Question

Calculate the $\left[{OH}^{-}\right]$ is $0.01M$ aqueous solution of NaOH $(K_{b}for{OCN}^{-}={10}^{-10})$:

• A. ${10}^{-6}M$
• B. ${10}^{-7}M$
• C. ${10}^{-8}M$
• D. None of these

Answer 1

Answer: (A)

${10}^{-6}M$

$K_{a\left(HOCN\right)}=\frac{{10}^{-14}}{{10}^{-10}}={10}^{-4}$
$K_h=K_{b\left(OC{N}^{-}\right)}={10}^{-10}$
$\alpha =\sqrt{\frac{K_h}{C}}=\sqrt{\frac{{10}^{-10}}{0.01}}={10}^{-4}$
$\therefore \alpha <<0.1$; $\therefore K_h=c{\alpha }^2$
$\left[O{H}^{-}\right]=c\alpha ;$
$\left[O{H}^{-}\right]=0.01\times {10}^{-4}={10}^{-6}M$