The solubility of Sb2S3 , in water is 1.0× 10-5 mol/L at 298 K . What will be its solubility product?

Question

The solubility of Sb2S3 , in water is 1.0× 10-5 mol/L at 298 K . What will be its solubility product? (A)  108× 10-25  (B)  1.0× 10-25  (C)  114× 10-25  (D)  126× 10-24

Tardigrade Admin · Accepted Answer

undersets mol/LSb2S3 <=> underset2s2Sb3+ + underset3s3S2- Solubility product (Ksp) = [Sb3+][S2-]3 =(2 s)2(3 s)3=108 s5 =108 ×(1.0 × 10-5)5=108 × 10-25

Q. The solubility of $S b_{2} S_{3}$ , in water is $1.0 \times 1 0^{- 5}$ $m o l / L$ at $298 K$ . What will be its solubility product?

$108 \times 1 0^{- 25}$

$1.0 \times 1 0^{- 25}$

$114 \times 1 0^{- 25}$

$126 \times 1 0^{- 24}$

$s m o l / L S b_{2} S_{3} 2 s 2 S b^{3 +} + 3 s 3 S^{2 -}$
Solubility product $(K_{s p}) = [S b^{3 +}] [S^{2 -}]^{3}$
$= (2 s)^{2} (3 s)^{3} = 108 s^{5}$
$= 108 \times (1.0 \times 1 0^{- 5})^{5} = 108 \times 1 0^{- 25}$

Q. The solubility of Sb2​S3​ , in water is 1.0×10−5 mol/L at 298K . What will be its solubility product?

108×10−25

1.0×10−25

114×10−25

126×10−24