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Chemistry Question on Equilibrium

At 25∘C25^{\circ} C, the dissociation constant of a base, BOHBOH, is 1.0×10−121.0 \times 10^{-12}. The concentration of hydroxyl ions in 0.01 M0.01\, M aqueous solution of the base would be

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

Base BOHBOH is dissociated as follows
BOH⇌B++OH−BOH \rightleftharpoons B ^{+}+ OH ^{-}
So, the dissociation constant of BOHB O H base
Kb=[B+][OH−][BOH]    ...(i)K_{b}=\frac{\left[B^{+}\right]\left[O H^{-}\right]}{[B O H]}\,\,\,\,...(i)
At equilibrium [B+]=[OH−]\left[B^{+}\right]=\left[ OH ^{-}\right]
∴Kb=[OH−]2[BOH]\therefore K_{b}=\frac{\left[ OH ^{-}\right]^{2}}{[ BOH ]}
Given that Kb=1.0×10−12 K_{b} =1.0 \times 10^{-12} and
[BOH]=0.01 M[B O H] =0.01\,M
Thus 1.0×10−12=[OH−]20.011.0 \times 10^{-12} =\frac{\left[ OH ^{-}\right]^{2}}{0.01}
[OH−]2=1×10−14\left[ OH ^{-}\right]^{2} =1 \times 10^{-14}
[OH−]=1.0×10−7 mol L−1\left[ OH ^{-}\right] =1.0 \times 10^{-7} \,mol\, L ^{-1}