Question

$pH$ of a mixture of $1M$ benzoic acid $(p{K}_a=4.2)$ and $1M$ sodium benzoate is $4.5$. In $300$ mL buffer, benzoic acid is:

• A. $200$ mL
• B. $150$ mL
• C. $100$ mL
• D. $50$ mL

Answer 1

The correct option is C $100$ mL

Concentration of benzoic acid, [benzoic acid] $=1M$ (initially)

concentration of benzoate, [benzoate] $=1M$ (initially)

Let's consider $V$ ml of benzoic acid be taken.

So $(300-V)$ ml of sodium benzoate is taken.

$\therefore$ ${\left[benzoicacid\right]}_{final}=\frac{1M\times Vml}{300ml}$

${\left[benzoate\right]}_{final}=\frac{1M\times (300-V)ml}{300ml}$

$\therefore$ $pH={pK}_a+log\frac{{\left[benzoate\right]}_{final}}{{\left[benzoicacid\right]}_{final}}$

$\Rightarrow 4.5=4.2+log\left\{\frac{(300-V)/300}{V/300}\right\}$

$\Rightarrow \frac{300-V}{V}={10}^{0.3}\approx 2$

$\Rightarrow V=100ml$

$\therefore$ $100ml$ of benzoic acid had been taken.

$\therefore$ Correct answer is C.