Question

Monochloroacetic acid (HC2H2ClO2) is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of Ka for monochloroacetic acid is 1.35 ✕ 10−3. Calculate the pH of a 0.31 M solution of monochloroacetic acid.

Answer 1

Answer: The pH of the solution is 1.703

Explanation:

We are given:

Concentration of monochloroacetic acid = 0.31 M

The chemical equation for the dissociation of monochloroacetic acid follows:

                  $HC_2H_2ClO_2\rightleftharpoons H^++C_2H_2ClO_2^-$

Initial:               0.31

At eqllm:        0.31-x             x            x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[H^+][C_2H_2ClO_2^-}}{[HC_2H_2ClO_2]}$

We are given:

$K_a=1.35\times 10^{-3}$

Putting values in above equation follows:

$1.35\times 10^{-3}=\frac{x\times x}{(0.31-x)}\\\\x=-0.021,0.0198$

Neglecting the negative value of 'x' because concentration cannot be negative.

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

$pH=-\log (0.0198)\\\\pH=1.703$

Hence, the pH of the solution is 1.703