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Question: At \[{25^\circ }C\], the dissociation constant of a base, BOH is\[1.0 \times {10^{ - 12}}\]. The con...

At 25∘C{25^\circ }C, the dissociation constant of a base, BOH is1.0×10−121.0 \times {10^{ - 12}}. The concentration of hydroxyl ions in 0.01M 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}}

Explanation

Solution

The base dissociates to give a metal cation and a hydroxyl anion. The dissociation constant is calculated by dividing the individual concentration of the dissociated ion by the concentration of the solution. At equilibrium condition the concentration of the cation and the hydroxyl anion becomes equal.

Complete answer:
Given,
Kb{K_b} of base is 1.0×10−121.0 \times {10^{ - 12}}
The concentration of aqueous solution is 0.01 M.
The base is the compound which dissociates into metal cation and hydroxyl anion.
The base BOH dissociates into B+{B^ + }and OH−O{H^ - }.
The dissociation reaction of base BOH into its constituent ion is shown below.
BOH→B++OH−BOH \to {B^ + } + O{H^ - }
Kb{K_b} is the base dissociation constant which measures the rate of dissociation of the base into its constituent ions.
The dissociation is calculated by dividing the concentration of the individual constituent ions by the total concentration of the base solution.
The dissociation constant of base BOH is given as shown below.
Kb=[B+][OH−][BOH]{K_b} = \dfrac{{[{B^ + }][O{H^ - }]}}{{[BOH]}}……..(i)
At equilibrium condition the concentration of cation is equal to the concentration of anion as shown below.
[B+]=[OH−][{B^ + }] = [O{H^ - }]
Thus, the is given as shown below.
Kb=[OH−]2[BOH]{K_b} = \dfrac{{{{[O{H^ - }]}^2}}}{{[BOH]}}……(ii)
To calculate the concentration of hydroxyl ion, substitute the value of Kb{K_b} and concentration of aqueous solution in the above equation.
⇒1.0×10−12=[OH−]20.01\Rightarrow 1.0 \times {10^{ - 12}} = \dfrac{{{{[O{H^ - }]}^2}}}{{0.01}}
⇒[OH−]2=1.0×10−12×0.01\Rightarrow {[O{H^ - }]^2} = 1.0 \times {10^{ - 12}} \times 0.01
⇒[OH−]2=1.0×10−14\Rightarrow {[O{H^ - }]^2} = 1.0 \times {10^{ - 14}}
⇒[OH−]=1.0×10−7\Rightarrow [O{H^ - }] = 1.0 \times {10^{ - 7}}
Thus, the concentration of hydroxyl ion is
1.0×10−7molL−11.0 \times {10^{ - 7}}mol{L^{ - 1}}.
Therefore, the correct option is D.

Note: Make sure that the dissociation constant is measured in this reaction when the reaction is at equilibrium condition where the concentration of anion and concentration of hydroxyl ion become the same.