At 25°C, if the concentration of Ag+ ion is 1.5 × 10-4 mol/L in the saturated solution of Ag2CrO4, then solubility product of Ag2CrO4 is

Question

At 25°C, if the concentration of Ag+ ion is 1.5 × 10-4 mol/L in the saturated solution of Ag2CrO4, then solubility product of Ag2CrO4 is (A) 6.25 × 10-18 (B) 5.1 × 10-16 (C) 1.69 × 10-12 (D) 7.2 × 10-14

Tardigrade Admin · Accepted Answer

AgCro4(s) leftharpoons2Ag+(aq)+CrO42-(aq) In the saturated solution of Ag2CrO4 [Ag+]=1.5×10-4 mol/L [CrO42-]=(1.5×10-4/2)=0.75×10-4 mol/L Ksp=[Ag+][CrO42-]=(1.5×10-4)2×0.75×10-4 =1.69×10-12 mol3L3

Q. At $2 5^{\circ} C$ , if the concentration of $A g^{+}$ ion is $1.5 \times 1 0^{- 4}$ mol/L in the saturated solution of $A g_{2} C r O_{4}$ , then solubility product of $A g_{2} C r O_{4}$ is

$6.25 \times 1 0^{- 18}$

$5.1 \times 1 0^{- 16}$

$1.69 \times 1 0^{- 12}$

$7.2 \times 1 0^{- 14}$

Q. At 25∘C, if the concentration of Ag+ ion is 1.5×10−4 mol/L in the saturated solution of Ag2​CrO4​, then solubility product of Ag2​CrO4​ is

6.25×10−18

5.1×10−16

1.69×10−12

7.2×10−14

AgCro4​(s)⇌2Ag+(aq)+CrO42−​(aq) In the saturated solution of Ag2​CrO4​ [Ag+]=1.5×10−4mol/L [CrO42−​]=21.5×10−4​=0.75×10−4 mol/L Ksp​=[Ag+][CrO42−​]=(1.5×10−4)2×0.75×10−4 =1.69×10−12mol3L3

Q. At $2 5^{\circ} C$ , if the concentration of $A g^{+}$ ion is $1.5 \times 1 0^{- 4}$ mol/L in the saturated solution of $A g_{2} C r O_{4}$ , then solubility product of $A g_{2} C r O_{4}$ is

$6.25 \times 1 0^{- 18}$

$5.1 \times 1 0^{- 16}$

$1.69 \times 1 0^{- 12}$

$7.2 \times 1 0^{- 14}$