contestada

noltoupe sonblod
Example: Ksp for cadmium carbonate is 1.8 * 10-14, calculate
the equilibrium solubility of cadmium carbonate in a solution at
constant pH of 11?
For H₂CO3
1- Ka = 4.45 * 10-7
2-ka =4.7*10-11

Respuesta :

onyebuchinnaji

The equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is 1.65 x 10⁻⁵ mol/dm³.

Equilibrium solubility of the compound

The equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is calculated as follows;

H₂CO3 ⇄ 2H⁺  + CO₃²⁻

x         :      2x        x

ksp = (2x)²x

ksp = 4x³

(1.8 x 10⁻¹⁴) = 4x³

(1.8 x 10⁻¹⁴) /4 = x³

4.5 x 10⁻¹⁵ = x³

x = (4.5 x 10⁻¹⁵)^¹/₃

x = 1.65 x 10⁻⁵ mol/dm³

Thus, the equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is 1.65 x 10⁻⁵ mol/dm³.

Learn more about equilibrium solubility here: https://brainly.com/question/23946616

#SPJ1

The equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is 1.65 x 10⁻⁵ mol/dm³.

What is solubility?

Solubility is defined as the maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature.

The equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is calculated as follows;

H₂CO3 ⇄ 2H⁺  + CO₃²⁻

x         :      2x        x

ksp = (2x)²x

ksp = 4x³

(1.8 x 10⁻¹⁴) = 4x³

(1.8 x 10⁻¹⁴) /4 = x³

4.5 x 10⁻¹⁵ = x³

x = [tex](4.5 X 10^{-15})^{\frac{1}{3} }[/tex]

x = 1.65 x 10⁻⁵ mol/dm³

Thus, the equilibrium solubility of cadmium carbonate in a solution at constant pH of 11 is 1.65 x 10⁻⁵ mol/dm³.

Learn more about equilibrium solubility here:

brainly.com/question/23946616

#SPJ1

Otras preguntas