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Using Ksp

Using Ksp. Chem 12A Mrs. Kay. Solubility equilibrium is any chemical equilibrium between solid and dissolved states of a compound at saturation . The equilibrium expresses the degree to which the solid is soluble in water Called solubility product or Ksp.

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Using Ksp

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  1. Using Ksp Chem 12A Mrs. Kay

  2. Solubility equilibrium is any chemical equilibrium between solid and dissolved states of a compound at saturation. • The equilibrium expresses the degree to which the solid is soluble in water • Called solubility product or Ksp

  3. AgCl (s) <===> Ag+(aq) + Cl¯ (aq) • Writing the equilibrium expression for Ksp is similar to Kc. • You only include (aq), not (s). Anything (s) or (l) is denoted as worth 1. • Ksp=[Ag+ ][Cl¯] • The concentrations of the two ions are EQUAL because of 1:1 ratio, so if we know its Ksp value we can solve for the concentration of ions easily.

  4. You treat the coefficients still as exponents Sn(OH)2 (s) <===> Sn2+ (aq) + 2 OH¯ (aq) Ksp = [Sn2+] [OH¯]2

  5. Write the chemical equation showing how the substance dissociates and write the Ksp expression 1) Na3PO4 2) BaSO43) CdS4) Cu3(PO4)25) CuSCN

  6. Calculations The coefficient does two things: 1) it puts a power on the 'x' which represents that particular ion and2) it puts a coefficient in front of the 'x' Example Problem: Calculate the molar solubility (in mol/L) of a saturated solution of Sn(OH)2 knowing its Ksp is 5.45 x 10¯27

  7. Solve • Sn(OH)2 <:===> Sn2+ + 2 OH¯ • Ksp = [Sn2+] [OH¯]2 • One Sn2+ makes two OH¯. That means that if 'x' Sn2+ dissolves, then '2x' of the OH¯ had to have dissolved, so 5.45 x 10¯27 = (x) (2x)2 4x3 = 5.45 x 10¯27 x=1.11 x 10¯9 mol/L

  8. Solve for Fe(OH)3 knowing its Ksp is 2.64 x 10¯39 • Iron(III) hydroxide • Fe(OH)3 <===> Fe3+ + 3OH¯ • Ksp = [Fe3+] [OH¯]3 • 2.64 x 10¯39 = (x) (3x)3 • 27 x4 = 2.64 x 10¯39 • x = 9.94 x 10¯11 M

  9. Solubility and Ksp • Solubility of a substance is the quantity that dissolves to form a saturated solution • It can be measured in g/L or mol/L • Ksp is the constant which describes the equilibrium between the ionic solid and its ions in a saturated solution.

  10. Flowchart Solubility of compound (g/L) Molar solubility of compound (mol/L) Molar conc. of ions (mol/L) Ksp

  11. Precipitation and Separation of Ions • When two solutions are mixed to form a precipitate, we can use Ksp to see if the precipitate will actually form, or if it is slightly soluble enough in the particular scenario to dissolve. • If Q = Ksp, equilibrium exists • If Q > Ksp, precipitate will occur • If Q < Ksp, solid will dissolve

  12. Practice: • Will a precipitate form if equal volumes of 3.0 x 10-3 M Pb(NO3)2 is added 5.0 x10-3 M Na2SO4 ? Figure out the double displacement reaction, then decide which is the insoluble/slightly soluble compound made. Determine the solubility product for that insoluble compound (Q) Compare it with the accepted value of Ksp

  13. Answer to #1 Pb(NO3)2 + Na2SO4  2NaNO3 + PbSO4 The insoluble compound is PbSO4, so PbSO4 Pb+2 + SO4-2 Q= [Pb+2][SO4-2] Q= [3.0 x 10-3 ][5.0 x10-3 ] Q= 1.5 x 10-5 so, Q>Ksp, so precipitate Ksp = 1.8 x 10-8 occurs Look it up on chart!

  14. Practice #2 • Will a precipitate form if 0.10 L of 3.0 x 10-3 M Pb(NO3)2 is added to 0.40L 5.0 x10-3 M Na2SO4? PbSO4 Pb+2 + SO4-2 • Total volume of solution is 0.10 + 0.40 = 0.50 L • Find concentrations involved:

  15. 0.10 L x 3.0 x 10-3 mol/L = 3.0 x 10-4 mol • 3.0 x 10-4 mol/ 0.50 L = 6.0 x 10-4 M • 0.40 L x 5.0 x10-3 mol/L = 2.0 x 10-3 mol • 2.0 x 10-3 mol/ 0.50 L = 4.0 x 10-3 M • Q= [6.0 x 10-4 ][4.0 x 10-3 ]=2.4 x 10-6 • Q > Ksp so precipitate will form

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