Silver chromate (Ag2CrO4) is a hazardous compound which may cause irritations in respiratory tracts and injuries in liver.

Question

Silver chromate (Ag₂CrO₄) is a hazardous compound which may cause irritations in respiratory tracts and injuries in liver. It will dissolve in water following the chemical equilibrium below.

Ag₂CrO₄(s) ⇌ 2Ag²(aq) + CrO₄^2-(aq)

It is known that the solubility product (K_sp) of silver chromate is 9.0 × 10^-12. at room temperature with appropriate units.

(a) Write the equilibrium expression for the equilibrium above and state the unit of the equilibrium constant in terms of mol dm^-3 .

(b) If 2.90g of silver chromate (Ag₂CrO₄) is added into pure water for dissolution, please evaluate the degree of ionization of silver chromate (Ag₂CrO₄) if the volume of the resulting solution is 100mL.

Answer 1

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄^2-(aq)

(a) Equilibrium expression: Keq = [Ag⁺]²[CrO₄^2-]

units are mol dm^-3 or M

(b) molar mass Ag2CrO4 = 331.7 g/mol

moles present = 2.90 g x 1 mol / 331.7 g = 0.00874 mols

Volume = 100 ml = 0.1 L

molar concentration = 0.00874 mols / 0.1 L = 0.0874 M

K = 9.0x10^-12 = [Ag⁺]²[CrO₄^2-]

Let [Cr₂O₄^2-] = x and then [Ag⁺] = 2x

9.0x10^-12 = (2x)²(x) = 4x³

x³ = 2.25x10^-12

x = [CrO₄^2-] = 1.31x10^-4 M

% ionization = 1.31x10^-4 M / 0.0874 M (x100%) = 0.15% ionized