Question

the solubility product constant of Ag2Cro4 and Agbr are 1.1×10^-12 and 5×10^-13 respectively calculate the ratio of the molarities of their saturated solutions ​

Answer 1

Given info : the solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10¯¹² and 5 × 10¯¹³ respectively.

To find : the ratio of the molarities of their saturated solution is..

solution : Ag2CrO4 => 2Ag⁺ + CrO₄¯

2s s

so solubility product = [Ag⁺]²[CrO₄¯]

⇒1.1 × 10¯¹² = (2s)²(s)

⇒1.1 × 10¯¹² = 4s³

⇒s³ = 0.275 × 10¯¹²

⇒s = 6.53 × 10¯⁵ M

similarly, AgBr => Ag⁺ + Br¯

s' s'

so solubility product = [Ag⁺][Br¯]

⇒5 × 10¯¹³ = (s')(s') = s²

⇒s'² = 0.5 × 10¯¹²

⇒s' = 0.707 × 10¯⁶ M = 7.07 × 10¯⁵ M

Therefore the ratio of their molarities of their saturated solutions = s/s' = 6.53 × 10¯⁵ M/7.07 × 10¯⁵ M = 92.3