Answers: 1
![pH=-log([H^+])](/tpl/images/0920/4064/79766.png)
![[H^+]=10^{-pH}=10^{-2.353}=4.436x10^{-3}M](/tpl/images/0920/4064/4c8f1.png)
![Ka=\frac{[H^+][A^-]}{[HA]}](/tpl/images/0920/4064/39962.png)
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The pH of 0.259 M solution of an unknown acid is measured to be 2.353. What is the Ka of the acid?...
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