The solubility product (Ksp: mol3 dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)

Question

The solubility product (Ksp: mol3 dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303 × R × 298/F = 0.059 V) (A)  1 × 10-15  (B)  4 × 10-15  (C)  1 × 10-12  (D)  4 × 10-12

Tardigrade Admin · Accepted Answer

The solubility equilibrium for MX2 is MX2 (s) leftharpoons M2+ (aq) + 2X- (aq) Solubility product, Ksp=][M2+][X-]2 = 10-5 × (2 × 10-5)2 =4 × 10-15 [∵ In saturated solution of MX2,[X-]=2[M2+]]

Q. The solubility product $(K_{s p} : m o l^{3} d m^{- 9})$ of $M X_{2}$ at $298 K$
based on the information available the given concentration cell is (take $2.303 \times R \times 298/ F = 0.059 V$ )

$1 \times 1 0^{- 15}$

$4 \times 1 0^{- 15}$

$1 \times 1 0^{- 12}$

$4 \times 1 0^{- 12}$

The solubility equilibrium for $M X_{2}$ is
$M X_{2} (s) ⇌ M^{2 +} (a q) + 2 X^{-} (a q)$
Solubility product, $K_{s p} =] [M^{2 +}] [X^{-}]^{2}$
$= 1 0^{- 5} \times (2 \times 1 0^{- 5})^{2} = 4 \times 1 0^{- 15}$
$[∵$ In saturated solution of $M X_{2}, [X^{-}] = 2 [M^{2 +}]]$

Q. The solubility product (Ksp​:mol3dm−9) of MX2​ at 298K based on the information available the given concentration cell is (take 2.303×R×298/F=0.059V)

1×10−15

4×10−15

1×10−12

4×10−12