Calcualting molar stability of aucl3 0/010 m mgcl2 solution

The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution. And for the AuCl₃ Ksp will be written as: Ksp = [Au³⁺][Cl⁻]³

AuCl₃(aq) ⇄ Au³⁺(aq) + 3Cl⁻(aq)

Let the solubility of the AuCl₃ in 0.013m solution of magnesium chloride is x, of Au³⁺ is x and of Cl⁻ is 3x. But we know that MgCl₂ is a strong electrolyte and it completely dissociates into its ion and will produce 2 moles of chloride ions. For this solution let we consider the volume is 1 litre then the concentration of chloride ions in MgCl₂ is 2(0.010) = 0.020M.

So, in MgCl₂ solution concentration of Cl⁻ becomes = 3x + 0.020.

Value of Ksp for AuCl₃ = 3.2 × 10⁻²⁵

On putting all values on the Ksp equation, we get

Ksp = (x)(3x + 0.020)³

Value of 3x is negligible as compared to the 0.020, so the equation becomes

3.2 × 10⁻²⁵ = (x)(0.020)³

x = 3.2×10⁻²⁵ / (0.020)³

x = 4 ×10⁻²⁰M

Hence the molar solubility of AuCl₃ in 0.010M MgCl₂ is 4 ×10⁻²⁰M.

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