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JEE Chemistry 2024 Question with Solution

The pH at which Mg(OH)2\text{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+\text{Mg}^{2+} ions is:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: KspK_{sp} of Mg(OH)2\text{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.

For the equilibrium

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

we use

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

Substituting the given concentration:

(0.10)[OH−]2=1×10−11(0.10)[\text{OH}^-]^2 = 1 \times 10^{-11}

So,

[OH−]2=1×10−110.10=1×10−10[\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10}

Taking square root,

[OH−]=1×10−5 M[\text{OH}^-] = 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\text{Mg(OH)}_2 begins to precipitate at pH = 9.

Direct Solubility Product Shortcut

Given: Ksp=1×10−11K_{sp} = 1 \times 10^{-11} and [Mg2+]=0.10 M[\text{Mg}^{2+}] = 0.10 \, \text{M}.

Find: The pH at the onset of precipitation.

At the point where precipitation just starts,

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

Thus,

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

Hence,

[OH−]=10−5[\text{OH}^-] = 10^{-5}

So the pOH is 55, and therefore the pH is 99.

This shortcut works because at the precipitation threshold, the ionic product is exactly equal to KspK_{sp}.

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\text{Mg(OH)}_2 releases two hydroxide ions. Always raise the ion concentration to its stoichiometric coefficient in the solubility product expression.

  • Confusing pOH with pH after finding [OH−]=10−5[\text{OH}^-] = 10^{-5}. This gives pOH = 5, not pH. Use pH+pOH=14\text{pH} + \text{pOH} = 14 to convert and obtain the required pH.

  • Dividing 10−1110^{-11} by 0.100.10 incorrectly. Since 0.10=10−10.10 = 10^{-1}, the result is 10−1010^{-10}, not 10−1210^{-12} or 10−1110^{-11}. Rewrite decimal concentrations in powers of ten before simplifying.

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