Question

0.01 M HA (aq.) is 2% ionized: $\left[O{H}^{-}\right]$ of solution is?

• A. $0.5\times {10}^{-10}$
• B. ${10}^{-8}$
• C. $2\times {10}^{-11}$
• D. $2\times {10}^{-12}$

Answer 1

Answer: (A)

$0.5\times {10}^{-10}$

$H{A}_{\left(aq\right)}\rightleftharpoons H_{\left(aq\right)}^{+}+A_{\left(aq\right)}^{-}$

$\Rightarrow \frac{\left[H^{+}\right]}{\left[HA\right]}\times 100=2$

$\Rightarrow \frac{\left[H^{+}\right]}{0.01}\times 100=2$

$\Rightarrow \left[H^{+}\right]=\frac{0.02}{100}=2\times {10}^{-4}M$

$\therefore \left[O{H}^{-}\right]=\frac{{10}^{-14}}{2\times {10}^{-4}}=0.5\times {10}^{-10}m$