Chapter 6, Problem 6D.18E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of 0.015 M aqueous ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ has to be calculated.

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.

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6D.18E

The $\text{pH}$ of 0.015 M aqueous ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ is 7.0.

Explanation of Solution

The proton transfer equilibrium reaction for  the base of ${\text{SO}}_3{}^{\text{2-}}$ is given below.

  ${\text{SO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}{}_{\text{(aq)}}{\text{+O}}^{\text{2-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[H}}_2{\text{SO}}_3\text{]}{\text{[O}}^{2-}\text{]}}{{\text{[SO}}_3{}^{2-}\text{]}}$

	${\text{SO}}_3{}^{2-}$	${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$	${\text{O}}^{\text{2-}}$
Initial concentration	0.015	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.015-x	x	x

The ${\text{K}}_{\text{b}}$ of ${\text{SO}}_3{}^{2-}$ value is calculated using given formula,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{1.5\times {10}^{-2}}\\ \\ =6.67\times {10}^{-13}\end{array}$

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.015-x}}$

The ${\text{K}}_{\text{b}}$ value is $6.67\times {10}^{-13}$ and it is substituted in above equation and then the value of x is calculated.

  $6.67\times {\text{10}}^{-13}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.015-x}}$

The above equation, assume that the x present in $\text{0}\text{.015-x}$ is very small than 0.015 then it can be negligible and as follows,

  $\begin{array}{l}\text{6}{\text{.67×10}}^{\text{-13}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.015}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.015×(6}{\text{.67×10}}^{\text{-13}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.015×(6}{\text{.67×10}}^{\text{-13}}\text{)}}\\ \\ \text{=}1.0{\text{×10}}^{\text{-7}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

The $\text{pOH}$ of the solution is calculated using the given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.0}\times {\text{10}}^{-7}\text{)}\\ \\ \text{=}7.0\end{array}$

The general equilibrium expression to find out the pH of the solution is given below,

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-7}\text{.0 }\\ \\ \text{ = 7}\text{.0}\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.015 M aqueous ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ is 7.0.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of 0.086 M aqueous $\text{NaF}$ has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.18E

The $\text{pH}$ of 0.086 M aqueous $\text{NaF}$ is 8.2.

Explanation of Solution

The proton transfer equilibrium reaction for  the base of ${\text{F}}^{\text{-}}$ is given below.

  ${\text{F}}^{-}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HF}}_{\text{(aq)}}{\text{+OH}}^{-}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{[HF]}{\text{[OH}}^{-}\text{]}}{{\text{[F}}^{-}\text{]}}$

	${\text{F}}^{\text{-}}$	$\text{HF}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.086	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.086-x	x	x

The ${\text{K}}_{\text{b}}$ of ${\text{F}}^{-}$ value is calculated using given formula,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{3.5\times {10}^{-4}}\\ \\ =2.86\times {10}^{-11}\end{array}$

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.086-x}}$

The ${\text{K}}_{\text{b}}$ value is $2.86\times {10}^{-11}$ and it is substituted in above equation and then the value of x is calculated.

  $2.86\times {\text{10}}^{-11}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.086-x}}$

The above equation, assume that the x present in $\text{0}\text{.086-x}$ is very small than 0.086 then it can be negligible and as follows,

  $\begin{array}{l}\text{2}{\text{.86×10}}^{\text{-11}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.086}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.086×(2}{\text{.86×10}}^{\text{-11}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.086×(2}{\text{.86×10}}^{\text{-11}}\text{)}}\\ \\ \text{=}1.57{\text{×10}}^{\text{-6}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

The $\text{pOH}$ of the solution is calculated using the given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.57}\times {\text{10}}^{-6}\text{)}\\ \\ \text{=}5.80\end{array}$

The general equilibrium expression to find out the pH of the solution is given below,

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-5}\text{.80 }\\ \\ \text{ = 8}\text{.2}\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.086 M aqueous $\text{NaF}$ is 8.2.