Question

At 25${}^{\circ }$C, the dissociation constant of a base, $BOH$, is ${10}^{12}$. The concentration of hydroxyl ions in the 0.01M aqueous solution of the base would be:

• A. ${10}^{6}mole/L$
• B. ${10}^{7}mole/L$
• C. $2\times {10}^{6}mole/L$
• D. ${10}^{5}mole/L$

Answer 1

The correct option is D ${10}^{5}mole/L$

Given $BOH\rightleftharpoons B^{+}+{OH}^{-}Kd={10}^{12}$

$\left[BOH\right]={10}^{-2}M$

$Kd=\frac{\left[B^{+}\right]\left[{OH}^{-}\right]}{\left[BOH\right]}$ Since value of $Kd$ is very much high,

$\downarrow$ hence complete dissociation

of $BOH$ takes place

$\therefore BOH\to \underset{}{B^{+}+{OH}^{-}}$

$\therefore {10}^{12}=\frac{{\left[OH\right]}^2}{0.01}$ $\leftarrow \left[B^{+}\right]=\left[{OH}^{-}\right]$

$\therefore \left[{OH}^{-}\right]={10}^{5}mole/L.$