Question

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionization constant is:

• A. $1.0\times {10}^{-3}$
• B. $1.0\times {10}^{-6}$
• C. $1.0\times {10}^{-8}$
• D. $1.0\times {10}^{-11}$

Answer 1

The correct option is D $1.0\times {10}^{-11}$

$\left(D\right)1.0\times {10}^{-11}$

$HA\rightleftharpoons H^{+}+A^{-1}$

at $t=0,0.1$ $0$ $0$

at $t=teq.c.c\alpha$ $c\alpha$ $c\alpha$

$\left[HA\right]=0.01m$

$\alpha =0.001$

now, $k\alpha =\frac{\left(c\alpha \right)\left(c\alpha \right)}{c(1-\alpha )}=\frac{c(\alpha {)}^2}{(1-\alpha )}$

$k\alpha =0.1\times ({10}^{-5}{)}^2=0.1\times {10}^{-10}$

$k\alpha =1.0\times {10}^{-11}$