JEE PYQ: Ionic Equilibrium Question 8
Question 8 - 2021 (25 Feb Shift 2)
Type: MCQ
The solubility of $\text{Ca(OH)}_2$ in water is:
[Given: The solubility product of $\text{Ca(OH)}_2$ in water $= 5.5 \times 10^{-6}$]
(a) $1.11 \times 10^{-6}$
(b) $1.77 \times 10^{-6}$
(c) $1.77 \times 10^{-2}$
(d) $1.11 \times 10^{-2}$
Show Answer
Answer: (c)
Solution
$\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-$
$s(2s + 10^{-7})^2 = 5.5 \times 10^{-7}$ (but $10^{-7}$ is negligible)
$4s^3 = 5.5 \times 10^{-6}$
$s^3 = \frac{5500}{4} \times 10^{-9}$
$s = (1375)^{1/3} \times 10^{-3} \approx 1.11 \times 10^{-2}$
BETA
