A mixture of 100 m mol of $\mathrm{Ca}(\mathrm{OH})_{2}$ and 2 g of sodium sulphate was dissolved in water and the volume was made upto 100 mL . The mass of calcium sulphate formed and the concentration of $\mathrm{OH}^{-}$in resulting solution, respectively are (Molar mass of $\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{Na}_{2} \mathrm{SO}_{4}$ and $\mathrm{CaSO}_{4}$ are 74,143 and $136 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively; $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ is $5.5 \times 10^{-6}$ )

(A) $1.9 \mathrm{~g}, 0.28 \mathrm{~mol} \mathrm{~L}^{-1}$

(B) $13.6 \mathrm{~g}, 0.28 \mathrm{~mol} \mathrm{~L}^{-1}$

(C) $1.9 \mathrm{~g}, 0.14 \mathrm{~mol} \mathrm{~L}^{-1}$

(D) $13.6 \mathrm{~g}, 0.14 \mathrm{~mol} \mathrm{~L}^{-1}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Equilibrium

A mixture of 100 m mol of

Ca (OH)_{2}

and 2 g of sodium sulphate was dissolved in water and the volume was made upto 100 mL . The mass of calcium sulphate formed and the concentration of

OH^{-}

in resulting solution, respectively are
(Molar mass of

Ca (OH)_{2}, Na_{2} SO_{4}

and

CaSO_{4}

are 74,143 and

136 g mol^{- 1}

respectively;

K_{sp}

Ca (OH)_{2}

5.5 \times 1 0^{- 6}

)

Answer: (A)

First count the sodium sulphate:

2

g over a molar mass of

143

is about

0.014

mol, that is

14

mmol, the limiting reagent against the

100

mmol of calcium hydroxide. The exchange

Ca (OH) X_{2} + Na X_{2} SO X_{4} CaSO X_{4} + 2 NaOH

then makes

14

mmol of calcium sulphate, of mass

14 \times 1 0^{- 3} \times 136 \approx 1.9

g, together with

28

mmol of sodium hydroxide. In the

100

mL made up, that hydroxide gives

\frac{28 mmol}{0.1 L} = 0.28

mol L

^{- 1}

, the sparingly soluble leftover calcium hydroxide adding a negligible amount more.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option