91Ó°ÊÓ

Firstly, we need to find the resultant pKa by subtracting the pKa value of product acid from reactant acid.

${\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right)$

Then we can calculate the equilibrium constant using the formulae,

Equilibrium constant is$\text{=}{\text{10}}^{-{\text{pKa}}_{\text{equi}}}$

$$\begin{gathered} \text{HCl}\text{+}{\text{H}}_{\text{2}}\text{O}→{\text{Cl}}^{\text{-}}\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}} \\ {\text{pK}}_{\text{a}}\text{=}\text{- 7}{\text{pK}}_{\text{a}}\text{=}\text{- 1.7} \end{gathered}$$

$$\begin{gathered} {\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right) \\ =-7+1.7 \\ =-5.3 \end{gathered}$$

And the equilibrium constant is${10}^{5.3}=1.99×{10}^5$.

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}→{\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+} \\ {\text{pK}}_{\text{a}}\text{=}4.8{\text{pK}}_{\text{a}}\text{=}\text{- 1.7} \end{gathered}$$

$\begin{array}{rcl}& & {\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right)\\ & =& 4.8+1.7\\ & =& 6.5\end{array}$

And the equilibrium constant is ${10}^{-6.5}=3.16×{10}^{-7}$.

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{NH}}_2+{\mathrm{H}}_2\mathrm{O}→{\mathrm{CH}}_3{\mathrm{NH}}_3^{+}+{\mathrm{OH}}^{-} \\ {\text{pK}}_{\text{a}}\text{=}14{\text{pK}}_{\text{a}}\text{=}10.\text{7} \end{gathered}$$

$$\begin{gathered} {\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right) \\ =14-10.7 \\ =3.3 \end{gathered}$$

And the equilibrium constant is${10}^{-3.3}=5.011×{10}^{-4}$ .

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{NH}}_3^{+}+{\mathrm{H}}_2\mathrm{O}→{\mathrm{CH}}_3{\mathrm{NH}}_2+{\mathrm{H}}_3{\mathrm{O}}^{+} \\ {\text{pK}}_{\text{a}}\text{=}10{\text{pK}}_{\text{a}}\text{=}\text{- 1.7} \end{gathered}$$$$\begin{gathered} {\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right) \\ =10+1.7 \\ =11.7 \end{gathered}$$

$$\begin{gathered} {\text{pK}}_a\text{equilibrium}{\text{= pK}}_a\left(\text{reactant acid}\right)\text{-}{\text{pK}}_a\left(\text{product acid}\right) \\ =10+1.7 \\ =11.7 \end{gathered}$$

And the equilibrium constant is ${10}^{-11.7}=1.99×{10}^{-12}$.