Chapter 18, Problem 18.76P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution has to be calculated for $\text{0}{\text{.175 M HY (K}}_{\text{a}}\text{ = 1}\text{.50}\times {10}^{-4})$.

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 $\text{0}{\text{.175 M HY (K}}_{\text{a}}\text{ = 1}\text{.50}\times {10}^{-4})$.

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

The balance equation is given below,

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

 $\text{0}{\text{.175 M HY (K}}_{\text{a}}\text{ = 1}\text{.50}\times {10}^{-4})$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.175 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.175-x		x	x

The initial concentration is $\text{0}\text{.175 M HZ}$.

  $\begin{array}{l}\text{HY (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ Y}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HY}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{ Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HY}\right]}=\text{ 1}\text{.50}\times {10}^{-4},\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{ Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HY}\right]}=\text{ 1}\text{.50}\times {10}^{-4}$

Let consider,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{\left[{\text{ Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HY}\right]}=\text{ 1}\text{.50}\times {10}^{-4}\\ \frac{x^2}{\left(0.175-x\right)}=\text{1}\text{.50}\times {10}^{-4}\end{array}$

Assume, x is negligible so $0.175–x\approx 0.175$

  $\begin{array}{c}\frac{x^2}{\left(0.175-x\right)}=\text{1}\text{.50}\times {10}^{-4}\\ \\ x^2=2.62\times {10}^{-5}\\ \\ x=5.12\times {10}^{-3}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{5.12\times {10}^{-3}}{0.175}\text{×100}\\ \text{Percent dissociated = 2}\text{.9}\%\end{array}$

Percentage of error is valid,

Therefore,

  $\begin{array}{l}\text{pH}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(}5.12\times {10}^{-3}\text{)}\\ \text{pH}\text{=}2.3\end{array}$

 $\text{pH}$ of $\text{0}\text{.175 M HY}$ is $2.3$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pOH}$ of the solution has to be calculated for $\text{0}{\text{.175 M HX (K}}_{\text{a}}\text{ = 2}\text{.00}\times {10}^{-2})$.

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 $\text{0}{\text{.175 M HX (K}}_{\text{a}}\text{ = 2}\text{.00}\times {10}^{-2})$.

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

The balance equation is given below,

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

 $\text{0}{\text{.175 M HX (K}}_{\text{a}}\text{ = 2}\text{.00}\times {10}^{-2})$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.175 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.175-x		x	x

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

  $\begin{array}{l}\text{HX (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ X}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{ X}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HX}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{ X}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HX}\right]}=\text{ 2}\text{.00}\times {10}^{-2},\end{array}$

Let consider,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{ X}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HX}\right]}=\text{ 2}\text{.00}\times {10}^{-2}$

Let consider,

  $\begin{array}{l}{\text{K}}_a\text{ =}\frac{\left[{\text{ X}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HX}\right]}=\text{ 2}\text{.00}\times {10}^{-2}\\ \\ \frac{x^2}{\left(0.175-x\right)}=\text{2}\text{.00}\times {10}^{-2}\end{array}$

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

  $\begin{array}{l}\frac{x^2}{\left(0.175-x\right)}=\text{2}\text{.00}\times {10}^{-2}\\ \\ x^2=3.5\times {10}^{-3}\\ \\ x=0.059\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{c}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{Percent dissociated = }\frac{0.059}{0.175}\text{×100}\\ \\ \text{Percent dissociated = 34}\%\end{array}$

The dissociation of $\text{HX}$ 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,

  $\begin{array}{l}\frac{x^2}{(0.175-x)}=\text{2}\text{.00}\times {10}^{-2}\\ x^2\text{=}(0.175-x)\times \text{2}\text{.00}\times {10}^{-2}\\ x^2+2.00\times {10}^{-2}x-3.5\times {10}^{-3}=0\end{array}$

  $\begin{array}{l}a=1,b=2.0\times {10}^{-2},c=-3.5\times {10}^{-3}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(2.0\times {10}^{-2})\pm \sqrt{{(2.0\times {10}^{-2})}^2-4(1)(-3.5\times {10}^{-3})}}{2(1)}\\ x=5.0\times {10}^{-2}\text{M}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

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

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{5.0\times {10}^{-2}}\\ {\text{K}}_{\text{b}}\text{ = 2}\text{.0}\times {10}^{-13}\end{array}$

Therefore,

  $\begin{array}{l}\text{pOH}\text{=}\text{-}\text{log}{\text{(OH}}^{-}\text{)}\\ \text{pOH}\text{=}\text{-}\text{log}\text{(2}\text{.0}\times {10}^{-13}\text{)}\\ \text{pOH}\text{=}12.69\end{array}$

 $\text{pOH}$ of $\text{0}\text{.175 M HX}$ is $12.69$.