An excess of Ag2CrO4(s) is added to a 5 × 103 M K2CrO4 solution. The concentration of Ag+ in the solution is closest to [ Solubility product for Ag2CrO4 = 1.1 × 10-12]

Question

An excess of Ag2CrO4(s) is added to a 5 × 103 M K2CrO4 solution. The concentration of Ag+ in the solution is closest to [ Solubility product for Ag2CrO4 = 1.1 × 10-12] (A) 2.2 × 10-10 M (B) 1.5 × 10-5 M (C) 10 × 10-6 M (D) 5.0 × 10-3 M

Tardigrade Admin · Accepted Answer

Given, solubility product of Ag2CrO4 =11× 10-12 Concentration of CrO2-4 ions, [CrO4]2- = 5 × 10-3 M For reaction, Ag 2 CrO 4 leftharpoons 2 Ag ++ CrO 42- Ksp =[Ag2 +]2 [CrO2-4] 11× 10-12=[Ag2 +]2 [5×10-13] ∴ [Ag+]=1.48×10-5 M ≈ 1.5×10-5 M

Q. An excess of $A g_{2} C r O_{4} (s)$ is added to a $5 \times 1 0^{3} M K_{2} C r O_{4}$ solution. The concentration of $A g^{+}$ in the solution is closest to [ Solubility product for $A g_{2} C r O_{4} = 1.1 \times 1 0^{- 12}]$

$2.2 \times 1 0^{- 10} M$

$1.5 \times 1 0^{- 5} M$

$10 \times 1 0^{- 6} M$

$5.0 \times 1 0^{- 3} M$

Q. An excess of Ag2​CrO4​(s) is added to a 5×103MK2​CrO4​ solution. The concentration of Ag+ in the solution is closest to [ Solubility product for Ag2​CrO4​=1.1×10−12]

2.2×10−10M

1.5×10−5M

10×10−6M

5.0×10−3M

Given, solubility product of Ag2​CrO4​=11×10−12 Concentration of CrO42−​ ions, [CrO4​]2− =5×10−3M For reaction, Ag2​CrO4​⇌2Ag++CrO42−​ Ksp​=[Ag2+]2[CrO42−​] 11×10−12=[Ag2+]2[5×10−13] ∴[Ag+]=1.48×10−5M≈1.5×10−5M

Q. An excess of $A g_{2} C r O_{4} (s)$ is added to a $5 \times 1 0^{3} M K_{2} C r O_{4}$ solution. The concentration of $A g^{+}$ in the solution is closest to [ Solubility product for $A g_{2} C r O_{4} = 1.1 \times 1 0^{- 12}]$

$2.2 \times 1 0^{- 10} M$

$1.5 \times 1 0^{- 5} M$

$10 \times 1 0^{- 6} M$

$5.0 \times 1 0^{- 3} M$