A weak acid HA has degree of dissociation x. Which option gives the correct expression of…
- 0
- $\log (1+2 \mathrm{x})$
- $\log \left(\frac{1-x}{x}\right)$
- $\log \left(\frac{x}{1-x}\right)$
Solution

$\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right](\mathrm{x})}{\mathrm{C}(1-\mathrm{x})} \\ & {\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \frac{(1-\mathrm{x})}{\mathrm{x}}} \\ & \log \mathrm{H}^{+}=\log \mathrm{K}_{\mathrm{a}}+\log \left(\frac{1-\mathrm{x}}{\mathrm{x}}\right) \\ & -\log \mathrm{H}^{+}=-\log \mathrm{K}_{\mathrm{a}}-\log \frac{1-\mathrm{x}}{\mathrm{x}} \\ & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}-\log \frac{1-\mathrm{x}}{\mathrm{x}} \\ & \mathrm{pH}-\mathrm{pK}_{\mathrm{a}}=\log \frac{\mathrm{x}}{1-\mathrm{x}}\end{aligned}$
Asked in: JEE Main 2025 (28 Jan Shift 1)