Given that, the solubility product Ksp of AgCl is 1.8 × 10-10, the concentration of Cl- ions that must be exceeded before AgCl will precipitate from a solution containing 4 × 10-3 M Ag+ ions is

Question

Given that, the solubility product Ksp of AgCl is 1.8 × 10-10, the concentration of Cl- ions that must be exceeded before AgCl will precipitate from a solution containing 4 × 10-3 M Ag+ ions is (A) 4.5 × 10-8 M (B) 4 × 10-8 M (C) 1.8 × 10-8 M (D) 1 × 10-8 M

Tardigrade Admin · Accepted Answer

Ksp=[Ag+][Cl-] [Cl-]=(1.8×10-10/4×10-3) =4.5×10-8 M If concentration of Cl- exceeds 4.5×10-8 M then ionic product becomes greater than Ksp and precipitation takes place.

Q. Given that, the solubility product $K_{s p}$ of $A g Cl$ is $1.8 \times 1 0^{- 10}$ , the concentration of $C l^{-}$ ions that must be exceeded before $A g Cl$ will precipitate from a solution containing $4 \times 1 0^{- 3} M A g^{+}$ ions is

$4.5 \times 1 0^{- 8} M$

$4 \times 1 0^{- 8} M$

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

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

$K_{s p} = [A g^{+}] [C l^{-}]$
$[C l^{-}] = \frac{1.8 \times 1 0 ^{- 10}}{4 \times 1 0 ^{- 3}}$
$= 4.5 \times 1 0^{- 8} M$
If concentration of $C l^{-}$ exceeds $4.5 \times 1 0^{- 8} M$ then ionic product becomes greater than $K_{s p}$ and precipitation takes place.

Q. Given that, the solubility product Ksp​ of AgCl is 1.8×10−10, the concentration of Cl− ions that must be exceeded before AgCl will precipitate from a solution containing 4×10−3MAg+ ions is

4.5×10−8M

4×10−8M

1.8×10−8M

1×10−8M