Write the expression for each of the following in water:

(a)${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}$

(b)${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{COOH}}$

(c)${\mathbf{\text{HBrO}}}_{\mathbf{\text{2}}}$

An acid dissociation constant $\mathbf{\left(}{\mathit{K}}_{\mathbf{a}}\mathbf{\right)}$is a numerical representation of an acid's strength in solution. The dissociation constant${\mathbf{\text{K}}}_{\mathbf{\text{a}}}\mathbf{\text{=}}\frac{{\mathbf{\text{[A}}}^{\mathbf{\text{-}}}{\mathbf{\text{][H}}}^{\mathbf{\text{+}}}\mathbf{\text{]}}}{\mathbf{\text{[HA]}}}$is commonly stated as a quotient of the equilibrium concentrations (in$\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L}$).

(a)

To write the$K_a$ expression of ${\text{HNO}}_2$in water, we should write the balanced reaction first –

${\text{HNO}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+NO}}_{\text{2}}^{\text{-}}\text{(aq)}$

Then write the$K_a$ expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][NO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\end{array}$

Therefore, the expression is obtained as.${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][NO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}$

(b)

To write the$K_a$expression of ${\text{CH}}_{\text{3}}\text{COOH}$in water, we should write the balanced reaction first –

${\text{CH}}_{\text{3}}{\text{COOH(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{(aq)}$

Then write$K_a$ the expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COO}H\text{]}}\end{array}$

Therefore, the expression is obtained as.${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COO}H\text{]}}$

(c)

To write the$K_a$expression of${\text{HBrO}}_{\text{2}}$ in water, we should write the balanced reaction first –

${\text{HBrO}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+BrO}}_{\text{2}}^{\text{-}}\text{(aq)}$

Then write the $K_a$expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][BrO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HBrO}}_{\text{2}}\text{]}}\end{array}$

Therefore, the expression is obtained as .${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][BrO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HBrO}}_{\text{2}}\text{]}}$