At $25^{\circ} \mathrm{C}$, the dissociation constant of a base, $\mathrm{BOH}$, is $1.0 \times 10^{-12}$.…

At $25^{\circ} \mathrm{C}$, the dissociation constant of a base, $\mathrm{BOH}$, is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in $0.01 \mathrm{M}$ aqueous solution of the base would be:
  1. $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
  2. $1.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
  3. $2.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
  4. $1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$

Solution

$\begin{gathered} \mathrm{C}=0.01 \mathrm{M} \\ \mathrm{K}_b=1 \times 10^{-12} \text { at } 25^{\circ} \mathrm{C} \\ \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\ \mathrm{C} \quad 0 \quad 0 \end{gathered}$ at eq. $\mathrm{C}-\mathrm{C} \alpha \mathrm{C} \alpha \mathrm{C} \alpha$ $\begin{aligned} {\left[\mathrm{OH}^{-}\right] } & =\mathrm{C} \alpha \\ {\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_b \mathrm{C}}=\sqrt{1 \times 10^{-12} \times 10^{-2}} \end{aligned}$ $\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$ Related Theory The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the law of mass action which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts.

Asked in: NEET 2005

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