Using Ksp

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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

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AgCl (s) ltgt 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.
• KspAg Cl
• The concentrations of the two ions are EQUAL
  because of 11 ratio, so if we know its Ksp value
  we can solve for the concentration of ions easily.

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You treat the coefficients still as exponents

• Sn(OH)2 (s) ltgt Sn2 (aq) 2 OH (aq)
• Ksp Sn2 OH2

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Write the chemical equation showing how the
substance dissociates and write the Ksp expression

• 1) Na3PO4
• 2) BaSO43) CdS4) Cu3(PO4)25) CuSCN

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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 1027

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Solve

• Sn(OH)2 ltgt Sn2 2 OH
• Ksp Sn2 OH2
• 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 1027 (x) (2x)2
• 4x3 5.45 x 1027
• x1.11 x 109 mol/L

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Solve for Fe(OH)3 knowing its Ksp is 2.64 x 1039

• Iron(III) hydroxide
• Fe(OH)3 ltgt Fe3 3OH
• Ksp Fe3 OH3
• 2.64 x 1039 (x) (3x)3
• 27 x4 2.64 x 1039
• x 9.94 x 1011 M

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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.

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Flowchart
Solubility of compound (g/L)
Molar solubility of compound (mol/L)
Molar conc. of ions (mol/L)
Ksp
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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 gt Ksp, precipitate will occur
• If Q lt Ksp, solid will dissolve

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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

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Answer to 1

• Pb(NO3)2 Na2SO4 ? 2NaNO3 PbSO4
• The insoluble compound is PbSO4, so
• PbSO4 Pb2 SO4-2
• Q Pb2SO4-2
• Q 3.0 x 10-3 5.0 x10-3
• Q 1.5 x 10-5 so, QgtKsp, so precipitate
• Ksp 1.8 x 10-8 occurs

Look it up on chart!
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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 Pb2 SO4-2
• Total volume of solution is 0.10 0.40 0.50 L
• Find concentrations involved

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• 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 gt Ksp so precipitate will form