Chapter 11, Problem 11.18P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with $4.20mL$ of $0.438M{\text{ HClO}}_4$ has to be determined.

Concept introduction:

The term $pH$ refers to concentration of $H^{+}$ion in a solution.

$pH$ is determined by the formula –

$pH{\text{ = -log[H}}^{+}\text{]}$

Answer to Problem 11.18P

$pH$ of the solution made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with $4.20mL$ of $0.438M{\text{ HClO}}_4$ is determined to be $\underset{\_\_\_\_\_\_\_\_}{9.44.}$

Explanation of Solution

When $NaCN$ and $HCl{O}_4$ are made into aqueous solution, they are completely dissociated into ions as $N{a}^{+},C{N}^{-}$ and $H^{+},Cl{O}_4^{-}$ respectively.  The actual reaction takes place between active species which are $H^{+}\text{ and }C{N}^{-}$ ions.

$C{N}_{(aq)}^{-}+H_{(aq)}^{+}\stackrel{}{\to }HC{N}_{(aq)}$

Given data:

$\begin{array}{l}volume\text{ of NaCN}\text{= 50}\text{.00 mL}\\ \text{Molarity}\text{= 0}\text{.100 M}\\ {\text{volume of HClO}}_4=\text{ 4}\text{.20 mL}\\ \text{Molarity}\text{= 0}\text{.438M}\end{array}$

Volume and molarity of $C{N}^{-}$ is equivalent to that of volume and molarity of $NaCN$. Similarly volume and molarity of $H^{+}$ corresponds to that of $HCl{O}_4.$

Find the equivalence point. At equivalence point,

$\begin{array}{l}mol{\text{ of CN}}^{-}=mol{\text{ of H}}^{+}\\ (0.100M).(50.00mL)=\text{ (0}\text{.438M)}{\text{.V}}_e\\ V_e=\frac{(0.100M).(50.00mL)}{\text{(0}\text{.438M)}}\\ =\text{ 11}\text{.42 mL}\text{.}\end{array}$

Therefore volume of $H^{+}$ at equivalence point is $\text{11}\text{.42 mL}\text{.}$

$pH$ of the solution has to be determined not exactly at equivalence point but while before nearing the equivalence point as the volume of $0.438M{\text{ HClO}}_4$ that we are mixing with $NaCN$ is less than that of the volume of acid at equivalence point ( $V_e$ ) .  At this stage the concentration of $C{N}^{-}$ is still left as it is not completely reacted.  $H^{+}\text{ and }C{N}^{-}$ ions tend to form buffer at this stage and their respective concentration is charted as follows –

Reaction:   $C{N}_{(aq)}^{-}+H_{(aq)}^{+}\stackrel{}{\to }HC{N}_{(aq)}$

Initial conc.     $1$ $0$ $0$

Change in conc.   $-4.20/11.42$       $4.20/11.42$ $4.20/11.42$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

Final conc.            $1-4.20/11.42$ $0$ $4.20/11.42$

From the standard data of $p{K}_a$ values, $p{K}_a$ for $HCN$ is $\mathrm{9.20.}$

Determining $pH$,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[C{N}^{-}\right]}{\left[HCN\right]}\\ =9.20+\log \left[\frac{1-\frac{4.20}{11.42}}{\frac{4.20}{11.42}}\right]\\ =\text{ 9}\text{.44}\end{array}$

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with $11.82mL$ of $0.438M{\text{ HClO}}_4$ has to be determined.

Concept introduction:

The term $pH$ refers to concentration of $H^{+}$ion in a solution.

$pH$ is determined by the formula –

$pH{\text{ = -log[H}}^{+}\text{]}$

Answer to Problem 11.18P

$pH$ of the solution made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with $11.82mL$ of $0.438M{\text{ HClO}}_4$ is determined to be $\underset{\_\_\_\_\_\_\_\_}{2.55.}$

Explanation of Solution

When $NaCN$ and $HCl{O}_4$ are made into aqueous solution, they are completely dissociated into ions as $N{a}^{+},C{N}^{-}$ and $H^{+},Cl{O}_4^{-}$ respectively.  The actual reaction takes place between active species which are $H^{+}\text{ and }C{N}^{-}$ ions.

$C{N}_{(aq)}^{-}+H_{(aq)}^{+}\stackrel{}{\to }HC{N}_{(aq)}$

Given data:

$\begin{array}{l}volume\text{ of NaCN}\text{= 50}\text{.00 mL}\\ \text{Molarity}\text{= 0}\text{.100 M}\\ {\text{volume of HClO}}_4=\text{ 11}\text{.82 mL}\\ \text{Molarity}\text{= 0}\text{.438M}\end{array}$

Volume and molarity of $C{N}^{-}$ is equivalent to that of volume and molarity of $NaCN$. Similarly volume and molarity of $H^{+}$ corresponds to that of $HCl{O}_4.$

Find the equivalence point. At equivalence point,

$\begin{array}{l}mol{\text{ of CN}}^{-}=mol{\text{ of H}}^{+}\\ (0.100M).(50.00mL)=\text{ (0}\text{.438M)}{\text{.V}}_e\\ V_e=\frac{(0.100M).(50.00mL)}{\text{(0}\text{.438M)}}\\ =\text{ 11}\text{.42 mL}\text{.}\end{array}$

Therefore volume of $H^{+}$ at equivalence point is $\text{11}\text{.42 mL}\text{.}$

$pH$ of the solution has to be determined not exactly at equivalence point but after reaching the equivalence point – that is post-equivalence region.  At this stage the concentration of $C{N}^{-}$ is nil.  $pH$ is determined by measuring the concentration of excess acid $HCl{O}_4$ beyond the equivalence point.

$\text{11}\text{.82 mL}$ of $HCl{O}_4$ is beyond the equivalence point which is $\text{11}\text{.42 mL}\text{.}$ thus volume of $HCl{O}_4$ at this stage is $\text{11}\text{.82 mL-11}\text{.42 mL = 0}\text{.40 mL}\text{.}$ therefore,

$\begin{array}{l}{\text{[H]}}^{+}=\left(\frac{0.40mL}{50mL+11.82mL}\right)\times 0.438M\\ =\text{ 0}\text{.00283 M}\end{array}$

Determining $pH$,

$\begin{array}{l}pH=-\log [H^{+}]\\ =-\log \left(0.00283\right)\\ =\text{ 2}\text{.55}\end{array}$

(c)

Interpretation Introduction

Interpretation:

$pH$ of the solution at equivalence point made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with  $0.438M{\text{ HClO}}_4$ has to be determined.

Concept introduction:

The term $pH$ refers to concentration of $H^{+}$ion in a solution.

$pH$ is determined by the formula –

$pH{\text{ = -log[H}}^{+}\text{]}$

Answer to Problem 11.18P

$pH$ of the solution at equivalence point made by mixing $50.00mL$ of $0.100M\text{ NaCN}$ with $0.438M{\text{ HClO}}_4$ is determined to be $\underset{\_\_\_\_\_\_\_\_}{5.15.}$

Explanation of Solution

When $NaCN$ and $HCl{O}_4$ are made into aqueous solution, they are completely dissociated into ions as $N{a}^{+},C{N}^{-}$ and $H^{+},Cl{O}_4^{-}$ respectively.  The actual reaction takes place between active species which are $H^{+}\text{ and }C{N}^{-}$ ions.

$C{N}_{(aq)}^{-}+H_{(aq)}^{+}\stackrel{}{\to }HC{N}_{(aq)}$

Known data:

$\begin{array}{l}volume\text{ of NaCN}\text{= 50}\text{.00 mL}\\ \text{Molarity}\text{= 0}\text{.100 M}\\ \text{Molarity}\text{= 0}\text{.438M}\end{array}$

Volume and molarity of $C{N}^{-}$ is equivalent to that of volume and molarity of $NaCN$. Similarly volume and molarity of $H^{+}$ corresponds to that of $HCl{O}_4.$

Find the equivalence point. At equivalence point,

$\begin{array}{l}mol{\text{ of CN}}^{-}=mol{\text{ of H}}^{+}\\ (0.100M).(50.00mL)=\text{ (0}\text{.438M)}{\text{.V}}_e\\ V_e=\frac{(0.100M).(50.00mL)}{\text{(0}\text{.438M)}}\\ =\text{ 11}\text{.42 mL}\text{.}\end{array}$

Therefore volume of $H^{+}$ at equivalence point is $\text{11}\text{.42 mL}\text{.}$

At equivalence point, $HCN$ dissociates and equilibrium is attained,

$HCN\stackrel{}{\rightleftharpoons }{H}^{-}+C{N}^{-}$

Thus, $[H^{+}]=[C{N}^{-}]=x,[HCN]=0.0184-x$

Dissociation constant $K_a$ for the above reaction is known to be $6.2\times {10}^{10}$ and hence,

$\begin{array}{l}{K}_a=\left[\frac{[H^{+}][C{N}^{-}]}{[HCN]}\right]=6.2\times {10}^{10}\\ 6.2\times {10}^{10}=\frac{x^2}{0.0184-x}\end{array}$

Solving for ‘x’

$x=[H^{+}]=7.1\times {10}^{-6}$

Determining $pH$,

$\begin{array}{l}pH=-\log [H^{+}]\\ =-\log (7.1\times {10}^{-6})\\ =\text{ 5}\text{.15}\end{array}$