6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

$\alpha=\frac{2}{100}=0.02$

$\because C_{0}=\frac{K_{a}}{\alpha^{2}}=\frac{10^{-4}}{0.02 \times 0.02}=\frac{1}{4}$

$\left[\mathrm{H}^{+}\right]=\mathrm{C}_{\mathrm{o}} \alpha$

$=\frac{1}{4} \times 0.02=0.005 \Rightarrow 5 \times 10^{-3}$

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$=-\log \left[5 \times 10^{-3}\right]$

$=-[\log 5-3 \log 10]$

$=2.3$