Chapter 17, Problem 32QP

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{H}}^{\text{+}}$, ${\text{CH}}_3\text{COOH}$, ${\text{CH}}_3{\text{COO}}^{-}$, ${\text{OH}}^{\text{-}}$, and ${\text{Na}}^{\text{+}}$ in a solution, made by mixing given volume of $0.167\text{M NaOH}$ with given volume of $0.100{\text{M CH}}_3\text{COOH}$, are to be calculated.

Concept introduction:

Acid–base titration is a technique to analyze the unknown concentration of an acid or base through the known concentration of an acid and the base.

The equivalence point is the point in the acid–base titration in a chemical reaction where the number of moles of the titrant and the unknown concentration of the analyte are equal and it is used to identify the unknown concentration of the analyte.

If an anion reacts with water, it is called anionic hydrolysis. If a cation reacts with water, it is called cationic hydrolysis.

Answer to Problem 32QP

Solution: The final concentration of $\left[{\text{H}}^{+}\right]\text{ is }3\times {10}^{-13}\text{M}$, the final concentration of $\left[{\text{Na}}^{+}\right]\text{ is }0.0835\text{M}$, the final concentration of $\left[{\text{OH}}^{-}\right]\text{ is }0.0335\text{M}$, the final concentration of $\left[{\text{CH}}_3\text{COOH}\right]\text{ is 8}.4\times {10}^{-10}\text{M,}$ and thefinal concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]\text{ is }0.0500\text{M}$.

Explanation of Solution

Given information: The concentration of $\text{NaOH}$ is $0.167\text{M }$.

The volume of $\text{NaOH}$ is $500\text{mL}$ or $0.500\text{L}$.

The concentration of ${\text{CH}}_3\text{COOH}$ is $0.100\text{M }$.

The volume of ${\text{CH}}_3\text{COOH}$ is $500\text{mL}$ or $0.500\text{L}$.

Moles of $\text{NaOH}$ ion are calculated using the expression as follows:

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of sol}\text{.}$

Substitute $0.167\text{M }$ for the concentration of $\text{NaOH}$ and $0.500\text{L}$ for the volume in the above expression as follows:

$\begin{array}{l}\text{Moles of NaOH}=0.167\text{M }\times 0.500\text{L}\\ \text{ = }\left(\frac{0.167\text{mol}}{1\text{L sol}}\times 0.500\text{L}\right)\\ =0.0835\text{mole}\text{.}\end{array}$

Moles of ${\text{CH}}_3\text{COOH}$ ions are calculated using the expression as follows:

${\text{Moles of CH}}_3\text{COOH}={\text{Concentration of CH}}_3\text{COOH}\times \text{Volume of sol}$.

Substitute, $0.100\text{M }$ for the concentration of ${\text{CH}}_3\text{COOH}$ and $0.500\text{L}$ for the volume in the above expressionas follows:

$\begin{array}{l}{\text{Moles of CH}}_3\text{COOH}=0.100\text{ M}\times 0.500\text{L}\\ \text{ =}\left(\frac{0.100\text{mol}}{1\text{L sol}}\times 0.500\text{L}\right)\\ =0.0500\text{mole}\text{.}\end{array}$

Summarize the moles at equilibrium, which are as follows.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+\text{NaOH}\left(aq\right)\rightleftharpoons {\text{CH}}_{\text{3}}\text{COONa}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{mol}\right)& 0.0500& 0.0835& 0\\ \text{Change}\left(\text{mol}\right)& -0.0500& -0.0500& +0.0500\\ \text{Final}\left(\text{mol}\right)& 0& 0.0355& 0.0500\end{array}\end{array}$

The resulting solution is not a buffer solution.

The volume of the resulting solution is calculated as follows: $\left(500\text{mL}+500\text{mL}\right)=1000\text{mL}=1\text{L}$.

Concentration of $\left[{\text{Na}}^{+}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{Na}}^{+}\right]=\frac{\text{Moles}\text{of}{\text{Na}}^{+}}{\text{Volume}\text{of}\text{sol}}$.

Substitute $\left(0.0335+0.0500\right)\text{mol}$ for moles of ${\text{Na}}^{+}$ and $1\text{L}$ for the volume of solution in the above expression as follows:

$\begin{array}{l}\text{Concentration}\text{of}\left[{\text{Na}}^{+}\right]=\frac{\left(0.0335+0.0500\right)\text{mol}}{1\text{L}}\\ =0.0835\text{M}\text{.}\end{array}$

Concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=\frac{\text{Moles}\text{of}{\text{CH}}_3{\text{COO}}^{-}}{\text{Volume}\text{of}\text{sol}}$.

Substitute $0.0500\text{mol}$ for the moles of ${\text{CH}}_3{\text{COO}}^{-}$ and $1\text{L}$ for the volume in the above expression as follows:

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=\left(\frac{0.0500\text{mol}}{1\text{L}}\right)\\ =0.0500\text{M}\text{.}\end{array}$

Concentration of $\left[{\text{OH}}^{-}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]=\frac{\text{Moles}\text{of}{\text{OH}}^{-}}{\text{Volume}\text{of}\text{sol}}$.

Substitute $0.0335\text{mol}$ for the moles of ${\text{OH}}^{-}$ and $1\text{L}$ for the volume in the above expression as follows:

$\begin{array}{l}\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]=\left(\frac{0.0335\text{mol}}{1\text{L}}\right)\\ =0.0335\text{M}\text{.}\end{array}$

Concentration of $\left[{\text{H}}^{+}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{H}}^{+}\right]\times \text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]={10}^{-14}$.

Substitute $0.0335\text{mol}$ for the moles of ${\text{OH}}^{-}$ in the above expression as follows:

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{H}}^{+}\right]=\left(\frac{1\times {10}^{-14}}{0.0335}\right)\\ =3\times {10}^{-13}\text{M}\text{.}\end{array}$

Consider $x$ to be a degree of dissociation.

The conjugate base undergoes anionic hydrolysis.

Summarize the concentration at equilibrium as follows:

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(M\right)& 0.0500& -& 0& 0.0335\\ \text{Change}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(M\right)& 0.0500-x& -& x& 0.0335+x\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}$.

Here, ${\text{K}}_{\text{b}}$ is the base dissociation constant, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of the acetate ions, $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide ions, and $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid.

Substitute $\left(0.0500-x\right)$ for the concentration of acetate ions, $\left(0.0335+x\right)$ for the concentration of hydroxide ions, $\left(x\right)$ for the concentration of acetic acid, and $5.6\times {10}^{-10}{\text{ for K}}_{\text{b}}$ in the above expression as follows:

$5.6\times {10}^{-10}=\frac{\left(x\right)\left(0.0335+x\right)}{\left(0.0500-x\right)}$.

The value of x is very small as compared to $\text{0}\text{.0335 and 0}\text{.0500}$. It can be neglected.

Solving further,

$\begin{array}{l}5.6\times {10}^{-10}=\frac{\left(x\right)\left(0.0335\right)}{\left(0.0500\right)}\\ x=8.4\times {10}^{-10}\end{array}$

Conclusion

The final concentrations are as follows:

$\begin{array}{c}\left[{\text{H}}^{+}\right]=3\times {10}^{-13}\text{M}\\ \left[{\text{Na}}^{+}\right]=0.0835\text{M}\\ \left[{\text{OH}}^{-}\right]=0.0335\text{M}\\ \left[{\text{CH}}_3\text{COOH}\right]=\text{ 8}.4\times {10}^{-10}\text{M}\\ \left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.0500\text{M}\end{array}$