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Q.
pH of a mixture of 1 M benzoic acid $(pK_a = 4.20)$ and 1 M sodium benzoate is 4.5. In 300 mL of buffer solution , volume of benzoic acid solution is
Equilibrium
Solution:
$pH=pK_{a}+log \frac{\text{[Salt]}}{\text{Acid}}$
$4.5=4.2+log \frac{\text{[Salt]}}{\text{[Acid]}}$
$0.3=log \frac{\text{[Salt]}}{\text{[Acid]}}$
$\frac{\text{[Salt]}}{\text{[Acid]}}$ =antilog $0.3 = 2$
Let $\upsilon mL$ of $1\, M$ benzoic acid is mixed with $\left(300-\upsilon\right)$ mL of $1M$ sodium benzoate
$\therefore \frac{300-\upsilon}{\upsilon}=2$
$300-\upsilon=2\,\upsilon$
$\upsilon=100\, mL$