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

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