Chapter 18, Problem 18.75P

(a)

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ has to be calculated for $\text{3}\text{.5}\times {\text{10}}^{-2}{\text{ M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$.

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 $\text{3}\text{.5}\times {\text{10}}^{-2}{\text{ M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

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

 $\text{3}\text{.5}\times {\text{10}}^{-2}{\text{ M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.035 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.035-x		x	x

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ \\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\\ \\ {\text{10}}^{\text{ - 4}\text{.89}}{\text{ = K}}_{\text{a }}\\ \\ 1.29\times {10}^{-5}{\text{ = K}}_{\text{a }}\end{array}$

The initial concentration is $\text{3}\text{.5}\times {\text{10}}^{-2}\text{ M HQ }$.

  $\begin{array}{l}\text{HQ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ Q}}^{-}\text{ (aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}=\text{ 1}\text{.29}\times {10}^{-5},\end{array}$

Let consider,

  $\begin{array}{c}{\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}=\text{ 1}\text{.29}\times {10}^{-5}\\ \\ \frac{x^2}{\left(0.035-x\right)}=\text{1}\text{.29}\times {10}^{-5}\end{array}$

Assume, x is negligible so $0.035–x\approx \text{ 0}\text{.035}$

  $\begin{array}{l}\frac{x^2}{\left(0.035-x\right)}=\text{1}\text{.29}\times {10}^{-5}\\ \\ x^2=4.51\times {10}^{-7}\\ \\ x=6.71\times {10}^{-4}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{Percent dissociated = }\frac{6.71\times {10}^{-4}}{0.035}\text{×100}\\ \\ \text{Percent dissociated = 1}\text{.92}\%\end{array}$

  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{= }6.71\times {10}^{-4}$

The $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ for $\text{3}\text{.5}\times {\text{10}}^{-2}{\text{ M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$ is $6.71\times {10}^{-4}$

(b)

Interpretation Introduction

Interpretation:

The $\left[{\text{OH}}^{-}\right]$ has to be calculated for $\text{0}{\text{.65 M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$.

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 $\text{0}{\text{.65 M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

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

$\text{0}{\text{.65 M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$ Solution.

Therefore,

Construct ICE table:

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

Initial concentration	0.65 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.65-x		x	x

The initial concentration is $\text{0}\text{.65 M HQ }$.

  $\begin{array}{l}\text{HQ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ Q}}^{-}\text{ (aq)}\\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}=\text{ 1}\text{.29}\times {10}^{-5},\end{array}$

Let consider,

  $\begin{array}{c}{\text{K}}_a\text{ =}\frac{\left[{\text{Q}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HQ}\right]}=\text{ 1}\text{.29}\times {10}^{-5}\\ \\ \frac{x^2}{\left(0.65-x\right)}=\text{1}\text{.29}\times {10}^{-5}\end{array}$

Assume, x is negligible so $0.65–x\approx \text{ 0}\text{.65}$

  $\begin{array}{l}\frac{x^2}{\left(0.65-x\right)}=\text{1}\text{.29}\times {10}^{-5}\\ \\ x^2=8.38\times {10}^{-6}\\ \\ x=2.89\times {10}^{-3}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{Percent dissociated = }\frac{2.89\times {10}^{-3}}{0.65}\text{×100}\\ \\ \text{Percent dissociated = 0}\text{.44}\%\end{array}$

  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{= }2.89\times {10}^{-3}$

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

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{2.89\times {10}^{-3}}\\ \\ {\text{K}}_{\text{b}}\text{ = }3.46\times {10}^{-12}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ for $\text{0}{\text{.65 M HQ (pK}}_{\text{a}}\text{= 4}\text{.89)}$ is $3.46\times {10}^{-12}$