Verified Solution • 2025–26 Session moderate

QJEEM20Cha4e6: JEE Main 2020 Chemistry - Class 11 JEEM20Cha4e6 Solution: The solubility product of Cr(OH) 3 at 298 K is 6.0 × 10 –31 .…

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 :

Multiple Choice Options

  • A (2.22 × 10–31)1/4
  • B (4.86 × 10–29)1/4
  • C (18 × 10–31)1/4
  • D (18 × 10–31)1/2

Step-by-Step Verified Solution

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 = (627×1031)14{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}

As [OH–] = 3S

= 3(627×1031)143{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}

= (18 × 10–31)1/4 M

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