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JIPMER 2007
Exam
JIPMER
Year
2007
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2007 · Chemistry · Ionic Equilibrium

At 25^ C , the dissociation constant of a base, BOH , is 1.0 × 10^...

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

Options

  1. A2.0×10−6 mol L−12.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}
  2. B1.0×10−5 mol L−11.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}
  3. C1.0×10−6 mol L−11.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}
  4. D1.0×10−7 mol L−11.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}

Correct answer

D. 1.0×10−7 mol L−11.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}

Step-by-step solution

Base BOHB O H is dissociated as follows BOH⇌B++OH−\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} So, the dissociation constant of BOHB O H base Kb=[B+][OH−][BOH]...(i)K_b=\frac{\left[B^{+}\right]\left[\mathrm{OH}^{-}\right]}{[B \mathrm{OH}]}...(i) At equilibrium [B+]=[OH−]∴Kb=[OH−]2[BOH]\left[B^{+}\right]=\left[\mathrm{OH}^{-}\right] \therefore K_b=\frac{\left[\mathrm{OH}^{-}\right]^2}{[B \mathrm{OH}]} Given that Kb=1.0×10−12K_b=1.0 \times 10^{-12} and BOH=0.01 M\mathrm{BOH} = 0.01 \mathrm{~M}Therefore, [OH−]2=1×10−14[OH−]=1.0×10−7 mol L−1\begin{aligned} \text{Therefore, } {\left[\mathrm{OH}^{-}\right]^2 } & =1 \times 10^{-14} \\ {\left[\mathrm{OH}^{-}\right] } & =1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$$
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