Chapter 18, Problem 18.74P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution has to be calculated for $\text{0}{\text{.075 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$.

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 $\text{0}{\text{.075 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$.

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

The balance equation is given below,

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

 $\text{0}{\text{.075 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.075 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.075-x		x	x

The initial concentration is $\text{0}\text{.075 M HZ}$.

  $\begin{array}{l}\text{HZ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ Z}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}=\text{ 2}\text{.55}\times {10}^{-4},\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}=\text{ 2}\text{.55}\times {10}^{-4}$

Let consider,

  $\frac{x^2}{(0.075-x)}=\text{2}\text{.55}\times {10}^{-4}$

Assume, x is negligible so $0.075–x\approx 0.075$

  $\begin{array}{l}\frac{x^2}{(0.075)}=\text{2}\text{.55}\times {10}^{-4}\\ x^2=1.91\times {10}^{-5}\\ x=4.37\times {10}^{-3}\end{array}$

Percent dissociation can be calculated by using following formula,

$\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{4.37\times {10}^{-3}}{0.075}\text{×100}\\ \text{Percent dissociated = }5.83\%\end{array}$

The dissociation of $\text{HZ}$ 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,

  $\begin{array}{l}\frac{x^2}{(0.075-x)}=\text{2}\text{.55}\times {10}^{-4}\\ x^2\text{=}(0.075-x)\times \text{2}\text{.55}\times {10}^{-4}\\ x^2+2.55\times {10}^{-4}x-1.9\times {10}^{-5}=0\end{array}$

  $\begin{array}{l}a=1,b=2.55\times {10}^{-4},c=-1.9\times {10}^{-5}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(2.55\times {10}^{-4})\pm \sqrt{{(2.55\times {10}^{-4})}^2-4(1)(-1.9\times {10}^{-5})}}{2(1)}\\ x=0.00425\text{M}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

Therefore,

  $\begin{array}{l}\text{pH}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(0}\text{.00425)}\\ \text{pH}\text{=}2.4\end{array}$

$\text{pH}$ of $\text{0}\text{.075 M HZ}$ is $2.4$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pOH}$ of the solution has to be calculated for $\text{0}{\text{.045 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$.

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 $\text{0}{\text{.045 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$.

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

The balance equation is given below,

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

 $\text{0}{\text{.045 M HZ (K}}_{\text{a}}\text{ = 2}\text{.55}\times {10}^{-4})$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.045 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.1-x		x	x

The initial concentration is $\text{0}\text{.045 M HZ}$.

  $\begin{array}{l}\text{HZ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ Z}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}=\text{ 2}\text{.55}\times {10}^{-4},\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{ Z}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HZ}\right]}=\text{ 2}\text{.55}\times {10}^{-4}$

Let consider,

  $\frac{x^2}{(0.045-x)}=\text{2}\text{.55}\times {10}^{-4}$

Assume, x is negligible so $0.045–x\approx 0.045$

  $\begin{array}{l}\frac{x^2}{(0.045)}=\text{2}\text{.55}\times {10}^{-4}\\ x^2=1.15\times {10}^{-5}\\ x=3.38\times {10}^{-3}\end{array}$

Percent dissociation can be calculated by using following formula,

$\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{3.38\times {10}^{-3}}{0.045}\text{×100}\\ \text{Percent dissociated = 7}\text{.52}\%\end{array}$

The dissociation of $\text{HZ}$ 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,

  $\begin{array}{l}\frac{x^2}{(0.045-x)}=\text{2}\text{.55}\times {10}^{-4}\\ x^2\text{=}(0.045-x)\times \text{2}\text{.55}\times {10}^{-4}\\ x^2+2.55\times {10}^{-4}x-1.15\times {10}^{-5}=0\end{array}$

  $\begin{array}{l}a=1,b=2.55\times {10}^{-4},c=-1.9\times {10}^{-5}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(2.55\times {10}^{-4})\pm \sqrt{{(2.55\times {10}^{-4})}^2-4(1)(-1.15\times {10}^{-5})}}{2(1)}\\ x=3.26\times {10}^{-3}\text{M}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

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

  $\begin{array}{c}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{1×10}}^{\text{-14}}}{\text{3}\text{.26×}{\text{10}}^{\text{-3}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{b}}\text{ = 3}\text{.06×}{\text{10}}^{\text{-12}}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}\left[{\text{OH}}^{-}\right]\\ \\ \text{pOH}\text{=}\text{-}\text{log}\left(3.06\times {10}^{-12}\right)\\ \text{pOH}\text{=}11.51\end{array}$

$\text{pOH}$ of $\text{0}\text{.075 M HZ}$ is $11.51$.