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

a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b.  Ksp AgBr = s²

c. 5 x 10⁻¹³  mol/L

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

im doing test rn im going with 2 but ill give answer if itss wrong,   : )