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Anvisha [2.4K]
2 years ago
7

Noltoupe sonblod

Chemistry
2 answers:
vaieri [72.5K]2 years ago
4 0

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

<h3>Equilibrium solubility of the compound</h3>

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: brainly.com/question/23946616

#SPJ1

Degger [83]2 years ago
3 0

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

<h3>What is solubility?</h3>

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 = (4.5 X 10^{-15})^{\frac{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:

brainly.com/question/23946616

#SPJ1

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