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

Given:

K_sp of Ag₂CrO₄ =1.1 × 10^–12

K_sp of AgBr = 5.0 × 10^–13

To find the ratio of molarities of Ag₂CrO₄ and AgBr saturated solutions, first, we will calculate their solubilities separately and then calculate the ratio. Hence,

Ionization of Ag₂CrO₄:

Ag₂CrO₄ ↔ 2Ag⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction:

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [CrO₄^2-]

∴ K_sp = [Ag⁺]² + [CrO₄^2-]

As K_sp of Ag₂CrO₄ =1.1 × 10^–12

Let ‘s’ be the solubility of Ag₂CrO₄

[Ag⁺] =2s

[CrO₄^2-] =s

∴ 1.1 × 10^–12 = (2s)²(s)

^⇒1.1 × 10^–12 = (4s)³

⇒s =

⇒s = 6.5× 10^-5

Ionization of AgBr:

AgBr↔ Ag⁺ + Br^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [Br^-]

∴ K_sp = [Ag⁺] + [Br^-]

As K_sp of AgBr =5.0 × 10^–13

Let s’ be the solubility of AgBr

∴ 5.0 × 10^–13 = (s)(s)

^⇒1.1 × 10^–12 = (s)²

⇒s =

s’ = 7.07× 10^-7

Now the ratio of their solubilities

⇒ = = 9.91

Thus, the ratio of molarities of their saturated solutions is 9.91