Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

1. \({F^ - }\)
2. \(NH_4^ + \)
3. \(AsO_4^{3 - }\)
4. \({\left( {C{H_3}} \right)_2}NH_2^ + \)
5. \(N{O_2}\)
6. \(H{C_2}O_4^ - \) (asabase)

Acids are something that releases \({H^ + }\) ion in the solution and bases are something that accepts \({H^ + }\) ion from the solution.

Strong acids have weak conjugate bases and vice-versa.

Strong bases have weak conjugate acids and vice-versa.

If \({K_a}\) is given and \({K_b}\) is to found, then the formula that can be used is,

\(^n{K_w} = {K_a} \cdot {K_b}\)

Here, \({K_a}\)is the acid ionization constant and \({K_b}\) is the base ionization constantof the acid's conjugate base.

Also, \({K_w}\) is the ionic water product,a constant.

\({{\rm{K}}_{\rm{w}}} = {10^{ - 14}}{M^2}\)

The conjugate acidfor \({F^ - }\) is \({\bf{HF}}\).

As, the \({K_a}\)value for \({\bf{HF}}\) is \(3.5 \times {10^{ - 4}}\).

Therefore,

\(\begin{aligned}{K_b} &= \frac{{{K_w}}}{{{K_a}}}\\{K_b} &= \frac{{{{10}^{ - 14}}}}{{3.5 \cdot {{10}^{ - 4}}}}\\{K_b} &= 2.9 \cdot {10^{ - 11}}\end{aligned}\)

The conjugate baseof \(NH_4^ + \) is \(N{H_3}\).

The \({K_b}\) value for \(N{H_3}\) is \(1.8 \times {10^{ - 5}}\).

Therefore,

\(\begin{aligned}{K_a} &= \frac{{{K_w}}}{{{K_b}}}\\{K_a} &= \frac{{{{10}^{ - 14}}}}{{1.8 \cdot {{10}^{ - 5}}}}\\{K_a} &= 5.6 \cdot {10^{ - 10}}\end{aligned}\)

Hence, the \({K_a}\)value for \({\rm{NH}}_4^ + \) is \(5.6 \times {10^{ - 10}}\).

The conjugate acidof \(AsO_4^{3 - }\) is \(HAsO_4^{2 - }\).

The \({{\rm{K}}_a}\)value for \(HAsO_4^{2 - }\) is \(5.1 \times {10^{ - 12}}\).

\(\begin{aligned}{K_b} &= \frac{{{K_w}}}{{{K_a}}}\\{K_b} &= \frac{{{{10}^{ - 14}}}}{{5.1 \cdot {{10}^{ - 12}}}}\\{K_b} &= 2.0 \cdot {10^{ - 3}}\end{aligned}\)

Hence, the \({K_b}\) value for \(AsO_4^{3 - }\) is \(2.0 \times {10^{ - 3}}\).

The conjugate basefor \({\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{NH}}_2^ + \) is \({\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{NH}}\).

The \({{\rm{K}}_b}\)value for \({\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{NH}}\)equals \(5.9 \times {10^{ - 4}}.\)

\(\begin{aligned}{K_a} &= \frac{{{K_w}}}{{{K_b}}}\\{K_a} &= \frac{{{{10}^{ - 14}}}}{{5.9 \cdot {{10}^{ - 4}}}}\\{K_a} &= 1.7 \cdot {10^{ - 11}}\end{aligned}\)

The \({K_a}\)value for \({\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{NH}}_2^ + \) equals \(1.7 \times {10^{ - 11}}\).

The conjugate acidfor \(N{O_2}\) is \(HN{O_2}\).

The \({K_a}\) value for \(HN{O_2}\) equals \(5.6 \times {10^{ - 4}}.\)

\(\begin{aligned}{K_a} &= \frac{{{K_w}}}{{{K_b}}}\\{K_a} &= \frac{{{{10}^{ - 14}}}}{{5.6 \times {{10}^{ - 4}}}}\\{K_a} &= 1.8 \times {10^{ - 11}}\end{aligned}\)

The \({K_b}\)value for \(N{O_2}\) equals \(1.8 \times {10^{ - 11}}\).

The conjugate acidfor \(H{C_2}O_4^ - \) is \({H_2}{C_2}{O_4}\).

The \({K_a}\) value for \({H_2}{C_2}{O_4}\) equals \(6.0 \times {10^{ - 2}}\).

\(\begin{aligned}{K_b} &= \frac{{{K_w}}}{{{K_a}}}\\{K_b} &= \frac{{{{10}^{ - 14}}}}{{6.0 \cdot {{10}^{ - 2}}}}\\{K_b} &= 1.7 \cdot {10^{ - 13}}\end{aligned}\)

Therefore, the \({K_b}\)value for \({\rm{H}}{{\rm{C}}_2}{\rm{O}}_4^ - \) is \(1.7 \times {10^{ - 13}}\).