Question

A monoprotic acid in $100M$ solution is $0.001\%$ ionized. The dissociation constant of this acid is:

• A. $1.0\times {10}^{-3}$
• B. $1.0\times {10}^3$
• C. $1.0\times {10}^{-8}$
• D. $1.0\times {10}^{-10}$

Answer 1

Answer: (C)

$1.0\times {10}^{-8}$

$HA\leftrightharpoons H^{+}+A^{-}$

% ionization $=0.001$ %

$\left[HA\right]=100M$

Ionization percent $=\frac{\left[H^{+}\right]}{\left[HA\right]}\times 100$

$0.001=\frac{\left[H^{+}\right]\times 100}{100}\left[H^{+}\right]=0.001K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$=\frac{0.001\times 0.001}{100}$

$=1\times {10}^{-8}$