JEE Main · 2019 · Shift-IhardIEQ-067

A mixture of 100 mmol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

A mixture of 100 mmol of Ca(OH)2\mathrm{Ca(OH)_2} and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. What is the mass of calcium sulphate formed and the concentration of OH−\mathrm{OH^-} in the resulting solution, respectively?

(Molar masses: Ca(OH)2=74\mathrm{Ca(OH)_2} = 74, Na2SO4=143\mathrm{Na_2SO_4} = 143, CaSO4=136 g mol−1\mathrm{CaSO_4} = 136\ \mathrm{g\ mol^{-1}} respectively; Ksp(Ca(OH)2)=5.5×10−6K_{sp}(\mathrm{Ca(OH)_2}) = 5.5 \times 10^{-6})

Options
  1. a

    1.9 g, 0.14 mol L−11.9\ \mathrm{g},\ 0.14\ \mathrm{mol\ L^{-1}}

  2. b

    13.6 g, 0.28 mol L−113.6\ \mathrm{g},\ 0.28\ \mathrm{mol\ L^{-1}}

  3. c

    1.9 g, 0.28 mol L−11.9\ \mathrm{g},\ 0.28\ \mathrm{mol\ L^{-1}}

    ✓
  4. d

    13.6 g, 0.14 mol L−113.6\ \mathrm{g},\ 0.14\ \mathrm{mol\ L^{-1}}

Correct Answerc

1.9 g, 0.28 mol L−11.9\ \mathrm{g},\ 0.28\ \mathrm{mol\ L^{-1}}

Detailed Solution

Step 1 — Moles of each: Moles Ca(OH)2\mathrm{Ca(OH)_2} = 100 mmol = 0.1 mol Moles Na2SO4\mathrm{Na_2SO_4} = 2/143=0.0142/143 = 0.014 mol

Step 2 — Reaction: Ca(OH)2+Na2SO4→CaSO4+2NaOH\mathrm{Ca(OH)_2 + Na_2SO_4 \rightarrow CaSO_4 + 2NaOH}

Na2SO4\mathrm{Na_2SO_4} is limiting (0.014 mol)

Moles CaSO4\mathrm{CaSO_4} formed = 0.014 mol Mass CaSO4\mathrm{CaSO_4} = 0.014×136=1.904≈1.90.014 \times 136 = 1.904 \approx 1.9 g ✓

Step 3 — Remaining Ca(OH)2\mathrm{Ca(OH)_2}: Moles remaining = 0.1−0.014=0.0860.1 - 0.014 = 0.086 mol in 100 mL [Ca(OH)2]=0.86 M[\mathrm{Ca(OH)_2}] = 0.86\ \mathrm{M}

But Ksp(Ca(OH)2)=5.5×10−6K_{sp}(\mathrm{Ca(OH)_2}) = 5.5 \times 10^{-6} — Ca(OH)2\mathrm{Ca(OH)_2} is sparingly soluble.

Ksp=[Ca2+][OH−]2=s(2s)2=4s3=5.5×10−6K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2 = s(2s)^2 = 4s^3 = 5.5 \times 10^{-6}

s=0.11 M⇒[OH−]=2s=0.22 Ms = 0.11\ \mathrm{M} \Rightarrow [\mathrm{OH^-}] = 2s = 0.22\ \mathrm{M}

With NaOH from reaction: [NaOH]=2×0.0140.1=0.28 M[\mathrm{NaOH}] = \frac{2 \times 0.014}{0.1} = 0.28\ \mathrm{M}

Answer: Option (3) — 1.9 g, 0.28 mol/L

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