Solubility Equilibria

1
Solubility Equilibria

• Recall the Solubility Rules in Chap. 4 (Table
  4.1)
• a) All NO3 salts are soluble.
• b) All alkali metal and NH4 salts are soluble
• c) Most halide or SO42 salts are soluble.
• Refer to the table for complete rules.
• It is well worthwhile to review the names,
  formulas, and charges of ions in Table 2.5 on
  page 66!
• Now we turn our attention to slightly soluble
  salts, those which exist in equilibrium with
  their ions.

2

• Consider the slightly soluble salt AgCl.
• We write the equilbrium for the dissolving
  process
• AgCl(s) ? Ag(aq) Cl(aq)
• The equilibrium constant for this process is
  called the solubility product constant, Ksp
• Ksp Ag Cl 1.6 x 1010
• Recall that pure solids do not appear in
  mass-action expressions.
• Ksp values are found in tables such as Table 15.4
  or Appendix 5.4 of our text.

3

• Remember that the coefficients from the balanced
  reaction become exponents in the mass-action
  expression.
• Ex Ag2CrO4(s) ? 2 Ag(aq) CrO42(aq)
• Ksp Ag2 CrO42 9.0 x 1012
• Now let us define the solubility, S, of a
  slightly soluble salt.
• S moles of compound dissolved in one liter of a
  saturated solution

4

• Now, how do we relate S to the Ksp ?
• Ex 1 Calculate the molar solubility of AgCl.
• Lets try the equilibrium table (I.C.E.)
  approach
• AgCl(s) ? Ag(aq) Cl(aq)
• Initial 0
  0
• Change S S S
• Equil. S
  S
• Ksp 1.6 x 1010 AgCl S S S2
• Then S 1.3 x 105 mol/L

5

• Ex 2 The solubility of Ag2CrO4 at 25C is 1.3 x
  104 mol/L. Determine the Ksp value.
• Ag2CrO4(s) ? 2 Ag(aq) CrO42(aq)
• Init. 0
  0
• Change S 2S
  S
• Equil. 2S
  S
• Ksp Ag2 CrO42 (2S)2 S 4 S3
• 4 (1.3 x 104 mol/L)3
• 8.8 x 1012

6
The Common-Ion Effect and Solubility

• Ex Calculate the solubility of AgCl in a 0.0010
  M NaCl solution.
• Consider complete reactions first
• NaCl(aq) ? Na(aq) Cl(aq)
• 0 0.0010
  0.0010
• Now the equilibrium
• AgCl(s) ? Ag(aq) Cl(aq)
• Init. 0
  0.0010
• Change S S
  S
• Equil. S
  0.0010S

7

• Ksp 1.6 x 1010 AgCl (S)(0.0010 S)
• (S) (0.0010)
• Remember that S is very small because Ksp is very
  small.
• Then S (1.6 x 1010) / 0.0010 1.6 x 107
  mol/L
• as compared to 1.3 x 105 mol/L in pure water
• Conclusion The presence of a common ion will
  lower the solubility of a slightly soluble salt.
• Now, let us examine how pH may affect solubility.

8

• Ex Calculate the solubility of Pb(OH)2 in a
  solution buffered at pH 12.00.
• The Ksp of Pb(OH)2 is 1.2 x 1015
• The solution is very basic, with the buffering
  providing a high and constant concentration of
  OH.
• pOH 14.00 12.00 2.00
• OH 102.00 0.010 M
• Now, how does this affect the Pb(OH)2 equilibrium?

9

• Pb(OH)2(s) ? Pb2(aq) 2 OH(aq)
• init.
  0 0.010
• change S S
  2S
• equil. S
  0.010 2S
• Ksp 1.2 x 1015 Pb2OH2
• (S)(0.010 2S)2 (S) (0.010)2
• S 1.2 x 1015 / (0.010)2 1.2 x 1011
  mol/L
• In pure water, Ksp (S) (2S)2 4 S3
• S (Ksp / 4)1/3 6.7 x 106 mol/L

10
When does a precipitate form?

• In the case of precipitation, the reaction
  quotient Q is called the ion product.
• Unsaturated solution all of solid solute
  dissolves
• Saturated solution solid is in equilibrium with
  dissolved ions no more solid can dissolve.
• For the general reaction
• MaXb (s) ? a M(aq) b X(aq)
• Ksp Ma Xb and Q has the same form

11

• Recall that if Q lt Ksp, the reaction will
  spontane-ously proceed to the right to reach
  equilibrium,
• i.e., the solution is unsaturated and more
  solute can dissolve.
• If Q Ksp, the reaction is at equilibrium,
• i.e., the solution is saturated and no more
  solute will dissolve.
• If Q gt Ksp, the reaction will spontaneously
  proceed to the left,
• i.e., precipitate will spontaneously form until
  equilibrium is established (Q Ksp).

12

• ex If 100 mL of 0.010 M Ca(NO3)2 is mixed with
  100 mL of 0.050 M NaOH, will a precipitate of
  Ca(OH)2 form?
• The Ksp of Ca(OH)2 1.3 x 106.
• Ca(OH)2 (s) ? Ca2(aq) 2 OH(aq)
• Q Ca2 OH2
• We need to first calculate the (diluted)
  concentrations of Ca2 and OH after mixing.

13

• The dilution equation is M1 V1 M2 V2
• After mixing, Ca2 (0.010 M)(100 mL) / 200 mL
• 0.0050 M
• and OH (0.050 M)(100 mL) / 200
  mL
• 0.025 M
• Then Q (0.0050)(0.025)2 3.1 x 106
• Since Q gt Ksp, solid Ca(OH)2 precipitates.

14
Separation of Ions from a Mixture by Selective
Precipitation (Fractional Precipitation)

• Suppose we have a solution containing Mg2 and
  Ca2, both at a concentration of 0.10 M.
• Is it possible to quantitatively (completely)
  sepa-rate them (precipitate one metal but not the
  other) by gradually adding F ? Assume that the
  addition of solid NaF does not change the volume
  of the solution.
• Ksp for MgF2 6.4 x 109
• Ksp for CaF2 4.0 x 1011

15

• There are several questions which we must answer
  in order to determine the viability of this
  procedure.
• What concentration of F must be present in order
  for each metal to precipitate? Recall that
  precipitation begins when Q Ksp.
• For MgF2 6.4 x 109 Mg2F2 (0.10)F2
• Solving, F 2.5 x 104 M
• For CaF2 4.0 x 1011 Ca2F2 (0.10)F2
• Solving, F 2.0 x 105 M
• Thus, CaF2 precipitates first as F is added.

16

• Now as F is added, CaF2 begins to precipitate
  when the F concentration reaches 2.0 x 105 M
  and it continues to precipitate until the F
  concentration reaches 2.5 x 104 M. At this
  point, MgF2 also begins to precipitate.
• 2) What if we stop adding F at this point, so
  that no MgF2 precipitates? How much of the Ca2
  remains dissolved at this point? If less than 1
  remains, the separation has been successful.

17

• Mathematically, we want to find out what
  concentration of Ca2 remains (dissolved) in
  equilibrium with 2.5 x 104 M of F.
• Ca2 Ksp / F2 4.0 x 1011 / (2.5 x
  104)2
• 6.4 x 104 M
• What of the original Ca2 remains dissolved at
  this point?
• 6.4 x 104 / 0.10 x 100 0.64
• Thus, 99.36 of the Ca2 has been precipi-tated
  before any Mg2, and the separation is successful.