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)`.