Question

Consider the insoluble salts \(\mathrm{JQ}, \mathrm{K}_{2} \mathrm{R}, \mathrm{L}_{2} \mathrm{~S}_{3}, \mathrm{MT}_{2},\) and \(\mathrm{NU}_{3} .\) They are formed from the metal ions \(\mathrm{J}^{+}, \mathrm{K}^{+}, \mathrm{L}^{3+}, \mathrm{M}^{2+},\) and \(\mathrm{N}^{3+}\) and the nonmetal ions \(\mathrm{Q}^{-}, \mathrm{R}^{2-}, \mathrm{S}^{2-}, \mathrm{T}^{-},\) and \(\mathrm{U}^{-}\). All the salts have the same \(K_{\mathrm{sp}}, 1 \times 10^{-10},\) at \(25^{\circ} \mathrm{C}\). (a) Which salt has the highest molar solubility? (b) Does the salt with the highest molar solubility have the highest solubility in \(\mathrm{g}\) salt \(100 \mathrm{~g}\) water? (c) Can the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water be determined from the information given? If yes, calculate the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water. If no, why not?

Short answer

Answer: L₂S₃ has the highest molar solubility (1.8x10^-3 M).

Step-by-step solution

(a) Determining the molar solubility for each salt

First, we will write the dissolution equation for each salt: JQ: J^+ + Q^- → JQ K2R: 2K^+ + R^2- → K₂R L2S3: 2L^3+ + 3S^2- → L₂S₃ MT₂: M^2+ + 2T^- → MT₂ NU₃: N^3+ + 3U^- → NU₃ Now, we will write the Ksp expression for each salt: JQ: Ksp = [J^+][Q^-] K2R: Ksp = [K^+]^2[R^2-] L2S3: Ksp = [L^3+]^2[S^2-]^3 MT₂: Ksp = [M^2+][T^-]^2 NU₃: Ksp = [N^3+][U^-]^3 Since Ksp is the same for all salts (1x10^-10), we can find the molar solubility for each salt: For JQ: let x be the molar solubility, Ksp = x * x = x^2, x = sqrt(Ksp) = sqrt(1x10^-10) = 1x10^-5 M For K₂R: let x be the molar solubility, Ksp = (2x)^2 * x = 4x^3, x = (Ksp/4)^(1/3) = (1x10^-10/4)^(1/3) ≈ 6.3x10^-4 M For L₂S₃: let x be the molar solubility, Ksp = (2x)^2 * (3x)^3 = 108x^5, x = (Ksp/108)^(1/5) ≈ 1.8x10^-3 M For MT₂: let x be the molar solubility, Ksp = x * (2x)^2 = 4x^3, x = (Ksp/4)^(1/3) = (1x10^-10/4)^(1/3) ≈ 6.3x10^-4 M For NU₃: let x be the molar solubility, Ksp = x * (3x)^3 = 27x^4, x = (Ksp/27)^(1/4) ≈ 8.6x10^-4 M Based on our calculations, L₂S₃ has the highest molar solubility (1.8x10^-3 M).

(b) Compare solubility in g/100g water

It is not possible to determine the solubility in grams of salt per 100 grams of water without knowing the molecular weights of the salts. We need the molecular weight of each salt to convert molar solubility into grams per liter, and divide it by the density of water to find grams of salt per 100g water.

(c) Determine if solubility in g/100g water can be calculated

We cannot determine the solubility of each salt in grams per 100 grams of water using the given information because the exercise does not provide the molecular weights of the salts. The molecular weight is required to convert molar solubility into grams per liter and further into grams of salt per 100g of water.