Chapter 19, Problem 19.129P

(a)

Interpretation Introduction

Interpretation:

Concentration of each species that is present like $\left[HClO\right]$, $\left[Cl{O}^{-}\right]$, $\left[H_3{O}^{+}\right]$, $\left[O{H}^{-}\right]$, $\left[N{a}^{+}\right]$ has to be determined. 

Concept Introduction:

Dissociation constant of acid:

An acid dissociation constant, $K_a$ quantitative measure of the strength of an acid in solution.  It is the equilibrium constant for a chemical reaction known as dissociation in the context of acid–base reactions. 

The chemical equation for acid dissociation can be written symbolically as:

$HA\rightleftarrows A^{-}+H^{+}$

The acid dissociation constant can be presented as:

$K_a=\frac{\left[A^{-}\right]\left[H^{+}\right]}{\left[HA\right]}$

Concentration:

$Concentration=\frac{No.ofmoles}{Volumeofsolution}$

Answer to Problem 19.129P

$\left[HClO\right]$, $\left[Cl{O}^{-}\right]$, $\left[H_3{O}^{+}\right]$, $\left[O{H}^{-}\right]$, $\left[N{a}^{+}\right]$ is calculated as $0.16M$, $0.10M$, $4.8\times {10}^{-8}M$, $2.1\times {10}^{-7}M$, and $0.10M$ respectively. 

Explanation of Solution

Given data:

$\begin{array}{l}Volumeofsolution=500mL=500\times {10}^{-3}L=0.500L\\ \\ MolofsolidNaOH=0.050\\ \\ MolofHClO=0.13\\ \\ K_{a}ofHCl=3.0\times {10}^{-8}\end{array}$

$NaOH$ is a strong base, so it dissociates completely as follows:

$\begin{array}{l}NaOH\left(s\right)\to N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\\ 0.050mol0.050mol0.050mol\end{array}$

The $O{H}^{-}$ ions from $NaOH$ will react with $HClO$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{MolesO{H}^{-}\left(aq\right)+HClO\left(aq\right)\to H_{2}O\left(l\right)+Cl{O}^{-}\left(aq\right)}\\ Initial0.0500.13-0\\ Change-0.050-0.050-+0.050\end{array}}\\ Equilibrium00.080-0.050\end{array}$

$\begin{array}{l}\left[HClO\right]=\frac{0.080mol}{0.500L}\\ \\ =0.16M.\\ \\ \left[Cl{O}^{-}\right]=\frac{0.050mol}{0.500L}\\ \\ =0.10M.\end{array}$

The balanced chemical equation of dissociation of $HClO$ in water can be written as:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{Concentration\left(M\right)HClO\left(aq\right)+H_{2}O\left(l\right)\rightleftarrows H_2{O}^{+}\left(aq\right)+Cl{O}^{-}\left(aq\right)}\\ Initial0.16-0.10\\ Change-x+x+x\end{array}}\\ Equilibrium0.16-xx0.10+x\end{array}$

As x is small, $\left[HClO\right]\approx 0.16M$ and $\left[Cl{O}^{-}\right]\approx 0.10M$.

$\begin{array}{l}{K}_a=\frac{\left[Cl{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[HClO\right]}\\ \\ \left[H_3{O}^{+}\right]=\frac{\left[HClO\right]K_a}{\left[Cl{O}^{-}\right]}\\ \\ =\frac{\left(0.16\right)\left(3.0\times {10}^{-8}\right)}{0.10}\\ \\ =4.8\times {10}^{-8}M.\end{array}$

Therefore, $\left[H_3{O}^{+}\right]$ is $4.8\times {10}^{-8}M$.

$\begin{array}{l}{K}_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]=\frac{K_w}{\left[H_3{O}^{+}\right]}\\ \\ =\frac{1.0\times {10}^{-14}}{4.8\times {10}^{-8}}\\ \\ =2.08333\times {10}^{-7}=2.1\times {10}^{-7}M.\end{array}$

Therefore, $\left[O{H}^{-}\right]$ is $2.1\times {10}^{-7}M$.

$\begin{array}{l}\left[N{a}^{+}\right]=\frac{0.050mol}{0.500L}\\ \\ =0.10M.\end{array}$

$\left[N{a}^{+}\right]$ is calculated as $0.10M$.

(b)

Interpretation Introduction

Interpretation:

The $pH$ of the solution has to be calculated. 

Concept Introduction:

$\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

Answer to Problem 19.129P

The $pH$ of the solution calculated as $7.32$.

Explanation of Solution

$\begin{array}{l}\left[H_3{O}^{+}\right]=4.8\times {10}^{-8}M\\ \\ \text{pH}\text{=}-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =-\log \left(4.8\times {10}^{-8}\right)\\ \\ =7.31875876=\mathrm{7.32.}\end{array}$

Therefore, the $pH$ of the solution calculated as $\text{7}\text{.32}$.

(c)

Interpretation Introduction

Interpretation:

The $pH$ after adding $0.0050molofHCl$ to the flask has to be determined. 

Concept Introduction:

$\text{pH}\text{=}-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

Answer to Problem 19.129P

The $pH$ after adding $0.0050molofHCl$ to the flask calculated as $7.25.$

Explanation of Solution

Given data:

$\begin{array}{l}Volumeofsolution=500mL=500\times {10}^{-3}L=0.500L\\ \\ MolofsolidNaOH=0.050\\ \\ MolofHClO=0.13\\ \\ MolofHCl=0.0050\\ \\ K_{a}ofHCl=3.0\times {10}^{-8}\end{array}$

If $0.0050molofHCl$ is added then the $Cl{O}^{-}$ will react with the $H^{+}$ to form $HClO$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{Moles{H}^{+}\left(aq\right)+Cl{O}^{-}\left(aq\right)\to HClO\left(aq\right)}\\ Initial0.00500.0500.080\\ Change-0.0050-0.0050+0.0050\end{array}}\\ Equilibrium00.0450.085\end{array}$

$\begin{array}{l}\left[HClO\right]=\frac{0.085mol}{0.500L}\\ \\ =0.17M.\\ \\ \left[Cl{O}^{-}\right]=\frac{0.045mol}{0.500L}\\ \\ =0.090M.\end{array}$

The balanced chemical equation of dissociation of $HClO$ in water can be written as:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{Concentration\left(M\right)HClO\left(aq\right)+H_{2}O\left(l\right)\rightleftarrows H_2{O}^{+}\left(aq\right)+Cl{O}^{-}\left(aq\right)}\\ Initial0.17-0.090\\ Change-x+x+x\end{array}}\\ Equilibrium0.17-xx0.090+x\end{array}$

As x is small, $\left[HClO\right]\approx 0.17M$ and $\left[Cl{O}^{-}\right]\approx 0.090M$.

$\begin{array}{l}{K}_a=\frac{\left[Cl{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[HClO\right]}\\ \\ \left[H_3{O}^{+}\right]=\frac{\left[HClO\right]K_a}{\left[Cl{O}^{-}\right]}\\ \\ =\frac{\left(0.17\right)\left(3.0\times {10}^{-8}\right)}{0.090}\\ \\ =5.6667\times {10}^{-8}M.\end{array}$

$\begin{array}{l}\text{pH}\text{=}-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =-\log \left(5.6667\times {10}^{-8}\right)\\ \\ =7.246669779=\mathrm{7.25.}\end{array}$

Therefore, the $pH$ after adding $0.0050molofHCl$ to the flask calculated as $\text{7}\text{.25}\text{.}$