Answer:
[tex]pKa=3.58[/tex]
Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
[tex]pH=-log([H^+])[/tex]
[tex][H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}[/tex]
And the percent ionization is:
[tex]\% \ ionization=\frac{[H^+]}{[HA]}*100\%[/tex]
We compute the concentration of the acid, HA:
[tex][HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\[/tex]
[tex][HA]=6.03x10^{-4}[/tex]
Thus, the Ka is:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}[/tex]
So the pKa is:
[tex]pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58[/tex]
Regards.
