The Ksp of Q2 is slightly soluble salt containing cation of M2+ and anion of X2−, where Ksp = 4.5 × 10−9. The solubility expression of M2+ is given as [M2+] = s and [X2−] = 6s, where s is its solubility. What is the molar solubility of Q2 in a solution of NaX properly induced, 2S, then Na2X, respectively?

Answer

To solve for the molar solubility (s) of Q2, we first need to establish the equilibrium expression based on the dissolution of the salt: Q2(s) ⇌ M2+(aq) + 2X2-(aq) From this reaction, we can express the Ksp as follows: Ksp = [M2+][X2−]^2. Now substituting in the expressions from the problem: - Let the molar solubility of Q2 be s. - Thus, [M2+] will be s and [X2−] will be 2s (since for every mole of M2+, two moles of X2− are produced). Therefore, we write the equation for Ksp: Ksp = s(2s)^2 = s(4s^2) = 4s^3. We know that Ksp = 4.5 × 10−9, so we can now substitute this into the equation: 4.5 × 10−9 = 4s^3. To find s, we first solve for s^3: s^3 = (4.5 × 10−9) / 4 s^3 = 1.125 × 10−9. Now take the cube root of both sides: s = (1.125 × 10−9)^(1/3). Calculating the cube root gives: s ≈ 1.03 × 10−3 mol/L. Now, regarding the effect of the NaX solution: When NaX is dissolved in the solution, it adds X2- ions, which affects the solubility equilibrium of Q2 due to Le Chatelier's principle (increased concentration of products shifts the equilibrium left). This decreases the solubility of Q2. In a solution of Na2X, the concentration of X2- will be double that of the previous case (let's say it's now 2s or 2 × (1.03 × 10−3)), pushing the equilibrium further to the left, resulting in a lower molar solubility of Q2. In conclusion: Molar solubility of Q2 at equilibrium without additional salts = 1.03 × 10−3 mol/L. In the presence of NaX and Na2X, it will be lesser due to increased X2- ions from those soluble salts.