What is${\mathbf{\text{K}}}_{\mathbf{a}}$ for each compound? Use a calculator when necessary.

a.$$\begin{gathered} {\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{S}} \\ {\mathbf{\text{pK}}}_{\mathbf{\text{a}}}\mathbf{\text{=7.0}} \end{gathered}$$

b.$$\begin{gathered} {\mathbf{\text{ClCH}}}_{\mathbf{2}}\mathbf{\text{COOH}} \\ {\mathbf{\text{pK}}}_{\mathbf{a}}\mathbf{\text{=2.8}} \end{gathered}$$

c.$$\begin{gathered} \mathbf{\text{HCN}} \\ {\mathbf{\text{pK}}}_{\mathbf{a}}\mathbf{\text{=9.1}} \end{gathered}$$

When the rate of forward and backward reaction becomes equal, the reaction attains a state of equilibrium. At equilibrium, the concentration of the reactants and products remains the same.

For example, for a reaction in which an acid (HA) dissociates to form its respective ions i.e.,${\text{(H}}^{+}\text{)}$ and ${\text{(OH}}^{-}\text{)}$. The expression for equilibrium is written as:

${\text{HA⇌H}}^{+}{\text{+A}}^{-}$

The acid dissociation constant is the measure of the strength of an acid in terms of acid-base reactions in a solution. The dissociation constant is an equilibrium constant and is represented as ${\text{K}}_{\text{a}}$.

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

Or

${\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)}$

a. ${\text{H}}_2\text{S}$

${\text{pK}}_{\text{a}}\text{=7.0}$

$$\begin{gathered} {\text{K}}_{\text{a}}{\text{=10}}^{-({\text{pK}}_a\text{)}} \\ {\text{K}}_{\text{a}}{\text{=10}}^{-(7.0)} \\ {\text{K}}_{\text{a}}{\text{=10}}^{-7} \end{gathered}$$

Hence, ${\text{K}}_{\text{a}}$for ${\text{H}}_2\text{S}$is ${\text{10}}^{-7}$

b.role="math" localid="1649239522238" ${\text{ClCH}}_2\text{COOH}$

${\text{pK}}_{\mathrm{a}}\text{=2.8}$

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)} \\ {\text{K}}_{\text{a}}={10}^{-({\text{pk}}_{\text{a}}\text{)}} \\ {\text{K}}_{\text{a}}={10}^{-(2.8\text{)}} \\ {\text{K}}_{\text{a}}=0.158\times {10}^{-2} \end{gathered}$$

Hence, ${\text{K}}_{\text{a}}$for ${\text{ClCH}}_2\text{COOH}$is $0.158\times {10}^{-2}$

c. $\text{HCN}$

${\text{pK}}_{\mathrm{a}}\text{=9.1}$

role="math" localid="1649240017009" $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)} \\ {\text{K}}_{\text{a}}={10}^{-({\text{pk}}_{\text{a}}\text{)}} \\ {\text{K}}_{\text{a}}={10}^{-(9.1\text{)}} \\ {\text{K}}_{\text{a}}=0.8\times {10}^{-9} \end{gathered}$$

Hence, ${\text{K}}_{\text{a}}$for $\text{HCN}$is $0.8\times {10}^{-9}$