JEE MAIN - Chemistry (2020 - 9th January Evening Slot - No. 16)
The solubility product of Cr(OH)3 at 298 K is
6.0 × 10–31. The concentration of hydroxide ions
in a saturated solution of Cr(OH)3 will be :
(2.22 × 10–31)1/4
(4.86 × 10–29)1/4
(18 × 10–31)1/4
(18 × 10–31)1/2
Explanation
| Cr(OH)3 | ⇌ | Cr+3 | + | 3OH- |
|---|---|---|---|---|
| S | 3S |
Ksp = [Cr3+] [OH– ]3
$$ \Rightarrow $$ 6 × 10–31 = S × (3S)3
$$ \Rightarrow $$ 6 × 10–31 = 27 S4
S = $${\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$$
As [OH–] = 3S
= $$3{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$$
= (18 × 10–31)1/4 M
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