Find solubility product of a saturated solution of Ag2CrO4 in water at 298 K, if the emf of the cell. Ag|Ag+ (saturated Ag2CrO4 solution) || Ag+(0.1M)|Ag is 0.164 V at 298 K.

Question

Find solubility product of a saturated solution of Ag2CrO4 in water at 298 K, if the emf of the cell. Ag|Ag+ (saturated Ag2CrO4 solution) || Ag+(0.1M)|Ag is 0.164 V at 298 K.  (A)  2.287× 10-12mol3L-3  (B)  2.287× 10-13mol3L-3  (C)  2.287× 10-14mol3L-3  (D)  2.287× 10-15mol3L-3

Tardigrade Admin · Accepted Answer

Ecell=(0.059/1) log ([Ag+]RHS/[Ag+] LHS) =0.164=(0.059/1) log (0.1/[ textA textg+] textLHS) textA textg+(LHS)=1.66× 10-4M so, [CrO42-]=(1.66× 10-4/2) Ksp( textA textg text2 textCr textO text4)=[Ag+]2 [ textCrO42-] =(1.66× 10-4)2( (1.66× 10-4/2) ) =2.287× 10-12mol3L-3

Q. Find solubility product of a saturated solution of $A g_{2} C r O_{4}$ in water at 298 K, if the emf of the cell. $A g ∣ A g^{+}$ (saturated $A g_{2} C r O_{4}$ solution) || $A g^{+} (0.1 M) ∣ A g$ is $0.164 V$ at $298 K .$

$2.287 \times 10^{- 12} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 13} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 14} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 15} m o l^{3} L^{- 3}$

Q. Find solubility product of a saturated solution of Ag2​CrO4​ in water at 298 K, if the emf of the cell. Ag∣Ag+ (saturated Ag2​CrO4​ solution) || Ag+(0.1M)∣Ag is 0.164V at 298K.

2.287×10−12mol3L−3

2.287×10−13mol3L−3

2.287×10−14mol3L−3

2.287×10−15mol3L−3

Ecell​=10.059​log[Ag+]LHS[Ag+]RHS​ =0.164=10.059​log[Ag+]LHS0.1​ Ag+(LHS)=1.66×10−4M so, [CrO42−​]=21.66×10−4​ Ksp​(Ag2​CrO4​)=[Ag+]2[CrO42−​] =(1.66×10−4)2(21.66×10−4​) =2.287×10−12mol3L−3

Q. Find solubility product of a saturated solution of $A g_{2} C r O_{4}$ in water at 298 K, if the emf of the cell. $A g ∣ A g^{+}$ (saturated $A g_{2} C r O_{4}$ solution) || $A g^{+} (0.1 M) ∣ A g$ is $0.164 V$ at $298 K .$

$2.287 \times 10^{- 12} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 13} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 14} m o l^{3} L^{- 3}$

$2.287 \times 10^{- 15} m o l^{3} L^{- 3}$