The solubility product constant of Ag_2CrO_4 and AgBr are 1.1 × 10^–12 and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Name: The solubility product constant of Ag_2CrO_4 and AgBr are 1.1 × 10^–12 and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Uploaded: Thu, 2019-06-20 01:37
Description: The solubility product constant of Ag_2CrO_4 and AgBr are 1.1 × 10^–12 and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

7.68 The solubility product constant of Ag₂CrO₄ and AgBr is 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Answers (1)

silver chromate ( $Ag_2CrO_{4}$ )
Ionization of silver chromate

$Ag_2CrO_{4}\rightleftharpoons 2Ag^++CrO_{4}^{2-}$
Let " $s$ " be the solubility of $Ag_2CrO_{4}$
$[Ag^+] = 2s$
$[CrO_{4}^{2-}] = s$
$K_{sp}$ of $Ag_2CrO_{4}$ = $1.1\times 10^{-12}$