Question

The percent ionization of benzoic acid, 2.05 %, at a particular concentration is 25 °C. What was the concentration of benzoic acid in this solution? Ka of benzoic acid is 6.3 × 10-5

Answer 1

Answer : The concentration of benzoic acid in this solution was, 0.149 M

Explanation :

The balanced equilibrium reaction will be,

                          $C_6H_5COOH\rightleftharpoons C_6H_5COO^-+H^+$

initially conc.             c                     0                0

At eqm.                    c(1-α)                cα              cα

The expression for dissociation constant is:

$k_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}$

when α is very very small the, the expression will be,

$k_a=\frac{c^2\alpha^2}{c}=c\alpha^2$

Given:

α = 2.05 % = 0.0205

$K_a=6.3\times 10^{-5}$

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

$6.3\times 10^{-5}=c\times (0.0205)^2$

c = 0.149 M

Therefore, the concentration of benzoic acid in this solution was, 0.149 M