Q17 / 160 · AP EAMCET 2011 · Chemistry

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

  • A
    2
  • B
    11
  • C
    $10^{-3}$
  • Correct answer
    D
    3

Solution

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

Hence, the correct answer is option: 'D'.

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