Verified Solution • 2025–26 Session moderate QJEEM24Chb91a: JEE Main 2024 Chemistry - Class 11 JEEM24Chb91a Solution: The of an aqueous solution containing benzoic acid and sodium… The pH\mathrm{pH}pH of an aqueous solution containing 1M1 \mathrm{M}1M benzoic acid (pKa=4.20)\left(\mathrm{pK}_{\mathrm{a}}=4.20\right)(pKa=4.20) and 1M1 \mathrm{M}1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL300 \mathrm{~mL}300 mL of this buffer solution is _________ mL\mathrm{mL}mL . (given : log2=0.3\log 2=0.3log2=0.3 ) Step-by-Step Verified Solution amp; 1M Benzoic acid amp;+1M Sodium Benzoate amp;(Vaml)amp;(Vsml) Millimole amp;Va×1amp; Vs×1\begin{array}{ccc} & \text { 1M Benzoic acid } & +1 \mathrm{M} \text { Sodium Benzoate } \\& \left(\mathrm{V}_{\mathrm{a}} \mathrm{ml}\right) & \left(\mathrm{V}_{\mathrm{s}} \mathrm{ml}\right) \\\text { Millimole } & \mathrm{V}_{\mathrm{a}} \times 1 & \mathrm{~V}_{\mathrm{s}} \times 1\end{array} Millimole amp; 1M Benzoic acid amp;(Vaml)amp;Va×1amp;+1M Sodium Benzoate amp;(Vsml)amp; Vs×1 amp;pH=4.5amp;pH=pka+log[ salt ][ acid ]amp;4.5=4.2+log(VsVa)amp;VsVa=2..... (1)amp; Vs+Va=300..... (2)amp; Va=100 ml\begin{aligned}& \mathrm{pH}=4.5 \\& \mathrm{pH}=\mathrm{pka}+\log \frac{[\text { salt }]}{[\text { acid }]} \\& 4.5=4.2+\log \left(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}\right) \\& \frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}=2 \quad \text{..... (1)}\\& \mathrm{~V}_{\mathrm{s}}+\mathrm{V}_{\mathrm{a}}=300 \quad \text{..... (2)}\\& \mathrm{~V}_{\mathrm{a}}=100 \mathrm{~ml}\end{aligned}amp;pH=4.5amp;pH=pka+log[ acid ][ salt ]amp;4.5=4.2+log(VaVs)amp;VaVs=2..... (1)amp; Vs+Va=300..... (2)amp; Va=100 ml 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