Chapter 17, Problem 28PS

Assume you dissolve 0.235 g of the weak acid benzoic acid, C₆H₅CO₂H, in enough water to make 1.00 × 10² mL of solution and then titrate the solution with 0.108 M NaOH.

C₆H₅CO₂H(aq) + OH⁻(aq) ⇄ C₆H₅CO₂⁻(aq) + H₂O(ℓ)

1. (a) What was the pH of the original benzoic add solution?
2. (b) What are the concentrations of all of the following ions at the equivalence point: Na⁺, H₃O⁺, OH⁻, and C₆H₅CO₂⁻?
3. (c) What is the pH of the solution at the equivalence point?

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{COOH}$ and $\text{NaOH}$. The value of $\text{pH}$, of the original solution of ${\text{C}}_6{\text{H}}_5\text{COOH}$.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{COONa}\left(\text{aq}\right)$

$K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

Explanation of Solution

The $\text{pH}$ value before the titration can be calculated by using $K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$

Given:

Refer to the Apendix H in the textbook for the value of $K_{\text{a}}$.

The value of $K_{\text{a}}$ for benzoic acid is $6.3\times {10}^{-5}$.

The value of $K_{\text{w}}$ for water is $1.0\times {10}^{-14}$.

The amount of ${\text{C}}_6{\text{H}}_5\text{COOH}$ dissolved is $0.235\text{ g}$.

Molar mass of ${\text{C}}_6{\text{H}}_5\text{COOH}$ l is $122\text{ g}\cdot {\text{mol}}^{-1}$.

$\text{Number of moles}$ of ${\text{C}}_6{\text{H}}_5\text{COOH}$ are calculated by using following expression,

$\text{Number of moles = }\frac{\text{weight of solute }}{\text{molar mass of solute}}\left(\text{mol}\right)$

Substitute $0.235\text{ g}$ for $\text{weight of solute }$, $122\text{ g}\cdot {\text{mol}}^{-1}$ for $\text{molar mass of solute}$.

$\begin{array}{c}\text{Number of moles = }\frac{\text{0}\text{.235 g}}{122\text{ g}\cdot {\text{mol}}^{-1}}\\ =0.0019\text{ mol}\end{array}$

Therefore $\text{Number of moles}$ of ${\text{C}}_6{\text{H}}_5\text{COOH}$ are $0.0019\text{ mol}$.

The  volume of the solvent is $0.100\text{ L}$.

The initial concentration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ is calculated by using the expression,

$\text{Molarity}=\frac{\text{Number of moles}}{\text{volume of the solvent}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$

Substitute $0.0019\text{ mol}$ for $\text{Number of moles}$, $0.100\text{ L}$ for volume of the solvent.$\begin{array}{c}\text{Molarity}=\frac{0.0019\text{ mol}}{0.100\text{ L}}\\ =0.019\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore initial concentration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ is $0.019\text{ mol}\cdot {\text{L}}^{-1}$.

ICE table (1) gives the dissociation of ${\text{C}}_6{\text{H}}_5\text{COOH}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.019& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.019-x& & & & +x& & +x\end{array}$

There is an approximation, that the value of $x$ is very small as comparison to $0.019$ thus it can be neglected with respect to it.

Therefore, Concentration of benzoic acid left after reaction is $0.019\text{ mol}\cdot {\text{L}}^{-1}$.

The hydronium ion concentration is calculated by expression,

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$

Substitute $x$ for ${\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}$, $0.019$ for ${\left[\text{HA}\right]}_{\left(\text{eq}\right)}$ and $6.3\times {10}^{-5}$ for $K_{\text{a}}$.

$\begin{array}{l}6.3\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.019\right)}\\ 6.3\times {10}^{-5}=\frac{\left(x^2\right)}{\left(0.019\right)}\end{array}$

Rearrange for $x$,

$\begin{array}{c}x=\sqrt{\left(6.3\times {10}^{-5}\right)\left(0.019\right)}\\ x=1.09\times 10^{-3}\end{array}$

Therefore, concentration of hydronium ion is $1.09\times 10^{-3}$.

Calculate the $\text{pH}$ value by using following expression,

$\text{pH = }-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute $1.09\times 10^{-3}$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$.

$\begin{array}{c}\text{pH = }-\log \left(1.09\times {10}^{-3}\right)\\ =2.96\end{array}$

The value of $\text{pH}$ for the original solution of benzoic acid is $2.96$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ is to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{COOH}$ and $\text{NaOH}$. The concentration of ${\text{OH}}^{-}$,${\text{H}}_3{\text{O}}^{+}$,${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$ and ${\text{Na}}^{+}$ has to be calculated at equivalence point.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{COONa}\left(\text{aq}\right)$

$K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

Explanation of Solution

The concentration value of ${\text{OH}}^{-}$, benzoate ion, ${\text{H}}_3{\text{O}}^{+}$ and ${\text{Na}}^{+}$ is calculated below.

Given:

The concentration of $\text{NaOH}$ is $0.108\text{ mol}\cdot {\text{L}}^{-1}$.

The volume of $\text{NaOH}$ used to neutralize all the phenol is calculated.

Expression used for the neutralization is as follows,

${\text{C}}_{\text{1}}{\text{V}}_1={\text{C}}_{\text{2}}{\text{V}}_{\text{2}}$

Here,

• ${\text{C}}_{\text{1}}$ is the concentration of benzoic acid
• ${\text{V}}_{\text{1}}$ is the volume of benzoic acid.
• ${\text{C}}_{\text{2}}$ is the concentration of $\text{NaOH}$ used.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ used for the neutralization.

Substitute $0.019\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{1}}$, $0.100\text{ L}$ for ${\text{V}}_{\text{1}}$,$0.108\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{2}}$.

$\left(0.019\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.100\text{ L}\right)=\left(0.108\text{ mol}\cdot {\text{L}}^{-1}\right)\left({\text{V}}_{\text{2}}\right)$

Rearrange for ${\text{V}}_2$,

$\begin{array}{c}{\text{V}}_{\text{2}}=\frac{\left(0.019\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.100\text{L}\right)}{\left(0.108\text{ mol}\cdot {\text{L}}^{-1}\right)}\\ =0.0175\text{ L}\end{array}$

Therefore, volume of the $\text{NaOH}$ used is $0.0175\text{ L}$ or $17.5\text{ mL}$.

The calculation of moles is done by using the expression,

$\text{Number}\text{of moles}=\text{concentration}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\cdot \text{volume}\left(\text{L}\right)$

The ICE table (2) for the reaction between $\text{NaOH}$ and ${\text{C}}_6{\text{H}}_5\text{OH}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5\text{COONa}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0019& & 0.0019& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0019& & -0.0019& & & & +0.0019\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.0019\end{array}$

From ICE table (2),

Number of moles of benzoate ion produced after the reaction are $0.0019\text{ mol}$.

The total volume after the reaction is calculated as,

$\begin{array}{c}\text{total}\text{volume = }0.100\left(\text{L}\right)+0.\text{0175 L}\\ =0.1175\text{ L}\end{array}$

Therefore, total volume after reaction is $0.1175\text{ L}$.

Concentration calculations is done by using the expression,

$\text{concentration = }\frac{\text{Number}\text{of moles}}{\text{total volume}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$

Calculate the concentration of benzoate ion after reaction.

Substitute, $0.0019\text{ mol}$ for $\text{Number}\text{of moles}$ and $0.1175\text{ L}$ for volume.

$\begin{array}{c}\text{concentration = }\frac{0.0019}{\text{0}\text{.1175}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\\ =0.0161\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of benzoate ion after reaction is $0.0161\text{mol}\cdot {\text{L}}^{-1}$.

The ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$ ion produced will undergo hydrolysis in presence of water and the reaction equilibrium is written in ICE table (3).

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0161& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0161-x& & & & +x& & +x\end{array}$

From ICE table (3),

Concentration of benzoate ion left after reaction is $\left(0.0161-x\right)\text{ mol}\cdot {\text{L}}^{-1}$.

Concentration of benzoic acid produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of ${\text{OH}}^{-}$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

The value of $x$ is very small as comparison to $0.0161$ and thus it can be neglected.

Calculate the concentration of ${\text{OH}}^{-}$ by using the equation (3).

The expression of $K_{\text{b}}$ for benzoate ion from the ICE table (3) will be written as,

$K_{\text{b}}=\frac{{\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]}_{\left(\text{eq}\right)}{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\right]}_{\left(\text{eq}\right)}}$

Substitute, $0.158\times {10}^{-9}$ for $K_{\text{b}}$, $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]}_{\left(\text{eq}\right)}$, $\left(0.0161\right)$ for ${\left[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\right]}_{\left(\text{eq}\right)}$.

$0.158\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{\left(0.0161\right)}$

Rearrange for $x$,

$\begin{array}{c}x=\sqrt{\left(0.158\times {10}^{-9}\right)\left(0.0161\right)}\\ =\sqrt{0.0025\times {10}^{-9}}\\ =0.158\times {10}^{-5}\end{array}$

Therefore value of ${\text{OH}}^{-}$ concentration is $1.58\times 10^{-6}mol\cdot L^{-1}$.

The value of ${\text{C}}_6{\text{H}}_5\text{COOH}$ concentration is $1.58\times 10^{-6}mol\cdot L^{-1}$.

The concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ at equivalence point is $0.0161mol\cdot L^{-1}$.

The concentration of ${\text{Na}}^{+}$ at equivalence point is equal to the initial concentration of benzoate ion before undergo hydrolysis i.e. $0.0161mol\cdot L^{-1}$ .

Calculate the value of $\text{pOH}$ by using the expression,

$\text{pOH = }-\log \left[{\text{OH}}^{-}\right]$

Substitute, $1.58\times {10}^{-6}$ for $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\text{pOH = }-\log \left(1.58\times {10}^{-6}\right)\\ =-\left(-5.80\right)\\ =5.80\end{array}$

Therefore, the value of $\text{pOH}$ is $5.80$.

Thus, the value of $\text{pH}$ is calculated by using expression,

$\text{pH + pOH =}\text{14}$

Rearrange for $\text{pH}$,

$\text{pH =}\text{14}-\text{ pOH}$

Substitute, $5.80$ for $\text{pOH}$.

$\begin{array}{c}\text{pH =}\text{14}-5.80\\ \text{= }8.20\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $8.20$.

The $\text{pH}$ value is used to calculate the concentration of hydronium ion at equivalence point.

Calculate the hydronium ion concentration by using expression,

$\text{pH = }-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$

Rearrange for $\left[{\text{H}}_3{\text{O}}^{+}\right]$,

$\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\left(\text{pH}\right)}$

Substitute $8.60$ for $\text{pH}$.

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\left(8.20\right)}\\ =6.30\times 10^{-9}\end{array}$

Therefore, hydronium ion concentration at equivalence point is $6.30\times 10^{-9}mol\cdot L^{-1}$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{COOH}$ and $\text{NaOH}$. The value of $\text{pH}$ at equivalence point has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{COOH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{COONa}\left(\text{aq}\right)$

$K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

Explanation of Solution

The addition of $17.5\text{ mL}$ $\text{NaOH}$ will required to reach the equivalence point of titration. At equivalence point all the benzoic acid and hydroxide ions will be used only benzoate will be present. The hydrolysis of benzoate ion will produce a very small amount of ${\text{OH}}^{-}$ ions. The $\text{pH}$ calculation at equivalence point is given below.

Given:

The value of $\text{pH}$ is calculated by using expression,

$\text{pH + pOH =}\text{14}$

Rearrange for $\text{pH}$,

$\text{pH =}\text{14}-\text{ pOH}$

Substitute, $5.80$ for $\text{pOH}$.

$\begin{array}{c}\text{pH =}\text{14}-\text{5}\text{.80}\\ \text{= }8.20\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $8.20$.