Answers: 1
![\% ionization=\frac{[H^+]}{[HA]}*100\%](/tpl/images/0739/3957/49c2b.png)
![[H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M](/tpl/images/0739/3957/75e88.png)
![Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}](/tpl/images/0739/3957/ccddd.png)
![pH=-log([H^+])=-log(0.012)\\\\pH=1.92](/tpl/images/0739/3957/80a40.png)
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The weak acid HA is 4 % ionized (dissociated) in a 0.30 M solution. A. What is Ka for this acid?B. W...
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