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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2010 · Chemistry · Ionic Equilibrium

At 25^ C , the solubility product of Mg(OH)_2 is 1.0 × 10^ -11 . At...

At 25∘C25^{\circ} \mathrm{C}, the solubility product of Mg(OH)2\mathrm{Mg}(\mathrm{OH})_2 is 1.0×10−111.0 \times 10^{-11}. At which pH\mathrm{pH}, will Mg2+\mathrm{Mg}^{2+} ions start precipitating in the form of Mg(OH)2\mathrm{Mg}(\mathrm{OH})_2 from a solution of 0.001MMg2+0.001 \mathrm{M} \mathrm{Mg}^{2+} ions ?

Options

  1. A9
  2. B10
  3. C11
  4. D8

Correct answer

B. 10

Step-by-step solution

Mg2++2OH−⇌Mg(OH)2 Ksp=[Mg2+][OH−]2[OH−]=Ksp[Mg2+]=10−4∴pOH=4\mathrm{Mg}^{2+}+2 \mathrm{OH}^{-} \rightleftharpoons \mathrm{Mg}(\mathrm{OH})_2 \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}}=10^{-4} \therefore \mathrm{p}^{\mathrm{OH}}=4 and pH=10\mathrm{p}^{\mathrm{H}}=10
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