Answers: 2
![pH=-log([H^+])](/tpl/images/0737/8701/79766.png)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](/tpl/images/0737/8701/c6088.png)
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](/tpl/images/0737/8701/deb28.png)
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](/tpl/images/0737/8701/71dd1.png)
![[HA]=6.03x10^{-4}](/tpl/images/0737/8701/e1514.png)
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](/tpl/images/0737/8701/17698.png)
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An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?...
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