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Ionic Equilibrium question

2024 · 30 Jan · Shift 1 · Q25
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  5. /2024 · 30 Jan · Shift 1 · Q25

Ionic Equilibrium question

2024 · 30 Jan · Shift 1 · Q25

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH\mathrm{pH}pH at which Mg(OH)2[ Ksp=1×10−11]\mathrm{Mg}(\mathrm{OH})_2\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-11}\right]Mg(OH)2​[ Ksp​=1×10−11] begins to precipitate from a solution containing 0.10 M Mg2+0.10 \mathrm{~M} \mathrm{~Mg}^{2+}0.10 M Mg2+ ions is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 9

  1. Condition for precipitation to begin

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

    the solubility product is Ksp=[Mg2+][OH−]2=1×10−11.K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 = 1\times 10^{-11}.Ksp​=[Mg2+][OH−]2=1×10−11.

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

  2. Substitute the given magnesium ion concentration

    Given: [Mg2+]=0.10=10−1 M[\mathrm{Mg^{2+}}] = 0.10 = 10^{-1}\,\text{M}[Mg2+]=0.10=10−1M

    So, 10−11=(10−1)[OH−]210^{-11} = (10^{-1})[\mathrm{OH^-}]^210−11=(10−1)[OH−]2

  3. Calculate [OH−][\mathrm{OH^-}][OH−]

    [OH−]2=10−1110−1=10−10[\mathrm{OH^-}]^2 = \frac{10^{-11}}{10^{-1}} = 10^{-10}[OH−]2=10−110−11​=10−10

    [OH−]=10−5 M[\mathrm{OH^-}] = 10^{-5}\,\text{M}[OH−]=10−5M

  4. Find pOH

    pOH=−log⁡[OH−]=−log⁡(10−5)=5\mathrm{pOH} = -\log[\mathrm{OH^-}] = -\log(10^{-5}) = 5pOH=−log[OH−]=−log(10−5)=5

  5. Find pH

    pH=14−5=9\mathrm{pH} = 14 - 5 = 9pH=14−5=9

  6. Final answer

    The pH at which Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​ begins to precipitate is 9\boxed{9}9​

  7. Comparison with stored answer

    Stored correct answer = 999

    This matches the derived answer.

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