Question

A solution of benzoic acid (C6H5COOH) that is 0.20 M has a pH of 2.45. What is Ka for benzoic acid?

Answer 1

We first calculate for the H+ ion concentration from the pH of the solution of benzoic acid. We do as follows:

pH = -log[H+]
2.45 = -log [H+]
[H+] = 3.55x10^-3

We then use the initial concentration and the h+ ion concentration as follows for Ka calculation:
   C6H5COOH = C6H5COO-   +   H+
I       0.20                      0                    0                
C        -x                        x                    x
E      0.20-x                   x                   x

x = [H+] = 3.55x10^-3
[C6H5COO-] = 3.55x10^-3
[C6H5COOH] = 0.1965

Ka = 3.55x10^-3 (3.55x10^-3) / 0.1965 = 6.42x10^-5

Answer 2

The Ka for benzoic acid is $6.42\times10^-^5$

Calculation:

From the pH of the benzoic acid solution, we first compute the H+ ion concentration.

pH = $\bold{-log^[^H^+^]}$

2.45 = $\bold{-log^[^H^+^]}$

$[H^+]$ = $\bold{3.55\times10^-^3}$

For the Ka calculation, we use the initial concentration and the h+ ion concentration as follows:

$= ^[H^+^] = 3.55\times10^-^3$

$[C_6H_5COO-] = 3.55\times10^-^3$

$[C_6H_5COOH] = 0.1965$

$Ka = \dfrac{3.55\times10^-3 (3.55\times10^-^3) }{0.1965} = 6.42\times10^-^5$

The Ka for benzoic acid is $6.42\times10^-^5$

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