Chapter 5, Problem 5.31P

(a)

Interpretation Introduction

Interpretation:

$pH$ of $1.0\times {10}^{-4}\text{ M}{H}_{3}B{O}_{3\left(aq\right)}$ solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  $\begin{array}{l}HA+H_{2}O\stackrel{}{\to }{H}_3{O}^{+}+A^{-}\\ \\ Ka=\frac{[H_3{O}^{+}][A^{-}]}{[HA]}\end{array}$

Explanation of Solution

Given,

    $1.0\times {10}^{-4}\text{ M}{H}_{3}B{O}_{3\left(aq\right)}$ Acts as a monoprotic acid.

The reaction of $H_{3}B{O}_{3\left(aq\right)}$ with water is,

  $H_{3}B{O}_{3\left(aq\right)}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+H_{2}B{O}_{{}_{(aq)}}^{-}$

From the above reaction the equilibrium constant is,

    $Ka=\frac{[H_3{O}^{+}][H_{2}B{O}_3{}_{}^{-}]}{[H_{3}B{O}_3]}=\text{2}\text{.0}\times {10}^{-9}$

The ICE table is,

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{BO}}_{\text{3}\left(\text{aq}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}_{{}_{\text{(aq)}}}^{\text{+}}\text{+}{\text{H}}_{\text{2}}{\text{BO}}_{{}_{\text{(aq)}}}^{\text{-}}\\ \text{Initial}\text{1}{\text{.0×10}}^{\text{-4}}\text{M}\text{0}\text{0}\\ \text{Change}\text{1}{\text{.0×10}}^{\text{-4}}\text{M-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{1}{\text{.0×10}}^{\text{-4}}\text{M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}Ka=\frac{[H_3{O}^{+}][H_{2}B{O}_3{}_{}^{-}]}{[H_{3}B{O}_3]}=\text{2}\text{.0}\times {10}^{-9}\\ \text{2}\text{.0}\times {10}^{-9}=\frac{x^2}{\text{1}{\text{.0×10}}^{\text{-4}}\text{M-x}}=\frac{x^2}{\text{1}{\text{.0×10}}^{\text{-4}}\text{M}}\\ x^2=\text{2}\text{.0}\times {10}^{-9}\times \text{1}{\text{.0×10}}^{\text{-4}}\text{M}\\ =\sqrt{\text{2}\text{.0}\times {10}^{-13}}\\ =4.47\times {10}^{-7}\end{array}$

We know,

    $\begin{array}{l}pH=-\log [H_3{O}^{+}]\\ =-\log [4.47\times {10}^{-7}]\\ =6.3\end{array}$

Hence, the $pH$ of $1.0\times {10}^{-4}\text{ M}{H}_{3}B{O}_{3\left(aq\right)}$ solution is $6.3$.

(b)

Interpretation Introduction

Interpretation:

$pH$ of $\text{0}{\text{.015 M H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  $\begin{array}{l}HA+H_{2}O\stackrel{}{\to }{H}_3{O}^{+}+A^{-}\\ \\ Ka=\frac{[H_3{O}^{+}][A^{-}]}{[HA]}\end{array}$

Explanation of Solution

Given,

    $\text{0}{\text{.015 M H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$

The reaction of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ with water is,

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{H}}_2{\text{PO}}_{{}_{\text{4(aq)}}}^{-}Ka1=\text{7}\text{.6}\times {10}^{-3}\\ {\text{H}}_2{\text{PO}}_{{}_{\text{4(aq)}}}^{-}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+H{\text{PO}}_{{}_{\text{4(aq)}}}^{2-}Ka2=6.\text{2}\times {10}^{-8}\\ {\text{HPO}}_{{}_{\text{4(aq)}}}^{2-}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{PO}}_{{}_{\text{4(aq)}}}^{3-}Ka3=2.1\times {10}^{-13}\end{array}$

From the above reactions the equilibrium constants are,

    $\begin{array}{l}Ka1=\frac{[H_3{O}^{+}][H_{2}P{O}_4{}_{}^{-}]}{[H_{3}P{O}_4]}=7.6\times {10}^{-3}\\ Ka2=\frac{[H_3{O}^{+}][HP{O}_4{}_{}^{2-}]}{[H_{2}P{O}_4{}_{}^{-}]}=6.2\times {10}^{-8}\\ Ka3=\frac{[H_3{O}^{+}][P{O}_4{}_{}^{3-}]}{[HP{O}_4{}_{}^{2-}]}=2.1\times {10}^{-13}\end{array}$

The ICE tables are,

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{H}}_2{\text{PO}}_{{}_{\text{4(aq)}}}^{-}\\ \text{Initial}\text{0}\text{.015 M}\text{0}\text{0}\\ \text{Change}\text{0}\text{.015 M-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.015 M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}Ka1=\frac{[H_3{O}^{+}][H_{2}P{O}_4{}_{}^{-}]}{[H_{3}P{O}_4]}=7.6\times {10}^{-3}\\ 7.6\times {10}^{-3}=\frac{x^2}{\text{0}\text{.015 M-x}}=\frac{x^2}{\text{0}\text{.015 M}}\\ x^2=7.6\times {10}^{-3}\times \text{0}\text{.015 M}\\ =\sqrt{1.14\times {10}^{-4}}\\ =0.0107M\end{array}$

  $\begin{array}{l}{\text{H}}_2{\text{PO}}_{{}_{\text{4(aq)}}}^{-}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+H{\text{PO}}_{{}_{\text{4(aq)}}}^{2-}\\ \text{Initial}0.0107M\text{0}\text{0}\\ \text{Change}0.0107M\text{-x}\text{+x}\text{+x}\\ \text{Equilibrium}0.0107\text{M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}\text{Ka2}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HPO}}_{\text{4}}{}_{}^{\text{2-}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}{}_{}^{\text{-}}\text{]}}\text{=}\text{6}{\text{.2×10}}^{\text{-8}}\\ \text{6}{\text{.2×10}}^{\text{-8}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0107M-x}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0107}\text{M}}\\ {\text{x}}^{\text{2}}\text{=}\text{6}{\text{.2×10}}^{\text{-8}}\text{×0}\text{.0107}\text{M}\\ \text{=}\sqrt{\text{6}{\text{.63×10}}^{\text{-10}}}\\ \text{=}\text{2}{\text{.5×10}}^{\text{-5}}\text{M}\end{array}$

  $\begin{array}{l}{\text{HPO}}_{{}_{\text{4(aq)}}}^{2-}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{PO}}_{{}_{\text{4(aq)}}}^{3-}\\ \text{Initial}\text{2}{\text{.5×10}}^{\text{-5}}\text{M}\text{0}\text{0}\\ \text{Change}\text{2}{\text{.5×10}}^{\text{-5}}\text{M-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{2}{\text{.5×10}}^{\text{-5}}\text{M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}\text{Ka3}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][PO}}_{\text{4}}{}_{}^{\text{3-}}\text{]}}{{\text{[HPO}}_{\text{4}}{}_{}^{\text{2-}}\text{]}}\text{=}\text{2}{\text{.1×10}}^{\text{-13}}\\ \text{2}{\text{.1×10}}^{\text{-13}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{2}{\text{.5×10}}^{\text{-5}}\text{M-x}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{2}{\text{.5×10}}^{\text{-5}}\text{M}}\\ {\text{x}}^{\text{2}}\text{=}\text{2}{\text{.1×10}}^{\text{-13}}\text{×2}{\text{.5×10}}^{\text{-5}}\text{M}\\ \text{=}\sqrt{\text{5}{\text{.25×10}}^{\text{-18}}}\\ \text{=}\text{2}{\text{.29×10}}^{\text{-9}}\text{M}\end{array}$

From the above calculations, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}_{{}_{\text{(aq)}}}^{\text{+}}\text{=}\text{0}\text{.0107}\text{M}\text{+2}{\text{.5×10}}^{\text{-5}}\text{M+}\text{2}{\text{.29×10}}^{\text{-9}}\text{M}\\ \text{=}\text{0}\text{.0107}\text{M}\end{array}$

We know,

    $\begin{array}{l}pH=-\log [H_3{O}^{+}]\\ =-\log [\text{0}\text{.0107}\text{M}]\\ =1.97pH\end{array}$

Hence, the $pH$ of $\text{0}{\text{.015 M H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ solution is $1.97pH$.

(c)

Interpretation Introduction

Interpretation:

$pH$ of $\text{0}\text{.10 M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{3(aq})}$ solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  $\begin{array}{l}HA+H_{2}O\stackrel{}{\to }{H}_3{O}^{+}+A^{-}\\ \\ Ka=\frac{[H_3{O}^{+}][A^{-}]}{[HA]}\end{array}$

Explanation of Solution

Given,

    $\text{0}\text{.10 M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{3(aq})}$

The reaction of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3(aq})}$ with water is,

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{3(aq})}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{HSO}}^{-}{}_{\text{3(aq})}Ka1=1.5\times {10}^{-2}\\ {\text{HSO}}^{-}{}_{\text{3(aq})}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{SO}}^{2-}{}_{\text{3(aq})}Ka2=1.2\times {10}^{-7}\end{array}$

From the above reaction the equilibrium constant is,

    $\begin{array}{l}Ka1=\frac{[H_3{O}^{+}][H_{2}S{O}_3{}_{}^{-}]}{[{\text{H}}_{\text{2}}{\text{SO}}_3]}=1.5\times {10}^{-2}\\ Ka2=\frac{[H_3{O}^{+}][S{O}_3{}_{}^{-}]}{[HS{O}_3]}=1.2\times {10}^{-7}\end{array}$

The ICE table is,

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{3(aq})}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{HSO}}^{-}{}_{\text{3(aq})}\\ \text{Initial}\text{0}\text{.10 M}\text{0}\text{0}\\ \text{Change}\text{0}\text{.10 M-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.10 M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}Ka1=\frac{[H_3{O}^{+}][H_{2}S{O}_3{}_{}^{-}]}{[{\text{H}}_{\text{2}}{\text{SO}}_3]}=1.5\times {10}^{-2}\\ 1.5\times {10}^{-2}=\frac{x^2}{\text{0}\text{.10 M}\text{-x}}=\frac{x^2}{\text{0}\text{.10 M}}\\ x^2=1.5\times {10}^{-2}\times \text{0}\text{.10 M}\\ =\sqrt{1.5\times {10}^{-3}}\\ =0.038M\end{array}$

  $\begin{array}{l}{\text{HSO}}^{-}{}_{\text{3(aq})}+H_2{O}_{(l)}\stackrel{}{\to }{H}_3{O}_{{}_{(aq)}}^{+}+{\text{SO}}^{2-}{}_{\text{3(aq})}\\ \text{Initial}\text{0}\text{.10 M}\text{0}\text{0}\\ \text{Change}\text{0}\text{.10 M-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.10 M-x}\text{x}\text{x}\end{array}$

From the above table, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}Ka2=\frac{[H_3{O}^{+}][S{O}_3{}_{}^{-}]}{[HS{O}_3]}=1.2\times {10}^{-7}\\ 1.2\times {10}^{-7}=\frac{x^2}{0.038M\text{-x}}=\frac{x^2}{0.038M}\\ x^2=1.2\times {10}^{-7}\times 0.038M\\ =\sqrt{4.5\times {10}^{-9}}\\ =6.7\times {10}^{-5}M\end{array}$

From the above calculations, the concentration of $H_3{O}_{{}_{(aq)}}^{+}$ is,

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}_{{}_{\text{(aq)}}}^{\text{+}}\text{=}0.038M+6.7\times {10}^{-5}M\\ \text{=}0.038M\end{array}$

We know,

    $\begin{array}{l}pH=-\log [H_3{O}^{+}]\\ =-\log [0.038M]\\ =1.4pH\end{array}$

Hence, the $pH$ of $1.0\times {10}^{-4}\text{ M}{H}_{3}B{O}_{3\left(aq\right)}$ solution is $1.4pH$.