Solubility Equilibria
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Transcript of Solubility Equilibria
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Solubility Equilibria
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CAN YOU? / HAVE YOU? · Write a balanced chemical equation to represent equilibrium in a saturated solution. · Write a solubility product expression. · Answer questions about Ksp and various missing
concentrations using I.C.E. tables.
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There are 3 actions that affect solubility:
1. Nature of the solute and solvent“like dissolves like”
Polar / ionic solute dissolve in polar solvent.Non-polar dissolve in non-polar.
Even the most insoluble ionic solids are actually soluble in water to a limited extent
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2. Temperature
Solids in liquids: ↑ temperature - ↑ solubility.Gases in liquids: ↑ in temperature - ↓ solubility.
3. Pressure
Does not affect the solubility of (s)/(l).(g): ↑ pressure ↑ solubility.
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AaBb(s) aA+(aq) + bB¯(aq)
Ksp, called the solubility product constant.
Ksp = [A+]a[B-]b
Product of ion concentrations in a saturated solution.
Kc =
[A+]a[B-]b [AaBb]
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Write the dissociation and the product constant equation for the solubility of calcium hydroxide.
Ksp = [Ca2+][OH-]2
Ca(OH)2 (s)
Pb3(PO4)2(s) 3 Pb2+(aq) + 2 PO4
3-(aq)
Ksp = [Pb2+]3[PO43-]2
Write a solubility product expression for Pb3(PO4)2.
Ca2+(aq) + OH-
(aq)2
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At equilibrium, the [Ag+] = 1.3 x 10-5 M and the [Cl-] = 1.3 x 10-5 M, what is the Ksp of silver chloride?
Ksp = [Ag+][Cl-]
Ksp =(1.3 x 10-5)(1.3 x 10-5)Ksp = 1.7 x 10-10
AgCl (s) Ag+(aq) + Cl-
(aq)
*NOTE: Ksp has no units.
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SolubilityAnd
I.C.E. Tables(Yeah!)
Solubility - maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature.
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Calculate Ksp of lead (II) chloride if a 1.0 L saturated solution has of lead ions.
I --- 0 0C -x +x +2xE 0
Ksp = [Pb+2][Cl -]2
Ksp = [1.62 x 10 -2][ 3. 24 x 10 -2]2
Ksp = 1.70 x 10 -5
PbCl2(s) Pb2+(aq) + 2 Cl-
(aq)
1.62 x 10-2 M
2(1.62 x 10-2)
1.62 x 10-2 M
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The solubility of PbF2 is . What is the value of the solubility product constant?
PbF2(s) Pb2+(aq) + 2 F¯(aq)
0.466 g1 L 245.2 g
1 mol= 1.90 x 10-3 M PbF2
0.466 g/L
Pb – 207 + 2 (19) = 245g/mol
Ksp = [Pb2+][F-]2
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Ksp = [Pb2+][F-]2
Ksp = (1.90 x 10-3)(3.80 x 10-3)2
Ksp = 2.74 x 10-8
PbF2(s) Pb2+(aq) + 2 F¯(aq)
[E] 0 1.9 x 10-3 M
[I] 1.9 x 10-3 M 0 0
[C] - x + x + 2x
3.8 x 10-3 MSaturated – all solid reactant dissociates.
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Calculate Ksp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride.
I 0.0159 M 0 0C -x +x +2xE 0 0.0159 M 0.0318 M
Ksp = [Pb2+][Cl-]2
0.2207 g278.1g1 mol = 0.0159 M PbCl20.05 L
PbCl2(s) Pb2+(aq) + 2 Cl-
(aq)
Ksp = [0.0159][0.0318]2 = 1.61 x 10-5
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Ksp of magnesium hydroxide is 8.9 x 10-12. What are the [equilibrium] of ions in saturated solution?
I --- 0 0C -x +x +2xE 0 x 2x
Mg(OH)2 (s) Mg2+(aq)
+ 2 OH-(aq)
Ksp = [Mg2+][OH-]2
8.9 x 10-12 = [x][2x]2
8.9 x 10-12 = [x]4x2
8.9 x 10-12 = 4x3
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[Mg2+] = x = 1.3 x 10-4 mol/L
[OH-] = 2x = 2.6 x 10-4 mol/L
8.9 x 10-12 = 4x3
44
2.23 x 10-12 = x33√ 3√
1.3 x 10-4 = x
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Estimate the solubility in g/L of Ag2CrO4 if the Ksp is 1.1 x 10-12.
I x 0 0C -x +2x +xE 0
Ksp = [Ag+]2[CrO42-]
1.1 x 10 -12 = [2x]2[x] 1.1 x 10 -12 = 4x3
x = 6.50 x 10 -5 M
Ag2CrO4(s) 2 Ag+(aq) + CrO4
2-(aq)
1.30 x 10 -4 M 6.50 x 10 -5 M
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[Ag2CrO4]i = 6.50 x 10 -5 M
Ag2CrO4(s) 2 Ag+(aq) + CrO4
2-(aq)
6.5 x 10-5mol 330 g1 mol = 0.022 g/L
1 L
E 0 1.30 x 10 -4 M 6.50 x 10 -5 M
1 1
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Precipitation
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Compare value of Q, with given Ksp to determine if an aqueous solution is saturated or unsaturated.
Q = Ksp Saturated solution, no precipitate.
Q >Ksp Precipitate forms (“oversaturated”)
Q < Ksp Solution is unsaturated.
Qsp = [A+]a[B¯]b
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PbF2(s) Pb2+(aq) + 2 F¯(aq)
Pb – 207 + 2 (19) = 245g/mol
Ksp of lead (II) fluoride is 1.6 x 10-5. If 0.57 g are mixed with 1500 mL of water, is solution saturated?
0.57 g1 L 245.2 g
1 mol= 2.32 x 10-3 mol/L
Qsp = (2.32 x 10-3)(4.64 x 10-3)2
Qsp = 5.0 x 10-8Q >Ksp Precipitate forms
Ksp = [Pb2+][F-]2
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0.01 M NaCl 0.02 M Pb(NO3)2
Predict if there is a precipitate of PbCl2 if 100 mL each of of and are added together. Ksp of PbCl2 = 1.7 x 10 -5
PbF2(s) Pb2+(aq) + 2 Cl¯(aq)
Qsp = [Pb2+][Cl-]2
0.01 M NaCl 0.02 M Pb(NO3)2
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C1V1 = C2V2
(0.01 M)(0.1 L) = (C2)(0.2 L) = 0.005 M Cl-
(0.02 M)(0.1 L) = (C2)(0.2 L) = 0.01 M Pb2+
Qsp = [0.01][0.005]2 = 2.5 x 10 -7
Qsp< Ksp A precipitate will not form.
Because you are adding volumes, you must account for the new diluted concentrations.
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If 20.0 mL of 0.0010 M silver nitrate is mixed with 20.0 mL of 3.0 x 10-5 M potassium bromide, does silver bromide (Ksp = 5.0 x 10-13) precipitate? Assume
the volumes are additive.
AgBr(s) Ag+(aq) + Br¯(aq)
Ksp = [Ag+][Br-]
(0.001 M)(0.02 L) = (C2)(0.04 L) = 5.0 x 10-4 M Ag+
(2e-5 M)(0.02 L) = (C2)(0.04 L) = 1.5 x 10-5 M Br-
Qsp = [5.0 x 10-4][1.5 x 10-5] = 7.5 x 10 -9
Qsp> Ksp A precipitate will form.
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· Substances which are insoluble are actually slightly soluble.
· The solubility product, Ksp, describes the product of ion concentrations in saturated solutions.
· Solubility can be determined from the solubility product.
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CAN YOU? / HAVE YOU? · Write a balanced chemical equation to represent equilibrium in a saturated solution. · Write a solubility product expression. · Answer questions about Ksp and various missing
concentrations using I.C.E. tables.