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Ionic Equilibrium question

2025 · 4 Apr · Shift 2 · Q25
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Ionic Equilibrium question

2025 · 4 Apr · Shift 2 · Q25

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
xmgx \mathrm{mg}xmg of Mg(OH)2(\mathrm{Mg}(\mathrm{OH})_2(Mg(OH)2​( molar mass =58)=58)=58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K . The value of xxx is ‾\underline{\hspace{2cm}}​ mg. (Nearest integer) (Given : Mg(OH)2\mathrm{Mg}(\mathrm{OH})_2Mg(OH)2​ is assumed to dissociate completely in H2O\mathrm{H}_2 \mathrm{O}H2​O ]
Numerical answer
View written solutionFree

Correct answer: 3

  1. Given pH, find pOH

Since at 298 K298\,\text{K}298K, pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14 so for pH=10\text{pH} = 10pH=10, pOH=14−10=4\text{pOH} = 14 - 10 = 4pOH=14−10=4

  1. Find hydroxide ion concentration

[OH−]=10−pOH=10−4 mol L−1[\mathrm{OH}^-] = 10^{-\text{pOH}} = 10^{-4}\,\text{mol L}^{-1}[OH−]=10−pOH=10−4mol L−1

  1. Relate this to dissociation of Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​

Given complete dissociation: Mg(OH)2→Mg2++2OH−\mathrm{Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-}Mg(OH)2​→Mg2++2OH−

If the molar concentration of dissolved Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​ is ccc, then [OH−]=2c[\mathrm{OH}^-] = 2c[OH−]=2c

Hence, 2c=10−42c = 10^{-4}2c=10−4 c=5×10−5 mol L−1c = 5\times 10^{-5}\,\text{mol L}^{-1}c=5×10−5mol L−1

  1. Moles required in 1.0 L

Since volume =1.0 L=1.0\,\text{L}=1.0L, n=cV=5×10−5×1=5×10−5 moln = cV = 5\times 10^{-5}\times 1 = 5\times 10^{-5}\,\text{mol}n=cV=5×10−5×1=5×10−5mol

  1. Convert moles to mass

Molar mass of Mg(OH)2=58\mathrm{Mg(OH)_2}=58Mg(OH)2​=58 g/mol.

So mass required is m=nM=5×10−5×58=2.9×10−3 gm = nM = 5\times 10^{-5}\times 58 = 2.9\times 10^{-3}\,\text{g}m=nM=5×10−5×58=2.9×10−3g

Convert to mg: 2.9×10−3 g=2.9 mg2.9\times 10^{-3}\,\text{g} = 2.9\,\text{mg}2.9×10−3g=2.9mg

  1. Nearest integer

x≈3 mgx \approx 3\,\text{mg}x≈3mg

Final Answer: 3\boxed{3}3​

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