Question

Calculate the solubility (in g/l) of silver chromate in water at 25°c if the ksp for ag2cro4 is 1.1 × 10-12. 2.7 × 10-2 g/l 2.2 × 10-2 g/l 3.4 × 10-2 g/l 3.5 × 10-4 g/l

Answer 1

The solubility of silver chromate in water at 25°C is approximately 2.2 × 10^-2 g/L.

Step-by-step explanation:

The solubility of silver chromate (Ag2CrO4) in water at 25°C can be calculated using the solubility product constant (Ksp). The equation for the equilibrium constant is Ksp = [Ag+][CrO4^2-].

Since Ag2CrO4 has a 1:1 stoichiometry, the solubility of the compound is one-half the concentration of silver ions. Therefore, if the solubility of Ag2CrO4 is represented as S, then [Ag+] = 2S and [CrO4^2-] = S.

Substituting these values into the equilibrium constant equation, Ksp = (2S)(S) = 4S^2.

Given that the Ksp value for Ag2CrO4 is 1.1 × 10^-12, we can set up the following equation:

1.1 × 10^-12 = 4S^2

Solving for S, we find that the solubility of silver chromate in water at 25°C is approximately 2.2 × 10^-2 g/L.

Answer 2

Answer is: solubility of silver chromate is 2.2 × 10-2 g/l.

Chemical reaction (dissociation) of silver chromate in water:
Ag₂CrO₄(s) → 2Ag⁺(aq) + CrO₄²⁻(aq).
Ksp(Ag₂CrO₄) = [Ag⁺]²·[CrO₄²⁻].
[CrO₄²⁻] = x.
[Ag⁺] = 2[CrO₄²⁻] = 2x
1,1·10⁻¹² = (2x)² · x = 4x³.

x = ∛1,1·10⁻¹² ÷ 4.

x = 6,5·10⁻⁵ M.

solubility of silver chromate: 6,5·10⁻⁵ mol/L · 331,73 g/mol = 0,022 g/L.

Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.
[Ag⁺] is equilibrium concentration of silver cations.
[CrO₄²⁻] is equilibrium concentration of chromate anions.