Chapter 10, Problem 10.8P

Interpretation Introduction

Interpretation:

Titration reaction has to be written along with equivalence volume and $\text{pH}$ at indicated volumes has to be calculated and plot of $\text{pH}$ against $V_{\text{b}}$ has to be drawn.

Concept Introduction:

Dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$

Here,

$M_1$ denotes molarity of  $\text{HA}$ solution.

$V_1$ denotes volume of $\text{HA}$ solution.

$M_2$ denotes molarity of $\text{NaOH}$ solution.

$V_2$ denotes volume  of $\text{NaOH}$ solution.

Explanation of Solution

The dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Substitute $\text{0}\text{.05 M}$ for $M_1$, $\text{0}\text{.500 M}$ for $M_2$ , $\text{50 mL}$ for $V_1$ in equation (1).

  $\left(\text{0}\text{.050 M}\right)\left(\text{50}\text{.0 mL}\right)=\left(\text{0}\text{.500 M}\right)V_2$        (2)

Rearrange equation (2) to calculate value of $V_2$.

  $\begin{array}{c}{V}_2=\frac{\left(\text{0}\text{.050 M}\right)\left(\text{50}\text{.0 mL}\right)}{\left(\text{0}\text{.5 M}\right)}\\ =5.00\text{ mL}\end{array}$

So, equivalence volume is $5.00\text{ mL}$.

Equilibrium for titration reaction is given as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{p}{K}_{\text{b}}$ from $\text{p}{K}_{\text{b}}$ is given as follows:

  $\text{p}{K}_{\text{a}}=14-\text{p}{K}_{\text{b}}$        (3)

Substitute 9.00 for ${\text{pK}}_{\text{a}}$ in equation (3).

  $\begin{array}{c}\text{p}{K}_{\text{a}}=14-5\\ =9\end{array}$

Formula to calculate ${\text{pK}}_{\text{a}}$ is given as follows:

  $K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$        (3)

Substitute 4.00 for ${\text{pK}}_{\text{a}}$ in equation (3).

  $K_{\text{a}}={10}^{-4}$

Corresponding expression of $K_{\text{a}}$ is written as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$        (4)

ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{HA}& \rightleftharpoons & {\text{H}}^{+}& +& {\text{A}}^{-}\\ \text{Initial}& 0.05& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}& 0.05-x& & x& & x\end{array}$

Substitute $x$ to $\left[{\text{H}}^{+}\right]$ and $\left[{\text{A}}^{-}\right]$ , ${10}^{-4}$ for $K_{\text{a}}$ $0.05-x$ for $\left[\text{HA}\right]$ in equation (4)

  ${10}^{-4}=\frac{x^2}{0.05-x}$

Simplify to obtain the value of $x$ as $0.00218\text{ M}$. Thus $\left[\text{HA}\right]$ is calculated as follows:

  $\begin{array}{c}\left[\text{HA}\right]=\left(0.05-0.00218\right)\text{ M}\\ =\text{0}\text{.0478 M}\end{array}$

Until just before equivalence point there is a mixture of unionized $\text{HA}$ and ${\text{A}}^{-}$. $\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_a+\log \frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$        (5)

Substitute 4.00 for ${\text{pK}}_{\text{a}}$, $0.00218\text{ M}$ for $\left[{\text{A}}^{-}\right]$, $\text{0}\text{.0478 M}$ for $\left[\text{HA}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=4.00+\log \left(\frac{0.00218}{\text{0}\text{.0478}}\right)\\ =2.659\\ \approx 2.66\end{array}$

When volume of base added is $\text{1 mL}$,  new ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{HA}& +& {\text{OH}}^{-}& \rightleftharpoons & {\text{A}}^{-}\\ & 2.50& & 0.5& & 0\\ \text{Change}& -0.5& & -0.5& & +x\\ \text{Equilibrium}& 2.0& & 0& & 0.5\end{array}$

Substitute 4.00 for ${\text{pK}}_{\text{a}}$, $0.5\text{ M}$ for $\left[{\text{A}}^{-}\right]$, $\text{2}\text{.0 M}$ for $\left[\text{HA}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=4.00+\log \left(\frac{0.5}{2.0}\right)\\ =3.397\\ \approx 3.40\end{array}$

Similarly when volume of base added is $\text{2}\text{.5 mL}$, new ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{HA}& +& {\text{OH}}^{-}& \rightleftharpoons & {\text{A}}^{-}\\ \text{Initial}& 2.50& & 1.25& & 0\\ \text{Change}& -1.25& & -1.25& & +1.25\\ \text{Equilibrium}& 1.25& & 0& & 1.25\end{array}$

Substitute 4.00 for ${\text{pK}}_{\text{a}}$, $\text{1}\text{.25 M}$ for $\left[{\text{A}}^{-}\right]$ and $\left[\text{HA}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=4.00+\log \left(\frac{1.25}{1.25}\right)\\ =4.00\end{array}$

Similarly when volume of base added is $\text{4}\text{.00 mL}$, new ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{HA}& +& {\text{OH}}^{-}& \rightleftharpoons & {\text{A}}^{-}\\ \text{Initial}& 2.50& & 2.00& & 0\\ \text{Change}& 2.00& & -2.00& & +2.00\\ \text{Equilibrium}& 0.50& & 0& & 2.00\end{array}$

Substitute 4.00 for ${\text{pK}}_{\text{a}}$, $\text{0}\text{.50 M}$ for $\left[{\text{A}}^{-}\right]$, $\text{2}\text{.00 M}$ and $\left[\text{HA}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=4.00+\log \left(\frac{2.00}{0.50}\right)\\ =4.60\end{array}$

Similarly when volume of base added is $\text{4}\text{.90 mL}$, new ICE table is drawn as follows:

$\begin{array}{cccccc}& \text{HA}& +& {\text{OH}}^{-}& \rightleftharpoons & {\text{A}}^{-}\\ \text{Initial}& 2.50& & 2.45& & 0\\ \text{Change}& 2.45& & -2.45& & +2.45\\ \text{Equilibrium}& 0.05& & 0& & 2.45\end{array}$

Substitute 4.00 for ${\text{pK}}_{\text{a}}$, $\text{2}\text{.45 M}$ for $\left[{\text{A}}^{-}\right]$, $\text{0}\text{.05 M}$ and $\left[\text{HA}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=4.00+\log \left(\frac{2.45}{0.05}\right)\\ =5.69\end{array}$

When volume of base added is $\text{5}\text{.00 mL}$, value of $\left[{\text{A}}^{-}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{A}}^{-}\right]=\frac{\left(50.00\text{ mL}\right)\left(0.050\text{ M}\right)}{\left(50.00+5.00\right)\text{mL}}\\ =0.0454\text{ M}\end{array}$

New ICE table can be drawn as follows:

  $\begin{array}{cccccc}& {\text{A}}^{-}& \rightleftharpoons & \text{HA}& +& {\text{OH}}^{-}\\ \text{Initial}& 0.0454& & 0& & 0\\ \text{Change}& -y& & +y& & +y\\ \text{Equilibrium}& 0.0454-y& & y& & y\end{array}$

Expression of equilibrium constant for this reaction is given as follows:

  $K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$        (6)

Substitute ${10}^{-14}$ for $K_{\text{w}}$, and ${10}^{-4}$ for $K_{\text{a}}$ in equation (6).

  $\begin{array}{c}{K}_{\text{b}}=\frac{{10}^{-14}}{{10}^{-4}}\\ ={10}^{-10}\end{array}$

Therefore expression of equilibrium constant can be written as follows:

  ${10}^{-10}=\frac{y^2}{0.0454-y}$

Simplify to obtain the value of $y$ as $2.13\times {10}^{-6}$. Expression to calculate $\text{pH}$ is then given as follows:

  $\text{pH}=-\log \left(\frac{K_{\text{w}}}{y}\right)$        (7)

Substitute ${10}^{-14}$ for $K_{\text{w}}$, and $2.13\times {10}^{-6}$ for $y$ in equation (7).

  $\begin{array}{c}{10}^{-10}=\frac{y^2}{0.04537-y}\\ =2.129\times {10}^{-6}\end{array}$

When $\text{5}\text{.10 mL}$ of base is added it exceeds acid by $0.05\text{ M}$, and total volume is $\text{55}\text{.10 mL}$, thus excess $\left[{\text{OH}}^{-}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{0.05\text{ M}}{\text{55}\text{.10 mL}}\\ =9.0744\times {10}^{-4}\end{array}$

Formula to calculate $\text{pOH}$ is given as follows:

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (3)

Substitute $9.0744\times {10}^{-4}\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(9.0744\times {10}^{-4}\right)\\ =3.04\end{array}$

Formula to calculate $\text{pH}$ from $\text{pOH}$ is given as follows:

  $\text{pH}=14-\text{pOH}$        (4)

Substitute 3.04 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-3.04\\ =10.95\end{array}$

Similarly when $\text{6}\text{.00 mL}$ of base is added it exceeds acid by $0.5\text{ M}$, and total volume is $\text{56}\text{.00 mL}$, thus excess $\left[{\text{OH}}^{-}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{0.5\text{ M}}{\text{56}\text{.00 mL}}\\ =8.92\times {10}^{-3}\end{array}$

Formula to calculate $\text{pOH}$ is given as follows:

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (3)

Substitute $8.92\times {10}^{-3}\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(8.92\times {10}^{-3}\right)\\ =2.04\end{array}$

Formula to calculate $\text{pH}$ from $\text{pOH}$ is given as follows:

  $\text{pH}=14-\text{pOH}$        (4)

Substitute 2.04 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-3.04\\ =11.95\end{array}$

Corresponding plot of $\text{pH}$ against $V_{\text{a}}$ is drawn as follows: