Question

In an exactly.200M solution of benzoic acid, a monoprotic acid, the [H+] = 3.55 × 10–3M. What is the Ka for benzoic acid?

Answer 1

6.3 *10^-5 I hope this helps

Answer 2

Answer : The value of $k_a$ for benzoic acid is, $6.4\times 10^{-5}$

Solution :

The balanced equilibrium reaction will be,

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

initial conc.          0.2 M                  0                0

at eqm.  $(0.2-0.00355)M$       $0.00355M$ $0.00355M$

The expression for dissociation constant for a benzoic acid will be,

$k_a=\frac{[H^+]\times [C_6H_5COO^-]}{[C_6H_5COOH]}$

Now put all the given values in this formula, we get the value of $k_a$

$k_a=\frac{(3.55\times 10^{-3})\times (3.55\times 10^{-3})}{(0.2-3.55\times 10^{-3})}=6.4\times 10^{-5}$

Therefore, the value of $k_a$ for benzoic acid is, $6.4\times 10^{-5}$