The solubility product of Cr(OH) at 298 K is 6×10 . The concentration of hydroxide ions in a saturated solution of Cr(OH) will be: 3 −31 3 −31 1/4 −31 1/2 −31 1/4 −29 1/4 sp 3 −31 4 − −31 1/4 (A) (18

Question

The solubility product of Cr(OH) at 298 K is 6×10 . The concentration of hydroxide ions in a saturated solution of Cr(OH) will be: 3 −31 3 −31 1/4 −31 1/2 −31 1/4 −29 1/4 sp 3 −31 4 − −31 1/4

(A) (18 × 10 )
(B) (18 × 10 )
(C) (2.22 × 10 )
(D) (4.86 × 10 )

Prashant Kotian · Accepted Answer

Correct answer: A) (18 × 10 ). K = s.(3s ) 6 × 10 = 27.s s = [OH ] = 3s = 3 × = (18 × 10 ) M Hence, answer is (a)

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