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JEE Chemistry 2024 Question with Solution

The pH at which Mg(OH)₂ [Ksp=1×10−11K_{sp} = 1 \times 10^{-11}] begins to precipitate from a solution containing 0.10 M0.10 \, \text{M} Mg²⁺ ions is:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: KspK_{sp} of Mg(OH)₂ is 1×10−111 \times 10^{-11} and [Mg2+]=0.10 M[\text{Mg}^{2+}] = 0.10 \, \text{M}.

Find: The pH at which precipitation begins.

Precipitation starts when the ionic product becomes equal to the solubility product.

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

So,

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

At the point where precipitation begins,

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

Substituting the given values,

(0.10)[OH−]2=1×10−11(0.10)[\text{OH}^-]^2 = 1 \times 10^{-11}[OH−]2=1×10−110.10=1×10−10[\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10}[OH−]=1×10−10=1×10−5 M[\text{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \, \text{M}

Now,

pOH=−log⁡10[OH−]=−log⁡10(1×10−5)=5\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1 \times 10^{-5}) = 5

Using

pH+pOH=14\text{pH} + \text{pOH} = 14

we get

pH=14−5=9\text{pH} = 14 - 5 = 9

Therefore, Mg(OH)₂ begins to precipitate at pH = 9.

Direct pOH Method

Given: Ksp=10−11K_{sp} = 10^{-11} and [Mg2+]=0.1 M[\text{Mg}^{2+}] = 0.1 \, \text{M}.

Find: The pH at the onset of precipitation.

Use directly:

[Mg2+][OH−]2=Ksp[\text{Mg}^{2+}][\text{OH}^-]^2 = K_{sp}0.1[OH−]2=10−110.1[\text{OH}^-]^2 = 10^{-11}[OH−]2=10−10[\text{OH}^-]^2 = 10^{-10}[OH−]=10−5[\text{OH}^-] = 10^{-5}

Hence,

pOH=5\text{pOH} = 5

and therefore,

pH=14−5=9\text{pH} = 14 - 5 = 9

The shortcut works because precipitation starts exactly when the ionic product equals KspK_{sp}, so the threshold [OH−][\text{OH}^-] can be found immediately from the given [Mg2+][\text{Mg}^{2+}]. The final answer is 9.

Common mistakes

  • Using Ksp=[Mg2+][OH−]K_{sp} = [\text{Mg}^{2+}][\text{OH}^-] instead of Ksp=[Mg2+][OH−]2K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2. This is wrong because Mg(OH)₂ releases two hydroxide ions. Always include the stoichiometric power on each ion concentration.

  • Forgetting to convert from pOH to pH. After finding [OH−]=10−5[\text{OH}^-] = 10^{-5}, the value obtained is pOH=5\text{pOH} = 5, not the pH. Use pH+pOH=14\text{pH} + \text{pOH} = 14 to get the required answer.

  • Dividing 10−1110^{-11} by 0.100.10 incorrectly. Since 0.10=10−10.10 = 10^{-1}, we get 10−11/10−1=10−1010^{-11}/10^{-1} = 10^{-10}. Handle powers of ten carefully before taking the square root.

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