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

The $\mathrm{pH}$ of $0.05 \mathrm{M}$ acetic acid is $\left(K_a=2 \times 10^{-5}ight)$
  1. 2
  2. 11
  3. $10^{-3}$
  4. 3

Solution

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

Asked in: JEE-TOPICTESTS-CHEMISTRY

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