Question

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

• A. $2.0\times {10}^{-6}mol{L}^{-1}$
• B. $1.0\times {10}^{-5}mol{L}^{-1}$
• C. $1.0\times {10}^{-6}mol{L}^{-1}$
• D. $1.0\times {10}^{-7}mol{L}^{-1}$

Answer 1

Answer: (B)

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