Chapter 20 The Solubility Product •Silver chloride, AgCl,is rather insoluble in water. •Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. Solubility Product Constants 2 Ag + Cl - (s) ⇌ Ag + (aq) + Cl - (aq) •The equilibrium constant expression for this dissolution is called a solubility product constant. •Ksp = solubility product constant •Ksp = [Ag + ][Cl - ] = 1.8 x 10 -10 Solubility Product Constants 3 •The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Examples: •Ag2S(s) ⇌ 2Ag + (aq) + S 2- (aq) •Ksp = [Ag + ] 2 [S 2- ] =1.0 x 10 -49 •Ca3(PO4)2(s) ⇌ 3Ca +2 (aq) + 2PO4 3- (aq) •Ksp = [Ca +2 ] 3 [PO4 -3 ] 2 = 1.0 x 10 -25 •CaNH4PO4(s) ⇌ Ca +2 (aq) NH4 + + PO4 3- (aq) •Ksp = [Ca 2+ ][NH4 + ][PO4 3- ] Solubility Product Constants 4 •All Ksp values are small. •All salts are only slightly soluble K sp Table 5 1 2 3 4 5
Transcript
Chapter 20
The Solubility Product
•Silver chloride, AgCl,is rather insoluble in water.•Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.
Solubility Product Constants
2
Ag+Cl-(s) ⇌ Ag+(aq) + Cl-(aq)
•The equilibrium constant expression for this dissolution is called a solubility product constant.•Ksp = solubility product constant•Ksp = [Ag+][Cl-] = 1.8 x 10-10
Solubility Product Constants
3
•The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Examples:•Ag2S(s) ⇌ 2Ag+
(aq) + S2-(aq)
•Ksp = [Ag+]2[S2-] =1.0 x 10-49
•Ca3(PO4)2(s) ⇌ 3Ca+2(aq) + 2PO4
3-(aq)
•Ksp = [Ca+2]3[PO4-3]2 = 1.0 x 10-25
•CaNH4PO4(s) ⇌ Ca+2(aq) NH4
+ + PO43-
(aq)
•Ksp = [Ca2+][NH4+][PO4
3-]
Solubility Product Constants
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•All Ksp values are small.•All salts are only slightly soluble
Ksp Table
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1
2
3
4
5
•One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25°C. Calculate the molar solubility of, and Ksp for, AgCl.
•Dissociation: AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
•Molar solubility:
•Ksp:
Determination of Solubility Product Constants
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0.00192 gAgCl
L x
1 molAgCl
143 gAgCl
= 1.34 x 10-5 molAgCl
L
Ksp = [Ag+][Cl-]
Ksp = [1.34 x 10-5][1.34 x 10-5]
Ksp =1.80 x 10-10
•One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2. Calculate the molar solubility of, and Ksp for, CaF2.•Dissociation: CaF2(s) ⇌ Ca2+
(aq) + 2F-(aq)
•Molar solubility:
•Ksp:
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Determination of Solubility Product Constants
0.0167 gCaF2
L x
1 molCaF2
78.1 gCaF2
= 2.14 x 10-4 molCaF2
L
Ksp = [Ca2+][F-]2
Ksp = [2.14 x 10-4][2 x 2.14 x 10-4]2
Ksp =3.92 x 10-11
or
Ksp = [x][2x]2 = 4x3 =4(2.14 x 10-4)3
Ksp =3.92 x 10-11
Write the solubility product expression for each of the following salts:(a) Mn3(AsO4)2
Given the solubility data for the following compounds, calculate their solubility product constants.(a) SrCrO4, strontium chromate, 1.2 mg/mL(b) BiI3, bismuth iodide, 7.7 x 10-3 g/L(c) Fe(OH)2, iron(II) hydroxide, 1.1 x 10-3 g/L(d) SnI2, tin(II) iodide, 10.9 g/L
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EOC #8, page 767Example 20-2
(a) SrCrO4, 1.2 x 10-3g ÷ 204 g/mol ÷ 1.0 x 10-3L = 5.9 x 10-3 M
Ksp = x2 = (5.9 x 10-3)2 = 3.5 x 10-5
(b) BiI3, (27x4) and Ksp = 7.7 x 10-19
(c) Fe(OH)2, (4x3) and Ksp = 6.9 x 10-15
(d) SnI2, (4x3) and Ksp = 1.01 x 10-4
•Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate.
•Dissociation: BaSO4(s) ⇌ Ba2+(aq) + SO4
2-(aq)
•Molar solubility:
Uses of Ksp
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Ksp = (x)(x) = x2 = 1.1 x 10-10
x= 1.1 x 10-10
x= 1.0 x 10-5 M
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•Calculate the mass of barium sulfate, BaSO4, in 1.00 L of saturated solution
•Dissociation: BaSO4(s) ⇌ Ba2+(aq) + SO4
2-(aq)
•Molar solubility: x = 1.0 x 10-5 (from previous slide)
•Mass:
Uses of Ksp
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234 gBaSO4
molBaSO4
x 1.0 x 10-5 molBaSO4
L x L = 2.3 x 10-3gBaSO4
FW x M x L = g from M =
g
FW
L
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
•The Ksp for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of saturated magnesium hydroxide solution.
•Dissociation: Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-
(aq)
•Molar solubility: Ksp = [Mg2+][OH-]2 = (x)(2x)2 = 4x3 = 1.5 x 10-11 ; x = 1.6 x 10-4 [molar solubility]
•[OH-] = 2 x molar solubility = 3.2 x 10-4
•pOH = 3.49 and pH = 10.51
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Uses of Ksp
•Calculate the concentration of calcium ion in a calcium phosphate, Ca3(PO4)2, in pure water. Ksp= 2.1 x 10-33.
•Dissociation: Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43-
•Molar solubility: Ksp = [Ca2+]3[PO43-]2 = (3x)3
(2x)2 = (27x3)(4x2) = 108x5 = 2.1 x 10-33 x = 1.1 x 10-7 [molar solubility]
•[Ca2+] = 3 x molar solubility = 3.3 x 10-7
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Uses of Ksp
Calculate molar solubilities, concentrations of ions, and solubilities in grams per liter for the following compounds at 25 °C:(a) CuCl, copper(I) chloride(b) Ba3(PO4)2, barium phosphate(c) PbI2, lead fluoride(d) Sr3(PO4)2, strontium phosphate
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EOC #16, page 790Example 20-3
(a) CuCl: x2 = Ksp = 1.9 x 10-7; x = 4.4 x 10-4 M (MS) (4.4 x 10-4 mol/L) x (99 g/mol) = 0.043 g/L(b) Ba3(PO4)2: MS = 6.5 x 10-7 (3.9 x 10-4 g/L)(c) PbI2: MS = 2.1 x 10-3 (0.51 g/L)(d) Sr3(PO4)2: MS = 2.5 x 10-7 (1.1 x 10-4 g/L)
•Solubility is decreased when a common ion is added.
•This is an application of Le Châtelier’s principle:
BaSO4 ⇌ Ba2+ + SO42-
•As SO42- (from Na2SO4, say) is added, the
equilibrium shifts away from the increase. •Therefore, BaSO4(s) is formed and precipitation occurs.
•As Na2SO4 is added to the system, the solubility of BaSO4 decreases.
Common Ion Effect
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•Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution. Compare this to the solubility of BaSO4 in pure water.
•Write equations to represent the equilibria:
Na2SO4 → 2 Na+ + SO42-
0.010 2(0.010) 0.010
BaSO4 ⇌ Ba+2 + SO42-
x x x
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Common Ion Effect
•Substitute the concentrations into the Ksp expression and solve for x.
Ksp = [Ba2+][SO42-] = 1.1 x 10-10
Ksp = (x)(0.010) = 1.1 x 10-10
MS = x = 1.1 x 10-8
•The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8 M.
•The molar solubility of BaSO4 in pure water is 1.0 x 10-5 M.
•900 times greater
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Common Ion Effect
Calculate the molar solubility of Ag2SO4 in 0.12 M K2SO4 solution.
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EOC #20, page 790Example 20-4
K2SO4 → 2K+ + SO42-
Ag2SO4 ⇌ 2Ag+ + SO42-
2x + (x + 0.12)Ksp = 1.7 x 10-5 = (2x)2 • 0.12; x = 6.0 x 10-3
MS = 6.0 x 10-3 mol Ag2SO4/L in 0.12 M K2SO4
Which has the greater molar solubility in a 0.125 M K2CrO4 solution: BaCrO4 or Ag2CrO4?
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EOC #26, page 791Example 20-4
BaCrO4: Ksp = x(0.125) = 2.0 x 10-10; x = 1.6 x 10-9
Ag2CrO4:Ksp = (2x)2(0.125) = 9.0 x 10-12; x = 4.2 x 10-6
Ag2CrO4 is more soluble
•We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form [PbSO4]?
K2SO4 → 2K+ + SO42-
Pb(NO3)2 → Pb2+ + 2NO3-
•Solution volumes are additive. Concentrations of the important ions are: 100 mL x 0.10 M ÷ 200 mL = 0.050 M Pb2+
100 mL x 0.010 M ÷ 200 mL = 0.0050 M SO42-
•Qsp = [Pb2+][SO42-] = (0.050)(0.0050) = 2.5 x 10-4
Ksp = 1.8 x 10-8 for PbSO4
Qsp > Ksp therefore precipitate forms
The Reaction Quotient
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•Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury(II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0 x 10-53. Ksp = [Hg2+][S2-]
•Ksp = [1.0 x 10-8][S2-] = 3.0 x 10-53
[S2-] = [3.0 x 10-53]÷ [1.0 x 10-8]= 3.0 x 10-45 MAdd enough S2-, in the form of Na2S, to just slightly exceed 3.0 x 10-45 M
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The Reaction Quotient
•A solution contains 0.020 M Ag+ and Pb2+. Add CrO4
2- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
•Solution•The substance whose Ksp is first exceeded precipitates first.
•The ion requiring the lesser amount of CrO42-
precipitates first.
Separating Salts
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A solution contains 0.020 M Ag+ and Pb2+. Add CrO4
2- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 and Ksp for PbCrO4 = 1.8 x 10-14
Solution:Calculate [CrO42-] required by each ion.
[CrO42-]to ppt. PbCrO4 = Ksp/[Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M[CrO4
2-]to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M
Qsp Ag2CrO4>Qsp PbCrO4 so PbCrO4 precipitates first.
Separating Salts
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How much Pb2+ remains in solution when Ag+ begins to precipitate?
SolutionWe know that [CrO4
2-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4.
What is the Pb2+ conc. at this point?Solution[Pb2+] = Ksp / [CrO4
2-] = 1.8 x 10-14 /2.3 x 10-8 = 7.8 x 10-7 M
Lead ion has dropped from 0.020 M to < 10-6M
Separating Salts
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99.9961% of Pb2+ ions precipitates before AgCrO4
begins to precipitate.
Separating Salts
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%unppt =[unppt]
[original] x 100%
%unppt =7.8 x 10-7
0.020 x 100% = 0.0039%
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Will a precipitate of PbCl2 form when 5.0 g of solid Pb(NO3)2 is added to 1.00 L of 0.010 M NaCl? Assume assume that the volume change is negligible.
Sodium bromide and lead nitrate are soluble in water. Will lead bromide precipitate when 1.03 g of NaBr and 0.332 g of Pb(NO3)2 are dissolved in sufficient water to make 1.00 L of solution?
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EOC #29, page 791Example 20-6
PbBr2 ⇌ Pb2+ +2Br-; Ksp = 6.3 x 10-6
[Pb2+] = 0.332 g ÷ 331 g/mol = 0.00100 M[Br-] = [NaBr] = 1.03 g ÷ 103 g/mol = 0.0100 M Qsp = (0.00100)(0.0100)2 = 1.00 x 10-7
Qsp < Ksp therefore there will be no precipitate
A solution is 0.0100 M in Pb2+ ions. If 0.103 mol of solid NaI is added to 1.00L of this solution (with negligible volume change), what percent of the Pb2+ ions remain in solution?
0.010 0.103Pb+2 + 2NaI → PbI2; PbI2 ⇌ Pb+2 + 2I-
Ksp = (x)(2x + (0.103 - 0.020))2 = 8.7 x10-9
[Pb+2] or x = 1.3 x 10-6 x<<0.083 or 0.103 - 0.020
%unppt = (1.3 x 10-6) ÷ (0.0100) x 100%%unppt = 0.013%
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EOC #32, page 791Example 20-7
• Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution.
•Look at a solution that contains 0.010 M each of Cu+, Ag+, and Au+
•We can precipitate them as chlorides
CuCl ⇌ Cu+ + Cl- Ksp = [Cu+][Cl-] = 1.9 x 10-7
AgCl ⇌ Ag+ + Cl- Ksp = [Ag+][Cl-] = 1.8 x 10-10
AuCl ⇌ Au+ + Cl- Ksp = [Au+][Cl-] = 2.0 x 10-13
Gold will precipitate first and copper will be last as we slowly add NaCl
Fractional Precipitation
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Concentration of Cl- required to precipitate each metal ion.
Fractional Precipitation
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For Au+ - Ksp =[Au+][Cl-] = 2.0 x 10-13
[Cl-]=2.0 x 10-13
[Au+]=
2.0 x 10-13
0.010 = 2.0 x 10-11 M
For Ag+ - Ksp =[Ag+][Cl-] = 1.8 x 10-10
[Cl-]=1.8 x 10-10
[Ag+]=
1.8 x 10-10
0.010 = 1.8 x 10-8 M
For Cu+ - Ksp =[Cu+][Cl-] = 1.9 x 10-7
[Cl-]=1.9 x 10-7
[Cu+]=
1.9 x 10-7
0.010 = 1.9 x 10-5 M
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• We have calculated the [Cl-] required • to precipitate AuCl, [Cl-] > 2.0 x 10-11 M• to precipitate AgCl, [Cl-] > 1.8 x 10-8 M• to precipitate CuCl, [Cl-] > 1.9 x 10-5 M
• We can calculate the amount of Au+ precipitated before Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before Cu+ begins to precipitate.
Fractional Precipitation
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• Calculate the percent of Au+ ions that precipitate before AgCl begins to precipitate.• Use the [Cl-] = 1.8 x 10-8 M to determine the
[Au+] remaining in solution just before AgCl begins to precipitate.
Fractional Precipitation
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For Au+ - Ksp =[Au+][Cl-] = 2.0 x 10-13
[Au+]=2.0 x 10-13
[Cl-]=
2.0 x 10-13
1.8 x 10-8 = 1.1 x 10-5 M unppt
%unppt =[unppt]
[original] x 100%
%unppt =1.1 x 10-5
0.010 x 100% = 0.11%
Therefore 99.89 % of the Au+ has precipitated
before AgCl begins to precipitate.
• Calculate the percent of Ag+ ions that precipitate before CuCl begins to precipitate.• Use the [Cl-] = 1.9 x 10-5 M to determine the
[Ag+] remaining in solution just before CuCl begins to precipitate.
Fractional Precipitation
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For Ag+ - Ksp =[Ag+][Cl-] = 1.8 x 10-10
[Ag+]=1.8 x 10-10
[Cl-]=
1.8 x 10-10
1.9 x 10-5 = 9.5 x 10-6 M unppt
%unppt =[unppt]
[original] x 100%
%unppt =9.5 x 10-6
0.010 x 100% = 0.095%
Therefore 99.905 % of the Ag+ has precipitated
before AgCl begins to precipitate.
To a solution that is 0.010 M in Cu+, 0.010 M in Ag+, and 0.010 M in Au, solid NaBr is added slowly. Assume that there is no volume change due to the addition of solid NaBr. (a) which compound will begin to precipitate first? (b) Calculate [Au+] when AgBr just begins to precipitate. What percentage of the Au+ has precipitated at this point? (c) Calculate [Au+] and [Ag+] when CuBr just begins to precipitate.
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EOC #36, page 791Example 20-8 & 20-9
(a) Both have similar ion ratio - PbCrO4 has the smaller Ksp and will ppt first.
(b) PbCrO4 will ppt when Qsp = [Pb2+][CrO42-] = Ksp
= 1.8 x 10-14 so [Pb2+] = Ksp ÷ [CrO42-] = 1.8 x
10-14 ÷ 0.050 = 3.6 x 10-13
(c) PbSO4 will ppt when Qsp = [Pb2+][SO42-] = Ksp =
1.8 x 10-8 so [Pb2+] = Ksp ÷ [SO42-] = 1.8 x 10-8 ÷
0.050 = 3.6 x 10-7
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EOC #38, page 791Example 20-8 & 20-9
A solution is 0.050 M in K2SO4 and 0.050 in K2CrO4. A solution of Pb(NO3)2 is added slowly without changing the volume appreciably. (a) Which salt, PbSO4 or PbCrO4, will precipitate first? (b) What is [Pb2+] when the salt in part (a) begins to precipitate? (c) What is [Pb2+] when the other salt begins to precipitate?
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If a solution is 2.0 x 10-5 M in Mn(NO3)2 and 1.0 x 10-3 M in aqueous ammonia, will Mn(OH)2 precipitate?
Mn(OH)2 ⇌ Mn2+ + 2OH- Ksp = 4.6 x 10-14
NH3 + H2O ⇌ NH4+ + OH- Kb = 1.8 x 10-5
[Mn2+] = [Mn(NO3)2] = 2.0 x 10-5
[OH-] = ionization of NH3
Kb = [NH3][OH-]÷ [NH3] = x2 ÷ (1.0 x 10-3); x = 1.3 x 10-4
Qsp = [Mn2+][OH-] = 3.4 x 10-13
Qsp>Ksp - a ppt will form36
EOC #46, page 791Example 20-10
SKIP and stop for test
•If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? For Mg(OH)2, Ksp = 1.5 x 10-11; Kb for NH3 = 1.8 x 10-5.
•Calculate Qsp for Mg(OH)2 and compare it to its Ksp. Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M. Aqueous ammonia is a weak base that we can calculate [OH-]
•Same type of problem as the last EOC[OH-] = 1.3 x 10-3 and [Mn2+] = 0.010Qsp = 1.7 x 10-8 Qsp>Ksp
Simultaneous Equilibria
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•How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2?
•Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.
Kb = [Mg2+][OH-] = 1.5 x 10-11
[OH-] = Kb ÷ [Mg2+][OH-] = 1.5 x 10-11 ÷ 0.010 = 3.9 x 10-5 M
Simultaneous Equilibria
38
•Using the maximum [OH-] that can exist in solution, we can calculate the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that [OH-] does not exceed 3.9 x 10-5 M.
(0.10 - 3.9 x 10-5) 3.9 x 10-5 3.9 x 10-5
NH3 + H2O ⇌ NH4+ + OH-
xM xM xM
NH4Cl → NH4+ + Cl-
(0.10 - 3.9 x 10-5) xM + 3.9 x 10-5 3.9 x 10-5
NH3 + H2O ⇌ NH4+ + OH-
x >> 3.9 x 10-5
Simultaneous Equilibria
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Simultaneous Equilibria
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Kb =[NH4
+][OH-][NH3]
=1.8 x 10-5
Kb =(x+(3.9 x 10-5))(3.9 x 10-5)
(0.10-(3.9 x 10-5))=1.8 x 10-5
(3.9 x 10-5)2 <<x so (x+(3.9x10-5))(3.9 x 10-5)= x(3.9 x 10-5)
Kb =x(3.9 x 10-5)
(0.10)=1.8 x 10-5
x=0.046 M =[NH4Cl]=0.046mol /LCHECK:Qsp =Ksp ∴at equilibrium
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If a solution is made 0.080 M in Mg(NO3)2, 0.075 M in aqueous ammonia, and 3.5 M in NH4NO3, will Mg(OH)2 precipitate? What is the pH of this solution?
Mg(OH)2 ⇌ Mg2+ + 2OH- Ksp = 1.5 x 10-11
NH3 + H2O ⇌ NH4+ + OH- Kb = 1.8 x 10-5
[Mg2+] = [Mg(NO3)2] = 0.080 M[OH-] = Kb x [base]/[salt] (see chapter 18 - buffers)
[OH-] = (1.8 x 10-5)(0.075 M)÷(3.5 M) = 3.9 x 10-7
Qsp = [Mg2+][OH-]2 = (0.080)(3.9 x 10-7)2 = 1.2 x 10-14
