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 :

Question

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

Tardigrade Admin · Accepted Answer

Ksp=27(s)4=6×10-31 ⇒ [3(s)]4=18×10-31 [OH-]=3(s)=[18×10-31]1 /4

Q. The solubility product of $C r (O H)_{3}$ at $298 K$ is $6.0 \times 1 0^{- 31} .$ The concentration of hydroxide ions in a saturated solution of $C r (O H)_{3}$ will be :

$(18 \times 1 0^{- 31})^{1/2}$

$(2.22 \times 1 0^{- 31})^{1/4}$

$(18 \times 1 0^{- 31})^{1/4}$

$(4.86 \times 1 0^{- 29})^{1/4}$

$K_{s p} = 27 (s)^{4} = 6 \times 1 0^{- 31}$
$\Rightarrow [3 (s)]^{4} = 18 \times 1 0^{- 31}$
$[O H -] = 3 (s) = [18 \times 1 0^{- 31}]^{1/4}$

Q. The solubility product of Cr(OH)3​ at 298K is 6.0×10−31. The concentration of hydroxide ions in a saturated solution of Cr(OH)3​ will be :

(18×10−31)1/2

(2.22×10−31)1/4

(18×10−31)1/4

(4.86×10−29)1/4