At 20° C the Ag +ion concentration in a saturated solution of Ag 2 CrO 4 is 1.5 × 10-4 mol / L. At 20° C the solubility product of Ag 2 CrO 4 would be

Question

At 20° C the Ag +ion concentration in a saturated solution of Ag 2 CrO 4 is 1.5 × 10-4 mol / L. At 20° C the solubility product of Ag 2 CrO 4 would be (A)  3.3750× 10-12  (B)  1.6875× 10-10  (C)  1.6875× 10-12  (D)  1.6875× 10-11

Tardigrade Admin · Accepted Answer

Given, 2[ Ag +]=1.5 × 10-4 mol / L [ Ag +]=0.75 × 10-4 [ Ag +]=[ CrO 42-]=0.75 × 10-4 mol / L Ag 2 CrO 4 leftharpoons 2 Ag ++ CrO 42- K s p=[ Ag +]2 ⋅[ CrO 42-] K s p=(1.5 × 10-4)2 ×(0.75 × 10-4) K s p=2.25 × 10-8 × 0.75 × 10-4 K s p=1.6875 × 10-12

Q. At $2 0^{\circ} C$ the $A g^{+}$ ion concentration in a saturated solution of $A g_{2} C r O_{4}$ is $1.5 \times 1 0^{- 4} m o l / L$ . At $2 0^{\circ} C$ the solubility product of $A g_{2} C r O_{4}$ would be

$3.3750 \times 1 0^{- 12}$

$1.6875 \times 1 0^{- 10}$

$1.6875 \times 1 0^{- 12}$

$1.6875 \times 1 0^{- 11}$

Q. At 20∘C the Ag+ion concentration in a saturated solution of Ag2​CrO4​ is 1.5×10−4mol/L. At 20∘C the solubility product of Ag2​CrO4​ would be

3.3750×10−12

1.6875×10−10

1.6875×10−12

1.6875×10−11

Given, 2[Ag+]=1.5×10−4mol/L [Ag+]=0.75×10−4 [Ag+]=[CrO42−​]=0.75×10−4mol/L Ag2​CrO4​⇌2Ag++CrO42−​ Ksp​=[Ag+]2⋅[CrO42−​] Ksp​=(1.5×10−4)2×(0.75×10−4) Ksp​=2.25×10−8×0.75×10−4 Ksp​=1.6875×10−12

Q. At $2 0^{\circ} C$ the $A g^{+}$ ion concentration in a saturated solution of $A g_{2} C r O_{4}$ is $1.5 \times 1 0^{- 4} m o l / L$ . At $2 0^{\circ} C$ the solubility product of $A g_{2} C r O_{4}$ would be

$3.3750 \times 1 0^{- 12}$

$1.6875 \times 1 0^{- 10}$

$1.6875 \times 1 0^{- 12}$

$1.6875 \times 1 0^{- 11}$