What will be the value of pH of $0.01 \mathrm{moldm}^{-3} \\mathrm{CH}_3\ \mathrm{COOH}\left(K_a=1

Question

What will be the value of pH of  $0.01 \mathrm{moldm}^{-3} \\mathrm{CH}_3\ \mathrm{COOH}\left(K_a=1.74 	imes 10^{-5}ight) ?$

Professor PC Thomas Classes · Answer

(a): $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} ightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$\ $K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}ight]\left[\mathrm{H}_3 \mathrm{O}^{+}ight]}{\left[\mathrm{CH}_3 \mathrm{COOH}ight]}$\ $K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}ight]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}ight]}$\ $\left(\because\left[\mathrm{CH}_3 \mathrm{COO}^{-}ight]=\left[\mathrm{H}_3 \mathrm{O}^{+}ight]ight)$\ $\left[\mathrm{H}_3 \mathrm{O}^{+}ight]^2=K_a\left[\mathrm{CH}_3 \mathrm{COOH}ight]$\ $\left[\mathrm{H}_3 \mathrm{O}^{+}ight]=\sqrt{K_a\left[\mathrm{CH}_3 \mathrm{COOH}ight]}$\ Given: $K_a=1.74 	imes 10^{-5},\left[\mathrm{CH}_3 \mathrm{COOH}ight]=0.01 \mathrm{~mol} \mathrm{dm}^{-3}$\ $\left[\mathrm{H}_3 \mathrm{O}^{+}ight]=\sqrt{1.74 	imes 10^{-5} 	imes 0.01}=\sqrt{1.74 	imes 10^{-7}}$\ $\left[\mathrm{H}_3 \mathrm{O}^{+}ight]=4.17 	imes 10^{-4}$\ $\mathrm{pH}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}ight]=-\log \left(4.17 	imes 10^{-4}ight)=3.379 \approx 3.4$