At equilibrium $\left[H^{+}=A^{-}\right]$
$K_{a} =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}=\frac{\left[H^{+}\right]^{2}}{[H A]}$
$\left[H^{+}\right]=\sqrt{K_{a}[H A]} =\sqrt{1 \times 10^{-5} \times 0.1}$
$= \sqrt{1 \times 10^{-6}}=1 \times 10^{-3}$
$\alpha =\frac{A \text {ctual ionisation}}{\text {Molar concentration}}$
$\%$ of acid dissociated $=10^{-2} \times 1.00$
$=1 \%$
$(a)\;\;60\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+40 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(b)\;\;55\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+45 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(c)\;\;75\; \mathrm{mL} \frac{\mathrm{M}}{5}\; \mathrm{HCl}+25 \;\mathrm{mL} \frac{\mathrm{M}}{5} \;\mathrm{NaOH}$
$(d)\;\;100\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+100 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
તેઓ પૈકી કોની $pH$ $1$ ને સમાન થશે ?