Question

Silver bromide (AgBr) has a solubility of 7.07 × 10^-7 mol/L

a. write the dissolution reaction of Abraham, including all states.
b. write the expression for ksp for AgBr
c. calculate the solubility product constant of AgBr​

Answer 1

a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b. Ksp AgBr = s²

c. 5 x 10⁻¹³ mol/L

Further explanation

Given

solubility AgBr = 7.07 x 10⁻⁷ mol/L

Required

The dissolution reaction

Ksp

The solubility product constant

Solution

a. dissolution reaction of AgBr

AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b. Ksp

Ksp AgBr = [Ag⁺] [Br⁻]

Ksp AgBr = (s) (s)

Ksp AgBr = s²

c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L