Question

What is the total concentration of an acetic acid solution (ch3cooh) with a ph of 3.5? acetic acid ka = 1.76 x 10-5 m?

Answer 1

The dissociation reaction of acetic acid is as follows:

$CH_{3}COOH\leftrightharpoons CH_{3}COO^{-}+H^{+}$

The acid dissociation constant K_{a} is $1.76\times 10^{-5}$.

Let the initial concentration of acid be A, and concentration of $CH_{3}COO^{-}$ and $H^{+}$ be zero.

After dissociation, concentration of acid becomes A-x and that of both $CH_{3}COO^{-}$ and $H^{+}$ becomes x.

Expression for acid dissociation constant will be:

$K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}$

pH of solution is 3.5, thus, concentration of hydrogen ion can be calculated as follows:

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

On rearranging,

$[H^{+}]=10^{-pH}=10^{-3.5}=0.0003162$

Since, $[CH_{3}COO^{-}]=[H^{+}]=x$

Thus,

$[CH_{3}COO^{-}]=0.0003162$

and, $x=0.0003162$

Putting the values, in expression for acid dissociation constant,

$1.76\times 10^{-5}=\frac{(0.0003162)(0.0003162)}{[CH_{3}COOH]_{initial}-0.0003162}$

On rearranging,

$[CH_{3}COOH]_{initial}=\frac{(0.0003162)\times (0.0003162)}{1.76\times 10^{-5}}+0.0003162=0.006$