Chapter 19, Problem 19.54P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the calculation has to be calculated for $\text{23}\text{.4}\text{mL}$ of $0.0390MHN{O}_2$.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

$\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

Acid ionization constant${\text{K}}_{\text{a}}$: 

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

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

Where $K_a$ acid ionization is constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

Answer to Problem 19.54P

The pH of a solution made by mixing $\text{23}\text{.4}\text{mL}$ of $0.0390MHN{O}_2$ is $7.76$

Explanation of Solution

Given,

The balance equation is given below,

${\text{K}}^{+}\text{(aq) +}{\text{OH}}^{-}\text{(aq)}\text{+}{\text{HNO}}_2\left(\text{aq}\right)\underset{}{\rightleftharpoons }{\text{K}}^{+}\text{(aq) +}{\text{NO}}_2\left(\text{aq}\right)\text{+}{\text{H}}_2\text{O(l)}$

Potassium ions on the product side are written as separate spices because they have no effect on $pH$ of the solution.

Next calculate the volume of $KOH$,

Volume of $KOH$ in $mL$,

$KOH=\left(\frac{0.0390molHN{O}_2}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(23.4mL\right)\left(\frac{1molKOH}{1molKOH}\right)\left(\frac{L}{0.0588molKOH}\right)\left(\frac{1mL}{{10}^{-3}L}\right)$

Determination the moles of $HN{O}_2$present,

$\begin{array}{c}MolesofHN{O}_2=\left(\frac{0.0390molHN{O}_2}{1mL}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(23.4mL\right)\\ \\ =0.0009126molHN{O}_2\end{array}$

The equivalent point, $\text{0}\text{.0009126}\text{mol}KOH$ will be added so, the moles acid = moles base.

The $KOH$ will react with an equal amount of the acid $0molHN{O}_2$ will remain and $\text{0}\text{.0009126}\text{mol}N{O}^{2-}$ will be formed.

ICE Table of equilibrium,

$\begin{array}{l}{\text{HNO}}_{\text{2}}\text{(aq) }\text{+}\text{KOH(aq)}\underset{}{\rightleftharpoons }{H}_{2}O\text{(l) +}{\text{NO}}_2^{-}\left(\text{aq}\right)\text{+}{K}^{+}\text{(aq)}\\ \text{Initial }\text{(M)}\text{0}\text{.0009126}\text{mol}\text{0}\text{.0009126}\text{mol }-\text{0}-\\ \text{Change}\left(\text{M}\right)-\text{0}\text{.0009126}\text{mol}-\text{0}\text{.0009126}\text{mol}-+\text{0}\text{.0009126}\text{mol}-\\ -----------------------------------------------------\\ \text{Final }00-+\text{0}\text{.0009126}\text{mol}-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(23.4+15.5204\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.0389204L\end{array}$

Concentration of $N{O}_2^{-}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\left(0.0009126molN{O}_2^{-}\right)\left(0.0389204L\right)\\ =0.023447858M\end{array}$

Next calculate $K_b$for $N{O}_2$

The equilibrium value for $HN{O}_2=7.1\times {10}^{-4}$

So,

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(7.1\times {10}^{-4}\right)}=1.40845\times {10}^{-11}$ 

Using a reaction table for the equilibrium reaction of $N{O}_2^{-}$

  $\begin{array}{l}{\text{NO}}_2^{-}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }HN{O}_2\text{ +}O{H}^{-}\\ \text{Initial }\text{(M)}:0.023447858M-0\text{0}\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ --------------------------------\\ \text{Equilibrium: }0.023447858M-x+x+x\end{array}$

Determination the hydroxide ion concentration from the $K_b$ and then determine the $pH$ from the $pOH$.

  $\begin{array}{l}{K}_b=1.40845\times {10}^{-11}=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.023447858-x\right]}=\frac{x^2}{\left[0.023447858\right]}\\ \left[O{H}^{-}\right]=x=5.7468\times {10}^{-7}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pOH=-\log \left(5.7468\times {10}^{-7}\right)\\ =6.240574\\ pH=14.00-pOH\\ =14.00-6.240574\\ =7.759426\\ pH=7.76\end{array}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the calculation has to be calculated for $17.3\text{mL}$ of $0.130M{H}_{2}C{O}_3$ with two equivalent point.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

Acid ionization constant${\text{K}}_{\text{a}}$: 

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

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

Where $K_a$ acid ionization is constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

  ${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

Answer to Problem 19.54P

The pH of a solution made by mixing $17.3\text{mL}$ of $0.130M{H}_{2}C{O}_3$ is $11.35$

Explanation of Solution

The balance equation is given below,

$\begin{array}{l}KOH\left(aq\right)+H_{2}C{O}_3\left(aq\right)\underset{}{\to }{\text{K}}^{+}\text{(aq)}+HC{O}_3^{-}\left(aq\right)+{\text{H}}_2\text{O(l)}\\ \\ KOH\left(aq\right)+HC{O}_3^{-}\left(aq\right)\underset{}{\to }{\text{K}}^{+}\text{(aq)}+C{O}_3^{2-}\left(aq\right)+{\text{H}}_2\text{O(l)}\end{array}$

Potassium ions on the product side are written as separate spices because they have no effect on the $pH$ of the solution.

Next calculate the volume of $KOH$,

Volume of $KOH$ in $mL$,

$\begin{array}{c}KOH=\left(\frac{0.130mol{H}_{2}C{O}_3}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(17.3mL\right)\left(\frac{1molKOH}{1mol{H}_{2}C{O}_3}\right)\left(\frac{L}{0.0588molKOH}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =38.248299\\ \\ =38.2mLofKOH\end{array}$

The solution requires an equal volume to reach second equivalence point $76.4mL$

Determination the moles of $HC{O}_3^{-}$produced,

$\begin{array}{c}Molesof{H}_{2}C{O}_3=\left(\frac{0.130mol{H}_{2}C{O}_3}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(17.3mL\right)\left(\frac{1molHC{O}_3^{-}}{1mol{H}_{2}C{O}_3}\right)\\ \\ =0.002249molHC{O}_3^{-}\end{array}$

An equal number of moles of $C{O}_3^{2-}$ will be present at the second equivalent point.

Determination the liters of solution present at the first equivalent point,

Volume

  $\begin{array}{l}=\left[\left(17.3+38.2482299\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.055548L\end{array}$

Determination the liters of solution present at the second equivalent point,

Volume

  $\begin{array}{l}=\left[\left(17.3+38.2482299+38.2482299\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.0937966L\end{array}$

Concentration of $HC{O}_3^{-}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.022249molHC{O}_3^{-}\right)}{\left(0.055548L\right)}\\ =0.0404875M\end{array}$

Concentration of $C{O}_3^{2-}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.022249molC{O}_3^{2-}\right)}{\left(0.0937966L\right)}\\ =0.023977M\end{array}$

Next calculate $K_b$for $HC{O}_3^{-}$

The equilibrium value for $K_a=4.5\times {10}^{-7}HC{O}_3^{-}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(4.5\times {10}^{-7}\right)}=2.222\times {10}^{-8}$ 

The equilibrium value for $K_a=4.7\times {10}^{-11}C{O}_3^{2-}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(4.7\times {10}^{-11}\right)}=2.1276596\times {10}^{-4}$

Determination the hydroxide ion concentration from the $K_b$ and then determine the $pH$ from $pOH$.

Calculation for first equilibrium point:

  $\begin{array}{l}{K}_b=2.222\times {10}^{-8}=\frac{\left[H_{2}C{O}_3\right]\left[O{H}^{-}\right]}{\left[HC{O}_3^{-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.0404875-x\right]}=\frac{x^2}{\left[0.0404875\right]}\\ \\ \left[O{H}^{-}\right]=x=2.999387\times {10}^{-5}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pOH=-\log \left(2.999387\times {10}^{-5}\right)\\ \\ =4.522967495\\ \\ pH=14.00-pOH\\ \\ =14.00-4.522967495\\ \\ =9.4770\\ \\ pH=9.48\end{array}$

Calculation for second equilibrium point:

  $\begin{array}{l}{K}_b=2.222\times {10}^{-8}=\frac{\left[HC{O}_3^{-}\right]\left[O{H}^{-}\right]}{\left[C{O}_3^{2-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.023977-x\right]}=\frac{x^2}{\left[0.023977\right]}\\ \\ \left[O{H}^{-}\right]=x=2.2584677\times {10}^{-3}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pOH=-\log \left(2.2586477\times {10}^{-3}\right)\\ \\ =2.6461515\\ \\ pH=14.00-pOH\\ \\ =14.00-2.6461515\\ \\ =11.3538\\ \\ pH=11.35\end{array}$