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

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.