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

ch3 [ch2]14cooh

explanation:

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.