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? 8.88 × 10–2 1.77 × 10–2 6.3 × 10–5 7.10 × 10–4

Answer 1

Answer : Option C) 6.3 X $10^{-5}$.

Explanation : Given is the concentration of benzoic acid is 0.200 M, now only one monoprotic acid ion is getting separated from it [$[ H^{+} ]$.

So, the reaction will be as,

$H C_{7} H_{6}O_{2}$   ⇔  $[ H^{+} ]$ + $C_{7} H_{6}O_{2}$ .

We can use the ICE method for calculation,
 The molar concentration of $H C_{7} H_{6}O_{2}$ is 0.200 M and the  [$[ H^{+} ]$ is 3.55 X $10^{-5}$.

So here on reactant side we get, (0.2 - 3.55 X $10^{-5}$) = 0.196
 on the product it remains as 3.55 X $10^{-5}$.

So, $K_{a}$ = (3.55 X $10^{-5}$) X (3.55 X $10^{-5}$) / 0.196

We will get $K_{a}$ as 6.42 X  $10^{-5}$ which is close to 6.3 X $10^{-5}$.