Question

Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: ×7.210−4gL solubility in 0.0120 M CoBr2 solution: ×5.210−7gL

Answer 1

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Step-by-step explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

Solubility in 0.0120 M CoBr₂ (S')

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M