The $\mathrm{pH}$ of $0.05 \mathrm{M}$ acetic acid is $\left(K_a=2 imes 10^{-5} ight)$

Question

The $\mathrm{pH}$ of $0.05 \mathrm{M}$ acetic acid is $\left(K_a=2 	imes 10^{-5}ight)$

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} ightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_3 \mathrm{COO}^{-} \ & K_a=\frac{\left[\mathrm{H}^{+}ight]\left[\mathrm{CH}_3 \mathrm{COO}^{-}ight]}{\left[\mathrm{CH}_3 \mathrm{COOH}ight]}=\frac{\left[\mathrm{H}^{+}ight]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}ight]} \ & {\left[\mathrm{H}^{+}ight] }=\sqrt{K_a\left[\mathrm{CH}_3 \mathrm{COOH}ight]} \ &=\sqrt{2 	imes 10^{-5} 	imes 0.05} \ &=\sqrt{10^{-6}}\end{aligned}$ $\begin{aligned} {\left[\mathrm{H}^{+}ight] } & =10^{-3} \ \because \quad \mathrm{pH} & =-\log \left[\mathrm{H}^{+}ight] \ & =-\log \left[10^{-3}ight] \ 	herefore \quad \mathrm{pH} & =3\end{aligned}$

The $\mathrm{pH}$ of $0.05 \mathrm{M}$ acetic acid is $\left(K_a=2 \times 10^{-5}\right)$

Solution