JEE PYQ: Ionic Equilibrium Question 20
Question 20 - 2020 (09 Jan Shift 2)
Type: MCQ
The solubility product of $\text{Cr(OH)}_3$ at 298 K is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\text{Cr(OH)}_3$ will be:
(a) $(2.22 \times 10^{-31})^{1/4}$
(b) $(18 \times 10^{-31})^{1/4}$
(c) $(18 \times 10^{-31})^{1/2}$
(d) $(4.86 \times 10^{-29})^{1/4}$
Show Answer
Answer: (a)
Solution
$\text{Cr(OH)}_3 \rightarrow \text{Cr}^{3+} + 3\text{OH}^-$
$K_{sp} = s \cdot (3s)^3 = 27s^4$
$6 \times 10^{-31} = 27s^4$
$s = \left(\frac{6}{27} \times 10^{-31}\right)^{1/4}$
$[\text{OH}^-] = 3s = 3 \times \left(\frac{6}{27} \times 10^{-31}\right)^{1/4} = (18 \times 10^{-31})^{1/4}$
BETA
