Question

The weak acid HA is 4 % ionized (dissociated) in a 0.30 M solution. A. What is Ka for this acid?B. What is the pH of this solution?

Answer 1

Answer:

$Ka=5x10^{-4}$

$pH=1.92$

Explanation:

Hello,

In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:

$\% ionization=\frac{[H^+]}{[HA]}*100\%$

$[H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M$

Therefore the Ka is computed by using the equilibrium expression:

$Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}$

And the pH:

$pH=-log([H^+])=-log(0.012)\\\\pH=1.92$

Regards.