The Ksp for Cr(OH)3 is 1.6 × 10-30. The molar solubiity of this compound in water is :

Question

The Ksp for Cr(OH)3 is 1.6 × 10-30. The molar solubiity of this compound in water is : (A) 4√1.6×10-30 (B) 4√1.6×10-30/ 27 (C) 1.6×10-30/ 27 (D) 2√1.6×10-30

Tardigrade Admin · Accepted Answer

Cr(OH)3(s) leftharpoons Cr3+(aq.) + 3OH-(aq.) S 3S 27 S4 = ksp S = ((KSP/27))1/ 4 = ((1.6×10-30/27))1/ 4

Q. The $K_{s p}$ for $C r (O H)_{3}$ is $1.6 \times 1 0^{- 30}$ . The molar solubiity of this compound in water is :

$4 1.6 \times 1 0^{- 30}$

$4 1.6 \times 1 0^{- 30} /27$

$1.6 \times 1 0^{- 30} /27$

$2 1.6 \times 1 0^{- 30}$

$C r (O H)_{3} (s) ⇌ C r^{3 +} (a q .) + 3 O H^{-} (a q .)$
$S 3 S < b r / > 27 S^{4} = k_{s p}$
$S = (\frac{K _{SP}}{27})^{1/4} = (\frac{1.6 \times 1 0 ^{- 30}}{27})^{1/4}$

Q. The Ksp​ for Cr(OH)3​ is 1.6×10−30. The molar solubiity of this compound in water is :

41.6×10−30​

41.6×10−30/27​

1.6×10−30/27

21.6×10−30​