1}} \end{align*} \]

d. Mass of Lead(II) Iodide Formed

Use the amount of PbI₂ formed from part (b) where we determined the amount of PbI₂ that could be created.

\[ \begin{align*} m(\mathrm{PbI_2}) &= n(\mathrm{PbI_2})_{\mathrm{formed}} ~ M(\mathrm{PbI_2}) \\[1.5ex] &= 0.24\bar{1}80~\mathrm{mol~PbI_2} ~ \left ( \dfrac{461.0\bar{0}~\mathrm{g}}{\mathrm{mol~PbI_2}} \right ) \\[1.5ex] &= 11\bar{1}.46~\mathrm{g} \\[1.5ex] &= 111~\mathrm{g} \end{align*} \]

e. Final Molar Concentration of K⁺

K⁺ is a spectator ion. Its concentration is its initial moles diluted by the total final volume.

\[ \begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{KNO_3})_{\mathrm{formed}} ~ r(\mathrm{K^+,KNO_3})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KI}) ~ r(\mathrm{KNO_3,KI})~ r(\mathrm{K^+,KNO_3})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KI}) ~ V(\mathrm{KI}) ~ r(\mathrm{KNO_3,KI})~ r(\mathrm{K^+,KNO_3})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{1.5\bar{6}~\mathrm{mol~KI}}{\mathrm{L}} \right ) \left ( \dfrac{310.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{2~\mathrm{mol~KNO_3}}{2~\mathrm{mol~KI}} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~K^+}}{\mathrm{mol~KNO_3}} \right ) \left ( \dfrac{1}{(275.0 + 310.0)~\mathrm{mL~solution}} \right ) \\[1.5ex] &= 0.82\bar{6}66~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.827~\mathrm{mol~L^{-1}} \end{align*} \]

f. Final Molar Concentration of NO₃^–

The dissolved K⁺ spectator ions are present due to the presence of KNO₃ and unreacted Pb(NO₃)₂, both soluble ionic compounds. Because all of the nitrate ions originate from the starting reactant, Pb(NO₃)₂, and remain in solution, we can determine the final nitrate ion concentration from the original molar concentration of Pb(NO₃)₂.

\[ \begin{align*} c(\mathrm{NO_3{^-}}) &= n(\mathrm{NO_3{^-}}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{NO_3{^-},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{NO_3{^-},Pb(NO_3)_2})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{1.2\bar{2}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{275.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\displaystyle 2~\mathrm{mol~NO_3{^{-}}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{(275.\bar{0} + 310.\bar{0})~\mathrm{mL~solution}} \right ) \\[1.5ex] &= 1.1\bar{4}70~\mathrm{mol~L^{-1}} \\[1.5ex] &= 1.15~\mathrm{mol~L^{-1}} \end{align*} \]

A student performs a titration to standardize a stock solution of NaOH. They weigh out a 1.250 g sample of pure KHP (potassium hydrogen phthalate, M(KHP) = 204.22 g mol⁻¹, a monoprotic acid), dissolve it in water, and find that it requires 75.0 mL of the NaOH solution to completely neutralize the acid. What is the molar concentration (in mol L⁻¹) of the stock NaOH solution? (KHP is a monoprotic acid).

Solution

What volume (in mL) of a 5.00 M hydrofluoric acid will completely react with 5.00 g of calcium hydroxide?

Solution

A 135 mL sample of 0.210 M aqueous magnesium chloride forms a precipitate when mixed with 325 mL 0.120 M aqueous sodium hydroxide.

How much (in g) precipitate is formed?
What is the molar concentration (in mol L⁻¹) of the magnesium(2+) ion?
What is the molar concentration (in mol L⁻¹) of the sodium(1+) ion?

Solution

Answer:

1.14 g
[Mg²⁺] = 0.0192 mol L⁻¹
[Na⁺] = 0.0848 mol L⁻¹

Concept: balancing equations; stoichiometry; molarity

The balanced equation for the precipitation reaction is:

\[ \begin{align*} \mathrm{MgCl_2(aq)} + 2~\mathrm{NaOH(aq)} &\longrightarrow 2~\mathrm{NaCl(aq)} + \mathrm{Mg(OH)_2(s)} \end{align*} \]

First, determine the limiting reactant by calculating the moles of the precipitate, Mg(OH)₂, that can be formed from each reactant.

\[ \begin{align*} n(\mathrm{Mg(OH)_2}) &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Mg(OH)_2,MgCl_2})\\[1.5ex] &= c(\mathrm{MgCl_2}) ~ V(\mathrm{MgCl_2}) ~ r(\mathrm{Mg(OH)_2,MgCl_2}) \\[1.5ex] &= \left ( \dfrac{0.21\bar{0}~\mathrm{mol~MgCl_2}}{\mathrm{L}} \right ) \left ( \dfrac{13\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 0.028\bar{3}5~\mathrm{mol} \\[3ex] n(\mathrm{Mg(OH)_2}) &= n(\mathrm{NaOH}) ~ r(\mathrm{Mg(OH)_2,NaOH})\\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{Mg(OH)_2,NaOH}) \\[1.5ex] &= \left ( \dfrac{0.12\bar{0}~\mathrm{mol~NaOH}}{\mathrm{L}} \right ) \left ( \dfrac{32\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{2~\mathrm{mol~NaOH}} \right )\\[1.5ex] &= 0.019\bar{5}00~\mathrm{mol} \end{align*} \]

Since NaOH produces the lesser amount of product, NaOH is the limiting reactant.

a. Mass of Precipitate

The mass of Mg(OH)₂ is determined by the moles formed from the limiting reactant.

\[ \begin{align*} m(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.019\bar{5}00~\mathrm{mol~Mg(OH)_2} \left ( \dfrac{58.3\bar{3}~\mathrm{g~Mg(OH)_2}}{\mathrm{mol}} \right )\\[1.5ex] &= 1.1\bar{3}74~\mathrm{g} \\[1.5ex] &= 1.14~\mathrm{g} \end{align*} \]

b. Final Concentration of Mg²⁺ (Excess Reactant)

\[ \begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Mg^{2+}})_{\mathrm{initial}} - n(\mathrm{Mg^{2+}})_{\mathrm{reacted}} \biggl\}~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ n(\mathrm{MgCl_2})_{\mathrm{initial}} ~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] ~-~ \biggr[ n(\mathrm{MgCl_2})_{\mathrm{final}}~ r(\mathrm{Mg^{2+},MgCl_2})~ \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ n(\mathrm{Mg^{2+}})_{\mathrm{initial}}~ r(\mathrm{Cl^-,MgCl_2}) \biggr] ~-~ \biggr[ n(\mathrm{Mg(OH)_2})~ r(\mathrm{MgCl_2,Mg(OH)_2})~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{MgCl_2})_{\mathrm{initial}}~ V(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] ~-~ \biggr[ n(\mathrm{Mg(OH)_2})~ r(\mathrm{MgCl_2,Mg(OH)_2})~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \left ( \dfrac{0.21\bar{0}~\mathrm{mol~MgCl_2}}{\mathrm{L}} \right ) \left ( \dfrac{13\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \biggr] - \\ &\phantom{=}~~~~~ \biggr[ \left ( 0.019\bar{5}00~\mathrm{mol~Mg(OH)_2} \right ) \left ( \dfrac{\mathrm{mol~MgCl_2}}{\mathrm{mol~Mg(OH)_2}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \biggr] \biggl\} \left ( \dfrac{1}{(13\bar{5} + 32\bar{5})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ \left ( 0.028\bar{3}50 - 0.019\bar{5}00 \right )~\mathrm{mol~Mg^{2+}} \biggl\} \left ( \dfrac{1}{46\bar{0}.~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ 0.0088\bar{5}00~\mathrm{mol~Mg^{2+}} \biggl\} \left ( \dfrac{1}{46\bar{0}.~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.019\bar{2}39~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.0192~\mathrm{mol~L^{-1}} \end{align*} \]

c. Final Concentration of Na⁺ (Spectator Ion)

\[ \begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{NaOH}) ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.12\bar{0}~\mathrm{mol~NaOH}}{\mathrm{L}} \right ) \left ( \dfrac{32\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Na^+}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{(13\bar{5} + 32\bar{5})~\mathrm{mL}} \right ) \\[1.5ex] &= 0.084\bar{7}82~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.0848~\mathrm{mol~L^{-1}} \end{align*} \]

A solution is prepared by dissolving 12.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL aliquot is taken and 50.00 mL of water is added. What is the molar concentration (in mol L^–1) of ammonium ions and sulfate ions in the final solution?

Solution

Answer:

[NH₄⁺] = 0.323 mol L^–1
[SO₄^2–] = 0.161 mol L^–1

Concept: stoichiometry; molarity; dilution

The strategy is to find the final concentration of each ion in the diluted solution. The final volume is the sum of the aliquot volume and the added water volume: 10.00 mL + 50.00 mL = 60.00 mL.

Final Concentration of NH₄⁺

\[ \begin{align*} c(\mathrm{NH_4{^+}}) &= n(\mathrm{NH_4{^+}})_{\textrm{dilute aliquot}} ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= n(\mathrm{NH_4{^+}})_{\textrm{stock solution}} ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{(NH_4)_2SO_4}) ~ M(\mathrm{(NH_4)_2SO_4})^{-1} ~ r(\mathrm{NH_4{^+},(NH_4)_2SO_4})\\[1.5ex] &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 12.\bar{8}~\mathrm{g~(NH_4)_2SO_4} \left ( \dfrac{\mathrm{mol~(NH_4)_2SO_4}}{132.1\bar{6}~\mathrm{g}} \right ) \left ( \dfrac{\displaystyle \mathrm{2~mol~NH_4{^+}}}{\mathrm{mol~(NH_4)_2SO_4}} \right ) \\[1.5ex] &\phantom{=(} \left ( \dfrac{1}{100.\bar{0}~\mathrm{mL~stock~solution}} \right ) \left ( \dfrac{10.\bar{0}~\mathrm{mL~aliquot}}{} \right ) \left ( \dfrac{1}{60.\bar{0}~\mathrm{mL~dilute~aliquot}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.32\bar{2}84~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.323~\mathrm{mol~L^{-1}} \end{align*} \]

Final Concentration of SO₄²⁻

\[ \begin{align*} c(\mathrm{SO_4{^{2-}}}) &= n(\mathrm{SO_4{^{2-}}})_{\textrm{dilute~aliquot}} ~ V(\mathrm{stock~solution})^{-1} \\[1.5ex] &= n(\mathrm{SO_4{^{2-}}})_{\textrm{stock~solution}} ~ V(\mathrm{stock~solution})^{-1}~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{(NH_4)_2SO_4}) ~ M(\mathrm{(NH_4)_2SO_4})^{-1} ~ r(\mathrm{SO_4{^{2-}},(NH_4)_2SO_4})\\[1.5ex] &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 12.\bar{8}~\mathrm{g~(NH_4)_2SO_4} \left ( \dfrac{\mathrm{mol~(NH_4)_2SO_4}}{132.16~\mathrm{g}} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~SO_4{^{2-}}}}{\mathrm{mol~(NH_4)_2SO_4}} \right ) \\[1.5ex] &\phantom{=(} \left ( \dfrac{1}{100.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{60.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.16\bar{1}42~\mathrm{mol~L^{-1}}\\[1.5ex] &= 0.161~\mathrm{mol~L^{-1}} \end{align*} \]

Alternatively…

Once the final concentration of one ion is known, the concentration of the other ion can be found directly from the salt’s 2:1 stoichiometry.

\[ \begin{align*} c(\mathrm{SO_4{^{2-}}}) &= c(\mathrm{NH_4^+}) ~ r(\mathrm{SO_4{^{2-}},NH_4{^+}}) \\[1.5ex] &= \left( \dfrac{0.32\bar{2}84~\mathrm{mol~NH_4{^+}}}{\mathrm{L}} \right) \left( \dfrac{1~\mathrm{mol~SO_4{^{2-}}}}{\displaystyle 2~\mathrm{mol~NH_4{^+}}} \right ) \\[1.5ex] &= 0.16\bar{1}42~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.161~\mathrm{mol~L^{-1}} \end{align*} \]

If 175 mL of a 0.125 M aqueous NaCl solution and 275 mL of a 0.575 M aqueous Na₂SO₄ solution are mixed, determine the molar concentrations (in mol L⁻¹) of the following:

chloride ions
sulfate ions
sodium ions

Solution

Answer:

[Cl⁻] = 0.0486 mol L⁻¹
[SO₄²⁻] = 0.351 mol L⁻¹
[Na⁺] = 0.767 mol L⁻¹

Concept: balancing equations; stoichiometry; molarity

First, we must determine if a reaction occurs by considering the potential products of a double displacement reaction. The reactants are NaCl and Na₂SO₄. The potential products would be the same two salts. According to solubility rules, all sodium compounds are soluble. Since no precipitate, gas, or weak electrolyte is formed, no reaction occurs and no limiting reactant exists.

The problem is a dilution calculation for each of the three spectator ions present in the mixture.

a. Final Molar Concentration of Cl⁻

\[ \begin{align*} c(\mathrm{Cl^-}) &= n(\mathrm{Cl^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl}) ~ r(\mathrm{Cl^-,NaCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.12\bar{5}~\mathrm{mol~NaCl}}{\mathrm{L}} \right ) \left ( \dfrac{17\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~NaCl}} \right ) \left ( \dfrac{1}{(17\bar{5} + 27\bar{5})~\mathrm{mL}} \right ) \\[1.5ex] &= 0.048\bar{6}11~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.0486~\mathrm{mol~L^{-1}} \end{align*} \]

b. Final Molar Concentration of SO₄²⁻

\[ \begin{align*} c(\mathrm{SO_4{^{2-}}}) &= n(\mathrm{SO_4^{2-}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Na_2SO_4}) ~ V(\mathrm{Na_2SO_4}) ~ r(\mathrm{SO_4^{2-},Na_2SO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.57\bar{5}~\mathrm{mol~Na_2SO_4}}{\mathrm{L}} \right ) \left ( \dfrac{27\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~SO_4{^{2-}}}}{\mathrm{mol~Na_2SO_4}} \right ) \left ( \dfrac{1}{(17\bar{5} + 27\bar{5})~\mathrm{mL}} \right ) \\[1.5ex] &= 0.35\bar{1}38~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.351~\mathrm{mol~L^{-1}} \end{align*} \]

c. Final Molar Concentration of Na⁺ (Common Ion)

The total moles of Na⁺ is the sum of the moles from both initial solutions.

\[ \begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{NaCl}) + n(\mathrm{Na_2SO_4}) \biggl\}~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{NaCl}) ~ V(\mathrm{NaCl}) ~ r(\mathrm{Na^+,NaCl}) ~ \biggr] ~+~ \biggr[ c(\mathrm{Na_2SO_4}) ~ V(\mathrm{Na_2SO_4}) ~ r(\mathrm{Na^+,Na_2SO_4}) ~ \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \left ( \dfrac{0.12\bar{5}~\mathrm{mol~NaCl}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{17\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Na^+}}{\mathrm{mol~NaCl}} \right ) \biggr] ~+~ \biggr[ \left ( \dfrac{0.57\bar{5}~\mathrm{mol~Na_2SO_4}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{27\bar{5}~\mathrm{mL}}{} \right ) \left ( \dfrac{2~\mathrm{mol~Na^+}}{\mathrm{mol~Na_2SO_4}} \right ) \biggr] \biggl\} \\ &\phantom{=}~~~~ \left ( \dfrac{1}{(17\bar{5} + 27\bar{5})~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ \left ( 0.02\bar{8}75 + 0.31\bar{6}25 \right )~\mathrm{mol~Na^+} \biggl\} \left ( \dfrac{1}{45\bar{0}.~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ 0.34\bar{5}00~\mathrm{mol~Na^+} \biggl\} \left ( \dfrac{1}{45\bar{0}.~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.76\bar{6}66~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.767~\mathrm{mol~L^{-1}} \end{align*} \]

What mass (in g) of iron(III) hydroxide precipitate can be produced by reacting a 62.0 mL 0.105 M aqueous iron(III) nitrate solution with a 125 mL 0.150 M aqueous sodium hydroxide solution?

Solution

A 2.50 g sample of phosphoric acid is added to a 150.0 mL 1.00 M sodium hydroxide solution to give a 151.5 mL mixture and the acid is completely neutralized. Determine the following:

[Na⁺] (in mol L⁻¹)
[PO₄³⁻] (in mol L⁻¹)
[OH⁻] (in mol L⁻¹)

Solution

Answer:

[Na⁺] = 0.990 mol L⁻¹
[PO₄³⁻] = 0.168 mol L⁻¹
[OH⁻] = 0.49 mol L⁻¹

Concept: balancing equations; stoichiometry; molarity; acid-base reaction; limiting reactant

\[ \begin{align*} \mathrm{H_3PO_4(aq)} + \mathrm{3~NaOH} \longrightarrow \mathrm{3~H_2O(l)} + \mathrm{Na_3PO_4(aq)} \end{align*} \]

Determine the limiting reagent.

\[ \begin{align*} n(\mathrm{Na_3PO_4}) &= n(\mathrm{H_3PO_4}) ~ r(\mathrm{Na_3PO_4,H_3PO_4}) \\[1.5ex] &= m(\mathrm{H_3PO_4}) ~ M(\mathrm{H_3PO_4})^{-1} ~ r(\mathrm{Na_3PO_4, H_3PO_4}) \\[1.5ex] &= 2.5\bar{0}~\mathrm{g~H_3PO_4} \left ( \dfrac{\mathrm{mol~H_3PO_4}}{\mathrm{98.0\bar{0}~g}} \right ) \left ( \dfrac{\mathrm{mol~Na_3PO_4}}{\mathrm{mol~H_3PO_4}} \right )\\[1.5ex] &= 0.025\bar{5}10~\mathrm{mol} \\[3ex] n(\mathrm{Na_3PO_4}) &= n(\mathrm{NaOH}) ~ r(\mathrm{Na_3PO_4,NaOH}) \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{Na_3PO_4,NaOH}) \\[1.5ex] &= \left ( \dfrac{1.0\bar{0}~\mathrm{mol~NaOH}}{\mathrm{L}} \right ) \left ( \dfrac{150.\bar{0}~\mathrm{mL~NaOH}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Na_3PO_4}}{\mathrm{3~mol~NaOH}} \right ) \\[1.5ex] &= 0.050\bar{0}00~\mathrm{mol} \end{align*} \]

Since H₃PO₄ produces fewer moles of product, H₃PO₄ is the limiting reactant.

The Na⁺ ions are spectators from the initial NaOH. Their total moles remain constant, but the volume changes.

\[ \begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{NaOH})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaOH})_{\mathrm{initial}} ~ V(\mathrm{NaOH})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{1.0\bar{0}~\mathrm{mol~NaOH}}{\mathrm{L}} \right ) \left ( \dfrac{150.\bar{0}~\mathrm{mL~NaOH}}{} \right ) \left ( \dfrac{\displaystyle 1~\mathrm{mol~Na^+}}{1~\mathrm{mol~NaOH}} \right) \left ( \dfrac{1}{151.\bar{5}~\mathrm{mL~solution}} \right ) \\[1.5ex] &= 0.99\bar{0}00~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.990~\mathrm{mol~L^{-1}} \end{align*} \]

All the H₃PO₄ is converted to PO₄^3– (as Na₃PO₄).

\[ \begin{align*} c(\mathrm{PO_4{^{3-}}}) &= n(\mathrm{PO_4{^{3-}}}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{Na_3PO_4}) ~ r(\mathrm{PO_4{^{3-}},Na_3PO_4}) ~ V(\mathrm{solution}^{-1})\\[1.5ex] &= 0.025\bar{5}10~\mathrm{mol~Na_3PO_4} \left ( \dfrac{\displaystyle \mathrm{mol~PO_4{^{3-}}}}{\mathrm{mol~Na_3PO_4}} \right ) \left ( \dfrac{1}{\mathrm{151.\bar{5}~\mathrm{mL}}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.16\bar{8}38~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.168~\mathrm{mol~L^{-1}} \end{align*} \]

The final moles of OH^– is the initial amount minus the amount that reacted with the acid.

\[ \begin{align*} c(\mathrm{OH^-}) &= n(\mathrm{OH^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{NaOH})_{\mathrm{initial}} - n(\mathrm{NaOH})_{\mathrm{consumed}} \biggl\} ~ r(\mathrm{OH^-,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{NaOH})~ V(\mathrm{NaOH}) \biggr] ~ - ~ \biggr[ n(\mathrm{Na_3PO_4}) ~ r(\mathrm{NaOH,Na_3PO_4}) \biggr] \biggl\} ~ r(\mathrm{OH^-,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \dfrac{1.0\bar{0}~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{150.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] ~-~ \biggr[ 0.025\bar{5}10~\mathrm{mol~Na_3PO_4} \left ( \dfrac{3~\mathrm{mol~NaOH}}{\mathrm{mol~Na_3PO_4}} \right ) \biggr] \biggl\} ~ \left ( \dfrac{\displaystyle \mathrm{mol~OH^-}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{\mathrm{151.\bar{5}~mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \biggl\{ \left ( 0.15\bar{0} ~-~ 0.076\bar{5}30 \right )~\mathrm{mol~NaOH} \biggl\} ~ \left ( \dfrac{\mathrm{\displaystyle mol~OH^-}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{\mathrm{151.\bar{5}~mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \left ( 0.07\bar{3}47~\mathrm{mol~NaOH} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~OH^-}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{\mathrm{151.\bar{5}~mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.4\bar{8}49~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.49~\mathrm{mol~L^{-1}} \end{align*} \]

A stock solution with a total volume of 1000.0 mL contains 35.0 g Mg(NO₃)₂. If you take a 20.0 mL aliquot and then dilute it with water to a total volume of 500.0 mL, what is the molar concentration (in mol L^–1) of Mg²⁺ and NO₃^– in the final solution?

Solution

Answer:

[Mg²⁺] = 9.43 × 10⁻³
[NO₃⁻] = 1.89 × 10⁻²

Concept: stoichiometry; molarity; dilution

The strategy is to find the final concentration of each ion in the diluted solution. This can be done in a single calculation chain by determining the initial moles of the salt in the stock solution, finding the fraction of those moles transferred in the aliquot, and then dividing by the final diluted volume.

Final Concentration of Mg²⁺

\[ \begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}})_{\textrm{dilute aliquot}} ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= n(\mathrm{Mg^{2+}})_{\textrm{stock solution}} ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} ~ r(\mathrm{Mg^{2+},Mg(NO_3)_2}) ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 35.\bar{0}~\mathrm{g~Mg(NO_3)_2} \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{148.3\bar{3}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{1000.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{20.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{500.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.0094\bar{3}84~\mathrm{mol~L^{-1}} \\[1.5ex] &= 9.43\times 10^{-3}~\mathrm{mol~L^{-1}} \end{align*} \]

Final Concentration of NO₃^–

The calculation for the nitrate ion follows the same logic, but uses the 2:1 mole ratio between NO₃^– and Mg(NO₃)₂.

\[ \begin{align*} c(\mathrm{NO_3{^-}}) &= n(\mathrm{NO_3{^-}})_{\textrm{dilute~aliquot}} ~ V(\mathrm{stock~solution})^{-1} \\[1.5ex] &= n(\mathrm{NO_3{^-}})_{\textrm{stock~solution}} ~ V(\mathrm{stock~solution})^{-1}~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} ~ r(\mathrm{NO_3{^-},Mg(NO_3)_2}) ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 35.\bar{0}~\mathrm{g~Mg(NO_3)_2} \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{148.3\bar{3}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{\displaystyle 2~mol~NO_3{^-}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{1000.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{20.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{500.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.018\bar{8}76~\mathrm{mol~L^{-1}} \\[1.5ex] &= 1.89 \times 10^{-2}~\mathrm{mol~L^{-1}} \end{align*} \]

Alternatively…

Once the final concentration of one ion is known, the concentration of the other ion can be found directly from the salt’s stoichiometry, as their ratio in the solution must be the same as in the chemical formula.

\[ \begin{align*} c(\mathrm{NO_3{^{-}}}) &= c(\mathrm{Mg^{2+}})_{\mathrm{dilute~aliquot}} ~ r(\mathrm{NO_3{^-},Mg^{2+}}) \\[1.5ex] &= \left ( \dfrac{0.0094\bar{3}84~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{\displaystyle 2~\mathrm{mol~NO_3{^{-}}}}{\displaystyle \mathrm{mol~Mg^{2+}}} \right ) \\[1.5ex] &= 0.018\bar{8}76~\mathrm{mol~L^{-1}} \\[1.5ex] &= 1.89 \times 10^{-2}~\mathrm{mol~L^{-1}} \end{align*} \]

Determine the molar concentrations (in mol L^–1) of the ions present in a solution created from mixing equal volumes of 0.50 M aqueous lead(II) nitrate and 0.50 M aqueous sodium chloride solutions? Assume that the volumes are precise to one decimal place in normalized scientific notation.

Solution

Answer:

[Pb²⁺] = 0.12 mol L⁻¹
[NO₃⁻] = 0.50 mol L⁻¹
[Na⁺] = 0.25 L⁻¹
[Cl⁻] = 0 mol L⁻¹

Concept: stoichiometry; molarity; dilution; limiting reactant

1. Balanced Equation and Initial Moles

The balanced equation for the precipitation reaction is:

\[ \begin{align*} \mathrm{Pb(NO_3)_2(aq)} + 2~\mathrm{NaCl(aq)} \longrightarrow \mathrm{PbCl_2(s)} + 2~\mathrm{NaNO_3(aq)} \end{align*} \]

This solution assumes a 1.0 L of each solution is mixed for a total volume of 2.0 L.

A Note on Assuming Volumes

Because the problem specifies mixing “equal volumes,” the final ion concentrations are independent of the actual starting volume chosen. The volume acts as a simple scaling factor that cancels out in the final moles/volume calculation. By choosing a convenient volume like 1.0 L, we make the mole calculations more straightforward without changing the final answer.

To determine which reactant is limiting, we calculate the moles of precipitate (PbCl₂) that could be formed from each of the initial reactants. The reactant that produces the smaller amount of product is the limiting reactant.

\[ \begin{align*} n(\mathrm{PbCl_2}) &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} ~ r(\mathrm{PbCl_2,Pb(NO_3)_2}) \\[1.5ex] &= \left ( \dfrac{0.5\bar{0}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~PbCl_2}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \\[1.5ex] &= 0.5\bar{0}0~\mathrm{mol} \\[3ex] n(\mathrm{PbCl_2}) &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl})_{\mathrm{initial}} ~ r(\mathrm{PbCl_2,NaCl}) \\[1.5ex] &= \left ( \dfrac{0.5\bar{0}~\mathrm{mol~NaCl}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~PbCl_2}}{2~\mathrm{mol~NaCl}} \right ) \\[1.5ex] &= 0.2\bar{5}0~\mathrm{mol} \end{align*} \]

NaCl is the limiting reactant.

The final concentration of the excess reactant ion, Pb²⁺, is calculated from the number of moles that remain in solution after the precipitation is complete. This is found by subtracting the moles that reacted from the initial moles, and then dividing by the total final volume of the mixture.

\[ \begin{align*} c(\mathrm{Pb^{2+}}) &= n(\mathrm{Pb^{2+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} - n(\mathrm{Pb(NO_3)_2})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Pb^{2+}, Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) \biggr] ~-~ \biggr[ n(\mathrm{PbCl_2}) ~ r(\mathrm{Pb(NO_3)_2,PbCl_2}) \biggr] \biggl\} ~ r(\mathrm{Pb^{2+}, Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \left ( \dfrac{0.5\bar{0}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L~Pb(NO_3)_2}}{} \right ) \biggr] ~-~ \biggr[ 0.2\bar{5}0~\mathrm{mol~PbCl_2} \left ( \dfrac{\mathrm{mol~Pb(NO_3)_2}}{\mathrm{mol~PbCl_2}} \right ) \biggr] \biggl\} ~ \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}}\right )\\[1.5ex] &= \left [ \left ( 0.5\bar{0} - 0.2\bar{5}0 \right ) ~\mathrm{mol~Pb(NO_3)_2} \right ] \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}}\right )\\[1.5ex] &= \left ( 0.2\bar{5}0~\mathrm{mol~Pb(NO_3)_2} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}}\right )\\[1.5ex] &= 0.1\bar{2}50~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.12~\mathrm{mol~L^{-1}} \end{align*} \]

The nitrate ion, NO₃^–, is a spectator ion, meaning it does not participate in the reaction. Its final concentration is determined by its initial moles diluted into the total final volume.

\[ \begin{align*} c(\mathrm{NO_3{^{-}}}) &= n(\mathrm{NO_3{^{-}}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} ~ r(\mathrm{NO_3{^{-}},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.5\bar{0}~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle 2~\mathrm{mol~NO_3{^{-}}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right ) \\[1.5ex] &= 0.5\bar{0}0~\mathrm{mol~L^{-1}} \end{align*} \]

Similarly, the sodium ion, Na⁺, is a spectator ion. Its final concentration is determined by its initial moles diluted into the total final volume.

\[ \begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.5\bar{0}~\mathrm{mol~NaCl}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~Na^+}}{\mathrm{mol~NaCl}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right )\\[1.5ex] &= 0.2\bar{5}0~\mathrm{mol~L^{-1}} \end{align*} \]

Because NaCl was identified as the limiting reactant, we assume that all of the chloride ions are consumed in the reaction to form the solid PbCl₂ precipitate.

\[ \begin{align*} c(\mathrm{Cl^-}) &= 0~\mathrm{mol~L^{-1}} \end{align*} \]

A precipitate forms when titanium(IV) chloride is added to water. Two water molecules react and form four HCl molecules. What is the identity of the precipitate?

What is the identity of the precipitate?
What is the molar concentration (in mol L^–1) of H⁺ ions if 3.00 g of TiCl₄ was added to enough water to give a 100.0 mL solution?

Solution

Answer:

TiO₂(s)
0.633 mol L⁻¹ H⁺(aq)

Concept: balancing equations; stoichiometry; molarity

a. Identity of the Precipitate

By taking an inventory of the atoms in the given reactants (TiCl₄ and 2 H₂O) and the known product (4 HCl), we can deduce that the unknown precipitate must contain 1 Ti atom and 2 O atoms. The precipitate is therefore TiO₂(s).

The complete balanced equation is: \[ \begin{align*} \mathrm{TiCl_4(s)} + 2~\mathrm{H_2O(l)} &\longrightarrow 4~\mathrm{HCl(aq)} + \mathrm{TiO_2(s)} \end{align*} \]

b. Molar Concentration of H⁺

\[ \begin{align*} c(\mathrm{H^+}) &= n(\mathrm{H^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{TiCl_4}) ~ r(\mathrm{HCl,TiCl_4}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{TiCl_4}) ~ M(\mathrm{TiCl_4})^{-1} ~ r(\mathrm{HCl,TiCl_4}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 3.0\bar{0}~\mathrm{g~TiCl_4} \left ( \dfrac{\mathrm{mol~TiCl_4}}{189.6\bar{7}~\mathrm{g}} \right ) \left ( \dfrac{4~\mathrm{mol~HCl}}{\mathrm{mol~TiCl_4}} \right ) \left ( \dfrac{\mathrm{mol~H^+}}{\mathrm{mol~HCl}} \right ) \biggr[ 100.\bar{0}~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr]^{-1} \\[1.5ex] &= 0.63\bar{2}67~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.633~\mathrm{mol~L^{-1}} \end{align*} \]

Chemist’s Note: Hydrolysis of Metal Halides

The vigorous reaction described in this problem is a classic example of the hydrolysis of a metal halide. “Hydrolysis” literally means “splitting by water” (hydro- + -lysis).

This reaction is common for halides containing small, highly charged metal cations, such as Ti⁴⁺, Al³⁺, and Sn⁴⁺. The high charge density of these metal ions strongly attracts the negative (oxygen) end of the polar water molecules. This M–O interaction is so strong that it weakens the O–H bonds within the water molecule, causing the water to “split” and release H⁺ ions. These protons then combine with the chloride ions (Cl^–) from the metal halide to form aqueous HCl.

The final result is the formation of a stable, insoluble metal oxide (like TiO₂) and a strong acid. This is why many metal halides, like titanium(IV) chloride and aluminum chloride, are powerful Lewis acids and must be stored in very dry conditions, as they will fume corrosively in moist air.

It is worth noting that while HCl(aq) is listed as a product of reaction, HCl(aq) is a strong acid and will not exist in water. It will react with water to create hydronium ions and hydroxide ions in equal amounts.

Determine the molar concentrations (in mol L⁻¹) the ions present in a solution created from mixing equal volumes of 0.75 M aqueous ammonium carbonate and 0.75 M aqueous potassium perchlorate. Assume that the volumes are precise to one decimal place in normalized scientific notation.

Solution

Answer:

[NH₄⁺] = 0.5 mol L⁻¹
[CO₃²⁻] = 0.38 mol L⁻¹
[K⁺] = 0.38 mol L^–1
[ClO₄⁻] = 0.38 mol L⁻¹

Concept: stoichiometry; molarity; dilution

The problem involves mixing two soluble ionic compounds: ammonium carbonate, (NH₄)₂CO₃, and potassium perchlorate, KClO₄. To determine if a reaction occurs, we must consider the potential products of a double displacement reaction: ammonium perchlorate (NH₄ClO₄) and potassium carbonate (K₂CO₃).

According to the solubility rules:

All ammonium (NH₄⁺) compounds are soluble.
All potassium (K⁺) compounds are soluble.

Since all possible products are also soluble in water, no precipitation occurs, and therefore there is no reaction. The final solution is simply a mixture of the four dissociated ions, each diluted by the presence of the other solution.

Choose a Starting Volume

This solution assumes 1.0 L of each solution is mixed, for a total final volume of 2.0 L. Since we are mixing equal volumes, the concentration of each initial salt is halved.

A Note on Assuming Volumes

Ammonium Ion [NH₄⁺]:

The initial salt is (NH₄)₂CO₃, which produces two moles of NH₄⁺.

\[ \begin{align*} c(\mathrm{NH_4{^+}}) &= n(\mathrm{NH_4{^+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{NH_4{^{+}},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{(NH_4)_2CO_3}) ~ V(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{NH_4{^{+}},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.7\bar{5}~\mathrm{mol~(NH_4)_2CO_3}}{\mathrm{L}} \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{2~mol~NH_4{^{+}}}}{\mathrm{mol~(NH_4)_2CO_3}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right ) \\[1.5ex] &= 0.7\bar{5}00~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.75~\mathrm{mol~L^{-1}} \end{align*} \]

Carbonate Ion [CO₃²⁻]:

The initial salt produces one mole of CO₃²⁻.

\[ \begin{align*} c(\mathrm{CO_3{^{2-}}}) &= n(\mathrm{CO_3{^{2-}}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{CO_3{^{2-}},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{(NH_4)_2CO_3}) ~ V(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{CO_3{^{2-}},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.7\bar{5}~\mathrm{mol~(NH_4)_2CO_3}}{\mathrm{L}} \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~CO_3{^{2-}}}}{\mathrm{mol~(NH_4)_2CO_3}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right ) \\[1.5ex] &= 0.3\bar{7}5~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.38~\mathrm{mol~L^{-1}} \end{align*} \]

Potassium Ion [K⁺]:

The initial salt KClO₄ produces one mole of K⁺.

\[ \begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KClO_4}) ~ r(\mathrm{K^+,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KClO_4}) ~ V(\mathrm{KClO_4}) ~ r(\mathrm{K^+,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.7\bar{5}~\mathrm{mol~KClO_4}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~K^+}}{\mathrm{mol~KClO_4}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right )\\[1.5ex] &= 0.3\bar{7}5~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.38~\mathrm{mol~L^{-1}} \end{align*} \]

Perchlorate Ion [ClO₄⁻]:

The initial salt KClO₄ produces one mole of ClO₄⁻.

\[ \begin{align*} c(\mathrm{ClO_4{^-}}) &= n(\mathrm{ClO_4{^-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KClO_4}) ~ r(\mathrm{ClO_4{^-},KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KClO_4}) ~ V(\mathrm{KClO_4}) ~ r(\mathrm{ClO_4{^-},KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.7\bar{5}~\mathrm{mol~KClO_4}}{\mathrm{L}} \right ) \left ( \dfrac{1.\bar{0}~\mathrm{L}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{mol~ClO_4{^-}}}{\mathrm{mol~KClO_4}} \right ) \left ( \dfrac{1}{2.\bar{0}~\mathrm{L}} \right )\\[1.5ex] &= 0.3\bar{7}5~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.38~\mathrm{mol~L^{-1}} \end{align*} \]

What volume (in mL) of each of the following bases will will completely react with 50.0 mL of 0.200 M HCl?

0.100 M NaOH
0.0500 M Sr(OH)₂
0.250 M KOH

Solution

Answer:

100. mL NaOH
100. mL Sr(OH)₂
40.0 mL KOH

Concept: balancing equations; stoichiometry; molarity; acid-base neutralization reaction

a. 0.100 M NaOH

The neutralization reaction is:

\[ \begin{align*} \mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)} \end{align*} \]

\[ \begin{align*} V(\mathrm{NaOH}) &= n(\mathrm{NaOH}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{NaOH,HCl}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{NaOH,HCl}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= \left ( \dfrac{0.20\bar{0}~\mathrm{mol~HCl}}{\mathrm{L}} \right ) \left ( \dfrac{50.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~NaOH}}{\mathrm{mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.10\bar{0}~\mathrm{mol~NaOH}} \right ) \\[1.5ex] &= 10\bar{0}.00~\mathrm{mL} \\[1.5ex] &= 100.~\mathrm{mL} \end{align*} \]

b. 0.0500 M Sr(OH)₂

Strontium hydroxide is a diacidic base, requiring two moles of HCl.

\[ \begin{align*} \mathrm{2~HCl(aq)} + \mathrm{Sr(OH)_2(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{SrCl_2(aq)} \end{align*} \]

\[ \begin{align*} V(\mathrm{Sr(OH)_2}) &= n(\mathrm{Sr(OH)_2}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{Sr(OH)_2,HCl}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{Sr(OH)_2,HCl}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= \left ( \dfrac{0.20\bar{0}~\mathrm{mol~HCl}}{\mathrm{L}} \right ) \left ( \dfrac{50.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Sr(OH)_2}}{\mathrm{2~mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.050\bar{0}~\mathrm{mol~Sr(OH)_2}} \right ) \\[1.5ex] &= 10\bar{0}.00~\mathrm{mL} \\[1.5ex] &= 100.~\mathrm{mL} \end{align*} \]

c. 0.250 M KOH

The neutralization reaction is:

\[ \begin{align*} \mathrm{HCl(aq)} + \mathrm{KOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KCl(aq)} \end{align*} \]

\[ \begin{align*} V(\mathrm{KOH}) &= n(\mathrm{KOH}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{KOH,HCl}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{KOH,HCl}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= \left ( \dfrac{0.20\bar{0}~\mathrm{mol~HCl}}{\mathrm{L}} \right ) \left ( \dfrac{50.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~KOH}}{\mathrm{mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.25\bar{0}~\mathrm{mol~KOH}} \right ) \\[1.5ex] &= 40.\bar{0}00~\mathrm{mL} \\[1.5ex] &= 40.0~\mathrm{mL} \end{align*} \]

Determine the molar concentration (in mol L⁻¹) of the salt produced by a reaction between a 250. mL 0.10 M aqueous HCl solution with a 150. mL 0.50 M aqueous KOH solution.

Solution

Answer: [KCl] = 0.0625 mol L⁻¹

Concept: stoichiometry; molarity; acid-base reaction; limiting reactant

This is a neutralization reaction between a strong acid (HCl) and a strong base (KOH), which react in a 1:1 ratio to produce water and a salt, potassium chloride (KCl).

\[ \begin{align*} \mathrm{HCl(aq)} + \mathrm{KOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KCl(aq)} \end{align*} \]

First, determine the limiting reactant by calculating the moles of the salt (KCl) that can be produced from each reactant.

\[ \begin{align*} n(\mathrm{KCl}) &= n(\mathrm{HCl}) ~ r(\mathrm{KCl,HCl}) \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{initial}) ~ r(\mathrm{KCl,HCl}) \\[1.5ex] &= \left ( \dfrac{0.1\bar{0}~\mathrm{mol~HCl}}{\mathrm{L}} \right ) \left ( \dfrac{25\bar{0}.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~KCl}}{\mathrm{mol~HCl}} \right ) \\[1.5ex] &= 0.02\bar{5}0~\mathrm{mol} \\[3ex] n(\mathrm{KCl}) &= n(\mathrm{KOH}) ~ r(\mathrm{KCl,KOH}) \\[1.5ex] &= c(\mathrm{KOH}) ~ V(\mathrm{initial}) ~ r(\mathrm{KCl,KOH}) \\[1.5ex] &= \left ( \dfrac{0.5\bar{0}~\mathrm{mol~KOH}}{\mathrm{L}} \right ) \left ( \dfrac{15\bar{0}.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~KCl}}{\mathrm{mol~KOH}} \right ) \\[1.5ex] &= 0.07\bar{5}0~\mathrm{mol} \end{align*} \]

Since HCl produces the lesser amount of product (0.0250 mol), HCl is the limiting reactant. The reaction will produce 0.0250 mol of KCl.

Now, calculate the final concentration of KCl, using the total final volume.

\[ \begin{align*} c(\mathrm{KCl}) &= n(\mathrm{KCl}) ~ V(\mathrm{final})^{-1} \\[1.5ex] &= 0.02\bar{5}0~\mathrm{mol~KCl} \left ( \dfrac{1}{\left (25\bar{0}. + 15\bar{0}. \right ) ~\mathrm{mL}}{} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.02\bar{5}0~\mathrm{mol~KCl} \left ( \dfrac{1}{40\bar{0}.~\mathrm{mL}}{} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.06\bar{2}5~\mathrm{mol~L^{-1}} \end{align*} \]

Lab Connection: Acid-Base Titration

This problem describes the calculation at the heart of an acid-base titration, one of the most fundamental techniques in analytical chemistry. In the lab, a chemist would add one solution (the titrant) from a long, graduated tube called a buret into a flask containing the other solution (the analyte).

To know when the neutralization is complete (the “equivalence point”), a tiny amount of a chemical indicator is added to the flask. The indicator, such as phenolphthalein, dramatically changes color at a specific pH, visually signaling the exact moment the reaction is finished. The volume of titrant added is then carefully read from the buret and used in a calculation just like the one above.

What volume (in mL) of a 0.150 M aqueous HNO₃ solution is required to neutralize a 50.0 mL 0.175 M aqueous Ba(OH)₂ solution?

Solution

A 120.0 mL 0.200 M aqueous potassium hydroxide solution is mixed with a 120.0 mL 0.200 M aqueous magnesium nitrate solution.

Write a balanced chemical equation for the reaction that occurs.
Determine the precipitate that forms (if any).
Determine the mass (in g) of precipitate that forms (if any).
Determine the molar concentration (in mol L^–1) of each ion in solution after the reaction goes to 100 % completion.

Solution

Answer:

2 KOH(aq) + Mg(NO₃)₂(aq) → 2 KNO₃(aq) + Mg(OH)₂(s)
Mg(OH)₂(s)
0.700 g Mg(OH)₂(s)
[K⁺] = 0.100 mol L⁻¹; [OH⁻] = 0 mol L⁻¹; [Mg²⁺] = 0.0500 mol L⁻¹; [NO₃⁻] = 0.200 mol L⁻¹

Concept: balancing equations; stoichiometry; molarity; precipitation; limiting reactant

a. Balanced Chemical Equation

\[ \begin{align*} \mathrm{2~KOH(aq)} + \mathrm{Mg(NO_3)_2(aq)} \longrightarrow \mathrm{2~KNO_3(aq)} + \mathrm{Mg(OH)_2(s)} \end{align*} \]

b. Precipitate

According to solubility rules, nitrates and potassium compounds are soluble, but magnesium hydroxide is insoluble. Therefore, the precipitate is Mg(OH)₂(s).

c. Mass of Precipitate

Determine the limiting reactant.

\[ \begin{align*} n(\mathrm{Mg(OH)_2}) &= c(\mathrm{KOH}) ~ V(\mathrm{initial}) ~ r(\mathrm{Mg(OH)_2,KOH}) \\[1.5ex] &= \dfrac{0.20\bar{0}~\mathrm{mol~KOH}}{\mathrm{L}} \left ( \dfrac{120.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{2~mol~KOH}} \right ) \\[1.5ex] &= 0.012\bar{0}00~\mathrm{mol}\\[3ex] n(\mathrm{Mg(OH)_2}) &= c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{initial}) ~ r(\mathrm{Mg(OH)_2,Mg(NO_3)_2}) \\[1.5ex] &= \dfrac{0.20\bar{0}~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{120.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 0.024\bar{0}00~\mathrm{mol} \end{align*} \]

KOH is limiting.

Determine the mass of the Mg(OH)₂ precipitate.

\[ \begin{align*} m(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.012\bar{0}00~\mathrm{mol~Mg(OH)_2}~ \left ( \dfrac{58.3\bar{3}~\mathrm{g~Mg(OH)_2}}{\mathrm{mol}} \right )\\[1.5ex] &= 0.69\bar{9}96~\mathrm{g} \\[1.5ex] &= 0.700~\mathrm{g} \end{align*} \]

d. Final Ion Concentrations

Molar concentration of K⁺ (spectator ion)

\[ \begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KOH}) ~ r(\mathrm{K^+,KOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KOH}) ~ V(\mathrm{KOH}) ~ r(\mathrm{K^+,KOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.20\bar{0}~\mathrm{mol~KOH}}{\mathrm{L}} \right ) \left ( \dfrac{120.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~K^+}}{\mathrm{mol~KOH}} \right ) \left ( \dfrac{1}{(120.\bar{0} + 120.\bar{0})~\mathrm{mL}} \right ) \\[1.5ex] &= 0.10\bar{0}00~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.100~\mathrm{mol~L^{-1}} \end{align*} \]

Molar concentration of OH⁻ (spectator ion)

Since KOH is the limiting reactant, we assume all the OH^– is consumed. The final concentration is approximately 0 mol L⁻¹.

\[ \begin{align*} c(\mathrm{OH^-}) &= 0 ~~~~\textrm{since KOH is limiting and all hydroxide precipitates as}~\mathrm{Mg(OH)_2} \end{align*} \]

Molar concentration of Mg²⁺ (excess reactant)

\[ \begin{align*} c(\mathrm{Mg^{2+}})_{\mathrm{final}} &= n(\mathrm{Mg^{2+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Mg(NO_3)_2})_{\mathrm{initial}} - n(\mathrm{Mg(NO_3)_2})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Mg^{2+},Mg(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{Mg(NO_3)_2}) \biggr] ~-~ \biggr[ n(\mathrm{Mg(OH)_2})_{\mathrm{formed}}~ r(\mathrm{Mg(NO_3)_2,Mg(OH)_2}) \biggr] \Biggl\} \\[1.5ex] &\phantom{=}~~~ r(\mathrm{Mg^{2+},Mg(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{\biggr[ \left ( \dfrac{0.20\bar{0}~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{120.\bar{0}~\mathrm{mL~Mg(NO_3)_2}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] ~- \\[1.5ex] &\phantom{=\Biggl\{~} ~\biggr[ \left ( \dfrac{0.012\bar{0}~\mathrm{mol~Mg(OH)_2}}{} \right ) \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{\mathrm{mol~Mg(OH)_2}} \right ) \biggr] \Biggl\} \\[1.5ex] &\phantom{=\biggr[} \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(120.\bar{0} + 120.\bar{0})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \left [ \left ( 0.024\bar{0} - 0.012\bar{0} \right ) ~\mathrm{mol~Mg(NO_3)_2} \right ] \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(120.\bar{0} + 120.\bar{0})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= \left [ 0.012\bar{0}~\mathrm{mol~Mg(NO_3)_2} \right ] \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(120.\bar{0} + 120.\bar{0})~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.050\bar{0}00~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.0500~\mathrm{mol~L^{-1}} \end{align*} \]

Molar concentration of NO₃⁻ (spectator ion)

\[ \begin{align*} c(\mathrm{NO_3{^-}}) &= n(\mathrm{NO_3{^-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Mg(NO_3)_2}) ~ r(\mathrm{NO_3{^-},Mg(NO_3)_2}) V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{Mg(NO_3)_2}) ~ r(\mathrm{NO_3{^-},Mg(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \left ( \dfrac{0.20\bar{0}~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{120.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{\displaystyle \mathrm{2~mol~NO_3{^-}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(120.\bar{0} + 120.\bar{0})~\mathrm{mL}} \right ) \\[1.5ex] &= 0.20\bar{0}00~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.200~\mathrm{mol~L^{-1}} \end{align*} \]