JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 4)
A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - $\mathrm{pK}_{\mathrm{a}}$)?
$\log \left(\frac{1-x}{x}\right)$
$0$
$\log (1+2 \mathrm{x})$
$\log \left(\frac{x}{1-x}\right)$
Explanation
$$\begin{aligned}
& \mathrm{HA} \rightleftharpoons \mathrm{H}^{\oplus}+\mathrm{A}^{\Theta} \\
& \mathrm{t}=0 \quad \mathrm{a} \\
& \mathrm{t}=\mathrm{t} \quad \mathrm{a}(1-\mathrm{x}) \quad \mathrm{ax} \quad \mathrm{ax} \\
& \mathrm{~K}_{\mathrm{a}}=(\mathrm{ax}) \frac{(\mathrm{x})}{1-\mathrm{x}} ;\left[\mathrm{H}^{+}\right]=\mathrm{ax} \\
& -\log \left(\mathrm{K}_{\mathrm{a}}\right)=-\log (\mathrm{ax})-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\
& \mathrm{pKa}=\mathrm{pH}-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\
& \mathrm{pH}-\mathrm{pKa}=\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right)
\end{aligned}$$
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