Question

What is the molar solubility of CdCO3 in a 0.055 M solution of NH3. Ksp(CdCO3) = 5.2 x 10−12, Kf([Cd(NH3)4]2+) = 1.3 x 107 What is the molar solubility of CdCO3 in a 0.055 M solution of NH3. Ksp(CdCO3) = 5.2 x 10−12, Kf([Cd(NH3)4]2+) = 1.3 x 107 1.9 x 10−3 6.2 x 10−10 2.5 x 10−5 2.7 x 10−4 6.7 x 10−5

Answer 1

Answer:

The molar solubility of CdCO3 is 2.5 *10^-5 M

Explanation:

Step 1: Data given

Molarity of NH3 = 0.055 M

Ksp(CdCO3) = 5.2 * 10^−12

Kf([Cd(NH3)4]2+) = 1.3 * 10^7

Step 2: The balanced equation

CdCO3 + 4NH3  → Cd(NH3)4 + CO3^2-

Step 3: Define the concentrations

Kf =  [Cd(NH3)4]^+2)/  [CdCO3] [NH3]^4

K = Ksp * Kf

K = 5.2 * 10^-12 * 1.3*10^7 = 0.0000676

The initial concentration of NH3 =0.055 M

The initial concentration of Cd(NH3)4^2+ = 0 M

The initial concentration of CO3^2- is 0M

For 4 moles NH3 consumed, we produce 1 mole Cd(NH3)4^2+ and 1 mole CO3^2-

This means there will be consumed 4X of NH3

There will be produced X of Cd(NH3)4^2+ and X of CO3^2-

The molarity of NH3 at the equilibrium will be: (0.055 - 4X)M

The molarity of Cd(NH3)4^2+ and  CO3^2- at the equilibrium will be X M

K = 0.0000676 = [Cd(NH3)4^2+][CO3^2-] / [NH3]^4

K = 0.0000676 = X² /(0.055 - 4x)^4    

X² = 0.0000676* ((0.055 - 4x)^4

X = 2.48 * 10^-5 M

The molar solubility of CdCO3 is 2.5 *10^-5 M