Answer:
[tex]Ka=5x10^{-4}[/tex]
[tex]pH=1.92[/tex]
Explanation:
Hello,
In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:
[tex]\% ionization=\frac{[H^+]}{[HA]}*100\%[/tex]
[tex][H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M[/tex]
Therefore the Ka is computed by using the equilibrium expression:
[tex]Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}[/tex]
And the pH:
[tex]pH=-log([H^+])=-log(0.012)\\\\pH=1.92[/tex]
Regards.
