Ionic EquilibriumHard
Question
The dissociation constant of a weak monoprotic acid is numerically equal to the dissociation constant of its conjugate base. What is the pH of 0.1M solution of this acid?
Options
A.7.0
B.6.0
C.8.0
D.4.0
Solution
$K_{a}(HA) = K_{b}\left( A^{-} \right) = \sqrt{Kw} = 10^{- 7}$
Now, $\left\lbrack H^{+} \right\rbrack = \sqrt{10^{- 7} \times 0.1} = 10^{- 4}\text{ M} \Rightarrow P^{H} = 4.0$
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