JEE PYQ: Ionic Equilibrium Question 28
Question 28 - 2019 (10 Jan Shift 1)
Type: MCQ
A mixture of 100 mmol of $\text{Ca(OH)}_2$ and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of $\text{OH}^-$ in resulting solution, respectively, are:
(Molar mass of $\text{Ca(OH)}_2$, $\text{Na}_2\text{SO}_4$ and $\text{CaSO}4$ are 74, 143 and 136 g $\text{mol}^{-1}$, respectively; $K{sp}$ of $\text{Ca(OH)}_2$ is $5.5 \times 10^{-6}$)
(a) 1.9 g, 0.28 mol $\text{L}^{-1}$
(b) 13.6 g, 0.28 mol $\text{L}^{-1}$
(c) 1.9 g, 0.14 mol $\text{L}^{-1}$
(d) 13.6 g, 0.14 mol $\text{L}^{-1}$
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Answer: 0
Solution
$\text{Na}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{NaOH}$
mmol of $\text{Na}_2\text{SO}_4 = \frac{2 \times 1000}{143} = 13.98$ mmol
mmol of $\text{CaSO}_4$ formed $= 13.98$ mmol
Mass of $\text{CaSO}_4 = 13.98 \times 10^{-3} \times 136 = 1.90$ g
mmol of NaOH formed $= 2 \times 13.98 = 28$ mmol
$\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-$
Value of $s$ will be negligible
$[\text{OH}^-] = \frac{0.028}{0.1} = 0.28$ mol $\text{L}^{-1}$
BETA
