Question

A monoprotic acid in $1.00M$ solution is $0.01\%$ ionised. The dissociation of this acid is:

• A. $1\times {10}^{-8}$
• B. $1\times {10}^{-4}$
• C. $1\times {10}^{-6}$
• D. $1\times {10}^{-5}$

Answer 1

The correct option is A $1\times {10}^{-8}$

$K=\frac{{\alpha }^{2}C}{1-\alpha }\alpha =\frac{0.01}{100}\approx 1\therefore K={\alpha }^{2}C={\left[\frac{0.01}{100}\right]}^2\times 1=1\times 10^{-8}$