The solubility product of silver chloride is 1.8 × 10-10 at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol/dm3 is

Question

The solubility product of silver chloride is 1.8 × 10-10 at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol/dm3 is (A) 2.4 × 10-9 (B) 3.6 × 10-8 (C) 0.9 ×10-10 (D) 1.8 × 10-8

Tardigrade Admin · Accepted Answer

Let solubility of AgCl in 0.01 M HCl=x mol L-1 ∴ [Ag+]=x mol L-1 [Cl-]=[HCl]=0.01 M=10-2 M [Cl-] from AgCl can be neglcctcd Ksp=[Ag+] [ Cl-] 1.8 × 10-10=x × 10-2 ⇒ x=1.8×10-8

Q. The solubility product of silver chloride is $1.8 \times 1 0^{- 10}$ at 298 K. The solubility of AgCl in 0.01 M HCl solution in $m o l / d m^{3}$ is

$2.4 \times 1 0^{- 9}$

$3.6 \times 1 0^{- 8}$

$0.9 \times 1 0^{- 10}$

$1.8 \times 1 0^{- 8}$

Let solubility of $A g Cl$ in $0.01 M H Cl = x m o l L^{- 1}$
$∴$ $[A g^{+}] = x m o l L^{- 1}$
$[C l^{-}] = [H Cl] = 0.01 M = 1 0^{- 2} M$
$[C l^{-}]$ from $A g Cl$ can be neglcctcd
$K_{s p} = [A g^{+}] [C l^{-}]$
$1.8 \times 1 0^{- 10} = x \times 1 0^{- 2}$
$\Rightarrow x = 1.8 \times 1 0^{- 8}$

Q. The solubility product of silver chloride is 1.8×10−10 at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol/dm3 is

2.4×10−9

3.6×10−8

0.9×10−10

1.8×10−8

Let solubility of AgCl in 0.01MHCl=xmolL−1 ∴ [Ag+]=xmolL−1 [Cl−]=[HCl]=0.01M=10−2M [Cl−] from AgCl can be neglcctcd Ksp​=[Ag+][Cl−] 1.8×10−10=x×10−2 ⇒x=1.8×10−8

Q. The solubility product of silver chloride is $1.8 \times 1 0^{- 10}$ at 298 K. The solubility of AgCl in 0.01 M HCl solution in $m o l / d m^{3}$ is

$2.4 \times 1 0^{- 9}$

$3.6 \times 1 0^{- 8}$

$0.9 \times 1 0^{- 10}$

$1.8 \times 1 0^{- 8}$