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Ionic Equilibrium question

2010 · Shift 0 · Q6
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Ionic Equilibrium question

2010 · Shift 0 · Q6

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
At 25°C, the solubility product of Mg(OH)2Mg(OH)_2Mg(OH)2​ is 1.0 ×\times× 10–11. At which pH, will Mg2+Mg^{2+}Mg2+ ions start precipitating in the form of Mg(OH)2Mg(OH)_2Mg(OH)2​ from a solution of 0.001 M Mg2+Mg^{2+}Mg2+ ions?
  1. A
    9
  2. B
    10
  3. C
    11
  4. D
    8
View written solutionFree

Correct answer: B

  1. Write the precipitation equilibrium

    Mg(OH)2(s)⇌Mg2++2OH−Mg(OH)_2(s) \rightleftharpoons Mg^{2+} + 2OH^-Mg(OH)2​(s)⇌Mg2++2OH−

    The solubility product expression is:

    Ksp=[Mg2+][OH−]2K_{sp} = [Mg^{2+}][OH^-]^2Ksp​=[Mg2+][OH−]2

  2. Condition for precipitation to start

    Precipitation begins when the ionic product becomes equal to KspK_{sp}Ksp​:

    [Mg2+][OH−]2=1.0×10−11[Mg^{2+}][OH^-]^2 = 1.0 \times 10^{-11}[Mg2+][OH−]2=1.0×10−11

    Given:

    [Mg2+]=0.001=10−3 M[Mg^{2+}] = 0.001 = 10^{-3}\,M[Mg2+]=0.001=10−3M

    So,

    10−3[OH−]2=10−1110^{-3}[OH^-]^2 = 10^{-11}10−3[OH−]2=10−11

  3. Calculate [OH−][OH^-][OH−]

    [OH−]2=10−8[OH^-]^2 = 10^{-8}[OH−]2=10−8

    [OH−]=10−4 M[OH^-] = 10^{-4}\,M[OH−]=10−4M

  4. Find pOH and pH

    pOH=−log⁡(10−4)=4pOH = -\log(10^{-4}) = 4pOH=−log(10−4)=4

    At 25∘C25^\circ C25∘C,

    pH+pOH=14pH + pOH = 14pH+pOH=14

    Hence,

    pH=14−4=10pH = 14 - 4 = 10pH=14−4=10

  5. Match with options

    • A: 9
    • B: 10
    • C: 11
    • D: 8

    Therefore, the correct option is B.

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