Question

Calculate the pH of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water sufficient to yield 1.00 L of solution. The Ka of benzoic acid is 6.30 × 10-5.

Answer 1

Answer : The pH of a solution is, 4.50

Explanation : Given,

$K_a=6.30\times 10^{-5}$

Concentration of benzoic acid (Acid) = 0.150 M

Concentration of sodium benzoate (salt) = 0.300 M

First we have to calculate the value of $pK_a$.

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.30\times 10^{-5})$

$pK_a=5-\log (6.30)$

$pK_a=4.20$

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=4.20+\log (\frac{0.300}{0.150})$

$pH=4.50$

Thus, the pH of a solution is, 4.50