Question

Using the solubility of a compound to calculate Rsp The solubility of Ni(OH), in water at 25°C is measured to be 4.9x10 U se this information to calculate K, for Ni(OH), Round your answer to 2 significant digits. 07.0 x 5 ?

Answer 1

Step 1: Explanation

The Ksp expression for a salt is the product of the concentrations of the ions, with each concentration raised to a power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium.

Step 2: Write the balanced decomposed reaction of Ni(OH)₂ and CaF₂ at equilibrium

(a) Ni(OH)₂(s) ⇌ Mg⁺² + 2 HO^-

(b) CaF₂(s) ⇌ Ca⁺² + 2F^-

Step 3: Calculate the Ksp of compound

(a) And now if we call the solubility of Mg(OH)₂ = S, then by definition, S = [Mg²⁺], and 2S = [⁻OH]

Ni(OH)₂(s) ⇌ Ni⁺² + 2 HO^-

Ksp = [Ni²⁺] × 2 [⁻OH]²

Ksp = S × ( 2S)² = 4S³

thus, Ksp = 4S³

if S = 4.9 × 10^-4 g/L

to get solubility in molar divide  it by molar mass of (Ni(OH)₂ =92.708 g/mol )

S = 4.9 × 10^-4 g/L / 92.708 g/mol =  5.285412262 × 10^-6 mol/L

on subsituting the value

Ksp = 4S³

Ksp = 4 × (  5.285412262 × 10^-6 )³ =  5.9 × 10^-16 mol/L

(b) And now if we call the solubility of CaF₂ = S, then by definition, S = [Ca²⁺], and 2S = [F⁻]

CaF₂ ⇌ Ca⁺² + 2 F^-

Ksp = [Ca²⁺] × 2 [F⁻]²

Ksp = S × ( 2S)² = 4S³

thus, Ksp = 4S³

if S = 0.017 g/L

to get solubility in molar divide  it by molar mass of CaF₂ = 78.07 g/mol )

S = 0.017 g/L / 78.07 g/mol = 2.177532983 × 10^-4 mol/L

on subsituting the value

Ksp = 4S³

Ksp = 4 × ( 2.177532983 × 10^-4 )³ = 4.1 × 10^-11 mol/L