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:
Ksp of Ag2CrO4 =1.1 × 10–12
Ksp 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+ + CrO42-
As we know that,
Ksp = [A+] [B-]
Where A and B are the ions dissolved
In the above reaction:
⇒[A+] = [Ag+]
⇒[B-] = [CrO42-]
∴ Ksp = [Ag+]2 + [CrO42-]
As Ksp of Ag2CrO4 =1.1 × 10–12
Let ‘s’ be the solubility of Ag2CrO4
[Ag+] =2s
[CrO42-] =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,
Ksp = [A+] [B-]
Where A and B are the ions dissolved
In the above reaction,
⇒[A+] = [Ag+]
⇒[B-] = [Br-]
∴ Ksp = [Ag+] + [Br-]
As Ksp 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