Question

in a 5.80E-2 M solution of a monoprotic acid HA, the acid is 23.9% dissociated. Calculate Ka for this acid.

Answer 1

First you must find
\[[H ^{+}] = \alpha \times cons of acid\]
\[[H ^{+}] = 0.239 \times 0.058 = 0.0138\]

Answer 2

Then \[[H ^{+}] =\sqrt{(Ka \times cons. of acid}\]
\[0.0138 =\sqrt{(Ka \times 0.058}\]
Ka = (0.0138)^2 / 0.058 = 3.31 x \[10^{-3}\]