Question

Calculate the pH of a solution prepared by dissolving 0.15 mol benzoic acid (C7H5O2H) and 0.30 mol of sodium benzoate (Na C7H5O2) in water sufficient to yield 1.00 L of solution. The Ka of benzoic acid is 6.50 x 10-5

Answer 1

Answer : The pH of a solution is, 4.5

Explanation : Given,

Moles of benzoic acid = 0.15 mol

Moles of sodium benzoate = 0.30 mol

Volume of solution = 1.00 L

The dissociation constant for benzoic acid = $K_a=6.5\times 10^{-5}$

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:

$pK_a=-\log (6.5\times 10^{-5})$

$pK_a=5-\log (6.5)$

$pK_a=4.2$

Now we have to calculate the pH of a solution.

Using Henderson Hesselbach equation :

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

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

$pH=4.2+\log [\frac{(\frac{0.30}{1.00L})}{(\frac{0.15}{1.00L})}]$

$pH=4.5$

Therefore, the pH of a solution is, 4.5