University:
Glendale Community College
Course:
CHM130 | Fundamental Chemistry
Academic year:

2019
Views:

313

Pages:

1
Author:

Ruben Owens

Solubility Key 1. Write the reaction and Ksp expression for barium sulfate. BaSO4(s) ' Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][SO42-] 2. The Ksp for calcium fluoride is 3.9 x 10-11. Calculate its solubility. Write the reaction for full credit. CaF2(s) ' Ca2+(aq) + 2 F-(aq) Ksp = (x)(2x)2 = 4x3 = 3.9 x 10-11 solve x = 2.1363 x 10-4 mol/L Now turn molar solubility into solubility 2.1363 x 10-4 mol/L (78.078 g/mol) = 0.017 g/mol 3. Calculate the molar solubility of calcium fluoride in a solution containing 0.010M calcium nitrate. Write the reaction for full credit. CaF2(s) ' Ca2+(aq) + 2 F-(aq) I 0.010 0 C x 2x E(.010+x) (2x) Kxp = 3.9 x 10-11 = (.010)(2x)2 = 0.04x2 x = 3.1 x 10-5 mol/L 4. Will a precipitate form if 0.10 L of 8.0 x 10-3M lead(II) nitrate is added to 0.40 L of 5.0 x 10-3M sodium sulfate? Write all balanced reactions for full credit. Pb(NO3)2(aq) + Na2SO4(aq) J 2 NaNO3(aq) + PbSO4(s) The precipitate ion molarities are needed to see if they exceed Ksp or not [Pb2+] = .10L (8.0 x 10-3mol/L) = .00080 moles / .50L = .0016M [SO42-] = .40L (5.0 x 10-3mol/L) = .0020 moles / .50L = .0040M PbSO4(s) ' Pb2+(aq) + SO42-(aq) Q = (.0016)(.0040) = 6.4 x 10-6 but Ksp = 1.8 x 10-8 so Q > Ksp so solid ppt forms