Ionic Equilibrium
32435
A monoprotic acid in \( 1.00 \,M \) solution is \( 0.01\%\) ionised. The dissociation constant of this acid is
1 \(1 \times {10^{ - 8}}\)
2 \(1 \times {10^{ - 4}}\)
3 \(1 \times {10^{ - 6}}\)
4 \({10^{ - 5}}\)
Explanation:
(a)\(K = \frac{{{\alpha ^2}C}}{{1 - \alpha }};\,\,\alpha = \frac{{0.01}}{{100}} \approx 1\,\,\,\therefore \,\,K = {\alpha ^2}C = {\left[ {\frac{{0.01}}{{100}}} \right]^2} \times 1\)
\( = 1 \times {10^{ - 8}}\).