Verified Solution • 2025–26 Session moderate QJEEM25Chff82: JEE Main 2025 Chemistry - Class 11 JEEM25Chff82 Solution: A weak acid HA has degree of dissociation x . Which option… A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - pKa\mathrm{pK}_{\mathrm{a}}pKa )? Multiple Choice Options A log(1+2x)\log (1+2 \mathrm{x})log(1+2x) B log(1−xx)\log \left(\frac{1-x}{x}\right)log(x1−x) C 000 D log(x1−x)\log \left(\frac{x}{1-x}\right)log(1−xx) Step-by-Step Verified Solution amp;HA⇌H⊕+AΘamp;t=0aamp;t=ta(1−x)axaxamp; Ka=(ax)(x)1−x;[H+]=axamp;−log(Ka)=−log(ax)−log(x1−x)amp;pKa=pH−log(x1−x)amp;pH−pKa=log(x1−x)\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}amp;HA⇌H⊕+AΘamp;t=0aamp;t=ta(1−x)axaxamp; Ka=(ax)1−x(x);[H+]=axamp;−log(Ka)=−log(ax)−log(1−xx)amp;pKa=pH−log(1−xx)amp;pH−pKa=log(1−xx) Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring