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Chemistry Question on Equilibrium

At 25∘25^\circ, the dissociation constant of a base, BOH is 1.0×10−121.0 \times 10^{-12}. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

A

2.0×10−6molL−12.0 \times 10^{-6} mol L^{-1}

B

1.0×10−5molL−11.0 \times 10^{-5} mol L^{-1}

C

1.0×10−6molL−11.0 \times 10^{-6} mol L^{-1}

D

1.0×10−7molL−11.0 \times 10^{-7} mol L^{-1}

Answer

1.0×10−7molL−11.0 \times 10^{-7} mol L^{-1}

Explanation

Solution

Base, BOH is dissociated as follows
\hspace15mm BOH \rightleftharpoons B^+ + OH^-
So, the dissociation constant of BOH base
\hspace20mm K_b = \frac {[B^+][OH^-]}{[BOH]} \, \, \, \, \, ..........(i)
At equilibrium [B+]=[OH−][B^+] = [OH^-]
\therefore \hspace20mm K_b = \frac{[OH^-]^2}{[BOH]}
Given that \hspace10mm K_b = 1.0 \times 10^{-12}
and \hspace20mm [BOH] = 0.01 M
Thus, \hspace10mm 1.0 \times 10^{-12} = \frac{[OH^-]^2}{0.01}
\hspace20mm [OH^-]^2 = 1 \times 10^{-14}
\hspace20mm [OH^-] = 1.0 \times 10^{-7} mol L^{-1}