Question

tated liquid media: 4 Lee calculation to show which of these two salts is more soluble in the stated licu MgBr2 with Kp = 3.7x10* dissolved in 0.01 Molar NaBr AICI3 with sp-2.810' dissolved in pure water.

Answer 1

Q4.

(1.) MgBr₂

The dissociation reaction is : MgBr₂ (s) $\rightleftharpoons$ Mg²⁺ (aq) + 2 Br^- (aq)

Let molar solubility of MgBr₂ = s

[Mg²⁺] = [MgBr₂] = s

[Br^-] = 2 * [MgBr₂] = 2s

Total Br^- concentration = (Br^- from MgBr₂) + (Br^- from NaBr)

Total Br^- concentration = 2s + 0.01 M

K_sp MgBr₂ = [Mg²⁺][Br^-]²

3.7 x 10^-8 = (s) * (2s + 0.01 M)²

Assuming s << 0.01 M

3.7 x 10^-8 = (s) * (0.01 M)²

s = (3.7 x 10^-8) / (0.01 M)²

s = 3.7 x 10^-4 M

molar solubility of MgBr₂ = s = 3.7 x 10^-4 M

(2.) AlCl₃

The dissociation reaction is : AlCl₃ (s) $\rightleftharpoons$ Al³⁺ (aq) + 3 Cl^- (aq)

Let molar solubility of AlCl₃ = s

[Al³⁺] = [AlCl₃] = s

[Cl^-] = 3 * [AlCl₃] = 3s

K_sp AlCl₃ = [Al³⁺][Cl^-]³

2.8 x 10^-16 = (s) * (3s)³

2.8 x 10^-16 = 27s⁴

s = (2.8 x 10^-16 / 27)^1/4

s = 5.7 x 10^-5 M

molar solubility of AlCl₃ = s = 5.7 x 10^-5 M

Since molar solubility of MgBr₂ is greater than molar solubility of AlCl₃, therefore, MgBr₂ is more soluble