Chapter 6, Problem 6D.17E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of 0.63 M aqueous ${\text{NaCH}}_{\text{3}}{\text{CO}}_{\text{2}}$ 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.17E

The $\text{pH}$ of 0.63 M aqueous ${\text{NaCH}}_{\text{3}}{\text{CO}}_{\text{2}}$ is 9.27.

Explanation of Solution

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

  ${\text{CH}}_3{\text{COO}}^{-}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{CH}}_3{\text{COOH}}_{\text{(aq)}}{\text{+OH}}^{-}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_3\text{COOH]}{\text{[OH}}^{-}\text{]}}{{\text{[CH}}_3{\text{COO}}^{-}\text{]}}$

	${\text{CH}}_3{\text{COO}}^{-}$	${\text{CH}}_3\text{COOH}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.63	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.63-x	x	x

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

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ \\ =5.56\times {10}^{-10}\end{array}$

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.63-x}}$

The ${\text{K}}_{\text{b}}$ value is $5.56\times {10}^{-10}$ and it is substituted in above equation and then the value of x is calculated.

  $5.56\times {\text{10}}^{-10}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.63-x}}$

The above equation, assume that the x present in $\text{0}\text{.63-x}$ is very small than 0.63 then it can be negligible and as follows,

  $\begin{array}{l}\text{5}{\text{.56×10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.10}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.63×(5}{\text{.56×10}}^{\text{-10}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.63×(5}{\text{.56×10}}^{\text{-10}}\text{)}}\\ \\ \text{=}1.87{\text{×10}}^{\text{-5}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.87}\times {\text{10}}^{-5}\text{)}\\ \\ \text{=}4.73\end{array}$

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-4}\text{.73 }\\ \\ \text{ = 9}\text{.27}\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.63 M aqueous ${\text{NaCH}}_{\text{3}}{\text{CO}}_{\text{2}}$ is 9.27.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of 0.65 M aqueous potassium cyanide has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.17E

The $\text{pH}$ of 0.65 M aqueous potassium cyanide is 11.56.

Explanation of Solution

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

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

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{[HCN]}{\text{[OH}}^{-}\text{]}}{{\text{[CN}}^{-}\text{]}}$

	${\text{CN}}^{\text{-}}$	$\text{HCN}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.65	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.65-x	x	x

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

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{4.9\times {10}^{-10}}\\ \\ =2.04\times {10}^{-5}\end{array}$

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.65-x}}$

The ${\text{K}}_{\text{b}}$ value is $2.04\times {10}^{-5}$ and it is substituted in above equation and then the value of x is calculated.

  $2.04\times {\text{10}}^{-5}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.65-x}}$

The above equation, assume that the x present in $\text{0}\text{.65-x}$ is very small than 0.65 then it can be negligible and as follows,

  $\begin{array}{l}\text{2}{\text{.04×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.65}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.65×(2}{\text{.04×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.65×(2}{\text{.04×10}}^{\text{-5}}\text{)}}\\ \\ \text{=}3.64{\text{×10}}^{\text{-3}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(3}\text{.64}\times {\text{10}}^{-3}\text{)}\\ \\ \text{=}2.44\end{array}$

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-2}\text{.44 }\\ \\ \text{ = 11}\text{.56}\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.65 M potassium cyanide is 11.56.