Question

Topic 12 Homework C2 Homework . Due on November 11, 2019 Homework. Unanswered Fill in the Blanks Now review the meaning of solubility and its relationship to Ksp and equilibrium as you did in Question 7. Consider these three salts with their Ksp values: BaCO3(s); Ksp = 5.1x10-9 Ag2 CrO4(s); Ksp = 1.1x10-12 Ca3(PO4)2; Ksp = 2.0x10-29 What is the solubility of each of these salts in grams/100 mL of solution? To solve this problem start by determining the solubility in moles per liter of solution. Then think about how you would convert molarity (moles of solute per liter of solution) to grams/liter of solution and then lastly, convert to solubility in grams of solute per 100 mL of solution. Solubility of grams/100 ml. Solubility of grams/100 ml. Solubility of grams/100 mL. 13 Fullscreen

Answer 1

(i) Ksp of BaCO₃ = [Ba²⁺][CO₃^2-]

i.e. 5.1*10^-9 = s*s

i.e. The solubility of BaCO₃(s) = 7.14*10^-5 mol/L = 7.14*10^-5 mol/L * 197 g/mol = 0.0141 g/L = 1.41*10^-3 g/100 mL

(ii) Ksp of Ag₂CrO₄ = [Ag⁺]²[CrO₄^2-]

i.e. 1.1*10^-12 = (2s)²*s

i.e. The solubility of Ag₂CrO₄(s) = 6.5*10^-5 mol/L = 6.5*10^-5 mol/L * 332 g/mol = 0.0216 g/L = 2.16*10^-3 g/100 mL

(iii) Ksp of Ca₃(PO₄)₂ = [Ca²⁺]³[PO₄^3-]²

i.e. 2*10^-29 = (3s)³*(2s)²

i.e. The solubility of Ca₃(PO₄)₂ (s) = 7.14*10^-7 mol/L = 7.14*10^-7 mol/L * 310 g/mol = 2.21*10^-4 g/L = 2.21*10^-5 g/100 mL