Question

An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?

Answer 1

Answer:

$pKa=3.58$

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

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

$[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}$

And the percent ionization is:

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

We compute the concentration of the acid, HA:

$[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%$

$[HA]=6.03x10^{-4}$

Thus, the Ka is:

$Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}$

So the pKa is:

$pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58$

Regards.