The solubility of two slightly soluble salts of m^2+ , ma and mz_2, are the same, 4 x 10^â4 mol/l.
(a) which has the larger numerical value for the solubility product constant?
(b) in a saturated solution of each salt in water, which has the higher concentration of m^2+?
(c) if you added an equal volume of a solution saturated in ma to one saturated in mz_2, what would be the equilibrium concentration of the cation, m^2+?

a) MZ₂

b) They have the same concentration

c) 4x10⁻⁴ mol/L

Explanation:

a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.

The salts solubilization reactions and their Kps values are:

MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²

MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³

Thus, the Kps of MZ₂ has a larger value.

b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.

c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:

8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.

the answer is

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