The Ksp of Ag2CrO4 is 1.1 × 10 - 12 at 298 K. The solubility (in mol/L) of Ag2 CrO4 in a 0.1 M AgNO3 solution is

Question

The Ksp of Ag2CrO4 is 1.1 × 10 - 12 at 298 K. The solubility (in mol/L) of Ag2 CrO4 in a 0.1 M AgNO3 solution is (A)  1.1 × 10 - 11  (B)  1.1 × 10 - 10  (C)  1.1 × 10 - 12  (D)  1.1 × 10 - 9

Tardigrade Admin · Accepted Answer

Let solubility of Ag2 CrO4 in presence of 0.1 M AgNO3 = x Ag2CrO4 leftharpoons underset2x2 Ag+ + undersetxCrO4 2 - AgNO3 leftharpoons underset0.1Ag+ + underset0.1 NO3- Total [ Ag+ ] = (2x + 0.1) M = 0.1 M as x < < < 0.1 M [CrO42 ] = x M Thus, [ Ag +]2 [ CrO42- ] = K sp (0 . 1)2 (x) = 1.1 × 10 - 12 ∵ x = 1. 1 × 10 - 10 M

Q. The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 1 0^{- 12}$ at $298 K$ . The solubility (in mol/L) of $A g_{2} C r O_{4}$ in $a 0.1 M A g N O_{3}$ solution is

$1.1 \times 1 0^{- 11}$

$1.1 \times 1 0^{- 10}$

$1.1 \times 1 0^{- 12}$

$1.1 \times 1 0^{- 9}$

Q. The Ksp​ of Ag2​CrO4​ is 1.1×10−12 at 298K. The solubility (in mol/L) of Ag2​CrO4​ in a0.1MAgNO3​ solution is

1.1×10−11

1.1×10−10

1.1×10−12

1.1×10−9

Let solubility of Ag2​CrO4​ in presence of 0.1M AgNO3​=x Ag2​CrO4​⇌2x2Ag+​+xCrO42−​​ AgNO3​⇌0.1Ag+​+0.1NO3−​​ Total [ Ag+]=(2x+0.1)M=0.1M as x<<<0.1M [CrO42​]=xM Thus, [Ag+]2[CrO42−​]=Ksp​ (0.1)2(x)=1.1×10−12 ∵x=1.1×10−10M

Q. The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 1 0^{- 12}$ at $298 K$ . The solubility (in mol/L) of $A g_{2} C r O_{4}$ in $a 0.1 M A g N O_{3}$ solution is

$1.1 \times 1 0^{- 11}$

$1.1 \times 1 0^{- 10}$

$1.1 \times 1 0^{- 12}$

$1.1 \times 1 0^{- 9}$