What is the${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ for each compound?

a.

b.

c.

The rate of forward and backward reaction in reversible reactions become equal at the equilibrium state. So, the concentration of the reactants and products remains the same at the state of equilibrium in a reaction.

The acid dissociation constant represented as ${\text{K}}_{\text{a}}$is a quantitative measure of the strength of an acid. The equilibrium expression is as follows:

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

Where,

$$\begin{gathered} \left[{\text{H}}^{+}\right]{\text{=ConcentrationofH}}^{+}\text{ions} \\ \left[{\text{B}}^{-}\right]=\mathrm{Concentration}\mathrm{of}{\text{B}}^{-}\text{ions} \\ \left[\text{HB}\right]=\mathrm{Concentration}\mathrm{of}\text{acidHB} \end{gathered}$$

The formula is as follows:

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

For the given compounds, the K_a values are known. So, the ${\text{pK}}_{\text{a}}$ is calculated as:

a. ${\text{K}}_a{\text{=4.7×10}}^{-10}$

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)} \\ {\text{pK}}_{\text{a}}{\text{=-log(4.7×10}}^{-10}\text{)} \\ {\text{pK}}_{\text{a}}\text{=-(-9.3)} \\ {\text{pK}}_{\text{a}}\text{=9.3} \end{gathered}$$

b. ${\text{K}}_{\text{a}}{\text{=2.3×10}}^{-5}$

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)} \\ {\text{pK}}_{\text{a}}{\text{=-log(2.3×10}}^{-5}\text{)} \\ {\text{pK}}_{\text{a}}\text{=-(-4.6)} \\ {\text{pK}}_{\text{a}}\text{=4.6} \end{gathered}$$

Hence, ${\text{pK}}_{\text{a}}$is 4.6.

c.${\text{K}}_{\text{a}}{\text{=5.9×10}}^{-1}$

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{=-log(K}}_{\text{a}}\text{)} \\ {\text{pK}}_{\text{a}}{\text{=-log(5.9×10}}^{-1}\text{)} \\ {\text{pK}}_{\text{a}}\text{=-(-0.23)} \\ {\text{pK}}_{\text{a}}\text{=0.23} \end{gathered}$$

Hence, ${\text{pK}}_{\text{a}}$is 0.23.