Chapter 18, Problem 18.77P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution has to be calculated for $\text{0}{\text{.55 M HCN (K}}_{\text{a}}\text{ = 6}\text{.2}\times {10}^{-10})$.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$                                                                                                   $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

Explanation of Solution

Given,

Solution has $\text{0}{\text{.55 M HCN (K}}_{\text{a}}\text{ = 6}\text{.2}\times {10}^{-10})$.

 $\text{HCN}$ is weak acid therefore it can donate the proton to the water.

The balance equation is given below,

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

 $\text{0}{\text{.55 M HCN (K}}_{\text{a}}\text{ = 6}\text{.2}\times {10}^{-10})$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.55 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.55-x		x	x

The initial concentration is $\text{0}\text{.55 M HCN}$.

  $\begin{array}{l}\text{HCN (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ CN}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{ CN}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HCN}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{ CN}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HCN}\right]}=\text{ 6}\text{.2}\times {10}^{-10},\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{ CN}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HCN}\right]}=\text{ 6}\text{.2}\times {10}^{-10}$

Let consider,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{\left[{\text{ CN}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HCN}\right]}=\text{ 6}\text{.2}\times {10}^{-10}\\ \frac{x^2}{(0.55-x)}=\text{6}\text{.2}\times {10}^{-10}\end{array}$

Assume, x is negligible so $0.55–x\approx 0.55$

  $\begin{array}{l}\frac{x^2}{(0.55-x)}=\text{6}\text{.2}\times {10}^{-10}\\ x^2=3.41\times {10}^{-10}\\ x=1.84\times {10}^{-5}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{1.84\times {10}^{-5}}{0.55}\text{×100}\\ \text{Percent dissociated = 3}\text{.35}\times {10}^{-3}\end{array}$

Percentage of error very less and it is valid,

Therefore,

  $\begin{array}{l}\text{pH}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(}1.84\times {10}^{-5}\text{)}\\ \text{pH}\text{=}4.73\end{array}$

 $\text{pH}$ of $\text{0}\text{.55 M HCN}$ is $4.73$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pOH}$ of the solution has to be calculated for $\text{0}{\text{.044 M HIO}}_{\text{3}}{\text{ (K}}_{\text{a}}\text{ = 0}\text{.16})$.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$                                                                                                   $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_a$ value is calculating by using following formula,

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

Explanation of Solution

Given,

Solution has $\text{0}{\text{.044 M HIO}}_{\text{3}}{\text{ (K}}_{\text{a}}\text{ = 0}\text{.16})$.

 ${\text{HIO}}_{\text{3}}$ is acid therefore it can donate the proton to the water.

The balance equation is given below,

  ${\text{HIO}}_{\text{3}}\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ IO}}_3^{-}\text{(aq)}$

 $\text{0}{\text{.044 M HIO}}_{\text{3}}{\text{ (K}}_{\text{a}}\text{ = 0}\text{.16})$ Solution.

Therefore,

Construct ICE table:

  ${\text{HIO}}_{\text{3}}\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ IO}}_3^{-}\text{(aq)}$

Initial concentration	0.044 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.044-x		x	x

The initial concentration is $\text{0}{\text{.044 M HIO}}_{\text{3}}$.

  $\begin{array}{l}{\text{HIO}}_{\text{3}}\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ IO}}_3^{-}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{IO}}_3^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HIO}}_{\text{3}}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{IO}}_3^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HIO}}_{\text{3}}\right]}=\text{ 0}\text{.16}\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{IO}}_3^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HIO}}_{\text{3}}\right]}=\text{ 0}\text{.16}$

Let consider,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{\left[{\text{IO}}_3^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HIO}}_{\text{3}}\right]}=\text{ 0}\text{.16}\\ \frac{x^2}{(0.044-x)}=0.16\end{array}$

Assume, x is negligible so $0.044–x\approx 0.044$

  $\begin{array}{l}\frac{x^2}{(0.044-x)}=0.16\\ x^2=7.04\times {10}^{-3}\\ x=0.0839\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{0.0839}{0.044}\text{×100}\\ \text{Percent dissociated = 190}\%\end{array}$

The dissociation of ${\text{HIO}}_{\text{3}}$ is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression can be solved using the quadratic formula.

The quadratic formula is given below,

  $\begin{array}{l}\frac{x^2}{(0.044-x)}=0.16\\ x^2\text{=}(0.044-x)\times 0.16\\ x^2+0.16x-7.04\times {10}^{-3}=0\end{array}$

  $\begin{array}{l}a=1,b=0.16,c=-7.04\times {10}^{-3}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(0.16)\pm \sqrt{{(0.16)}^2-4(1)(-7.04\times {10}^{-3})}}{2(1)}\\ x=0.03593\text{M}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

The ${\text{K}}_b$ value is calculating by using following formula,

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{0.03593}\\ {\text{K}}_{\text{b}}\text{ = 2}\text{.78}\times {10}^{-13}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}{\text{(OH}}^{-}\text{)}\\ \text{pOH}\text{=}\text{-}\text{log}\text{(2}\text{.78}\times {10}^{-13}\text{)}\\ \text{pOH}\text{=}12.55\end{array}$

 $\text{pOH}$ of $\text{0}{\text{.044 M HIO}}_{\text{3}}$ is $12.55$.