Answer:
For monobasic acid,
\[\alpha
=\sqrt{\frac{{{K}_{a}}}{C}=\sqrt{\frac{2.5\times {{10}^{-5}}}{0.08}}}\]
\[=0.0176\]
% Dissociation =
0.0176 x 100 = 1.76
\[pH=-\log
C\alpha =-\log (0.08\times 0.0176)\]
=2.85
You need to login to perform this action.
You will be redirected in
3 sec