equilibrium & solubility: the solubility product constant, ksp · 7/6/2010 · saturated...
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Equilibrium & Solubility:The Solubility Product Constant, Ksp
1 © Prof. Zvi C. Koren 20.07.2010
Pb2+ + 2NO3- + 2K+ + 2I- PbI2(s) + 2K+ + 2NO3
-
Pb2+ + 2I- PbI2(s)
Gross rxn.: Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
(recall double-replacement rxns)
Net-Ionic rxn.:
Precipitation Reactions
2 © Prof. Zvi C. Koren 20.07.2010
saturated
solution of
PbI2
PbI2(s) Pb2+ + 2I–
Solubility of a Salt = S, units of “moles/L”
Equilibrium
between:
dissolved
and
undissolved
(precipitate)PbI2 (s)
Pb2+ 2I–
Solubility of a Salt
3 © Prof. Zvi C. Koren 20.07.2010
Al2S3(s) 2Al3+ + 3S2-
K = [Al3+]2 [S2-]3sp
solubility product constant
Let x = solubility of salt: x moles salt dissolved / L
2x 3x
= (2x)2 (3x)3 = 108x5
For example,
two types of problems: Solubility Ksp
Note:
Don’t confuse “solubility” with “solubility product constant”
Solubility Product Constant, Ksp
4 © Prof. Zvi C. Koren 20.07.2010
5 © Prof. Zvi C. Koren 20.07.2010
Problem:
Calculate the solubility of magnesium fluoride in g/L.
Ksp = 6.4x10–9 @250C.
Solution:
Case 1: Ksp Solubility
1. Write the equilibrium dissolution rxn: MgF2(s) Mg2+ + 2F–
2. Write the Ksp expression: Ksp = [Mg2+][F–]2
3. Do the stoichometric “ICE-box” right on the equilibrium rxn.
Let x = solubility of the salt in moles/L.
x 2x
4. Calculate using Ksp expression: Ksp = (x)(2x)2
6.4x10–9 = 4x3
Solubility of MgF2 = x = 1.17x10–3 mole of MgF2 dissolves/L @250C.
in g/L: (1.17x10–3 mole of MgF2 /L)(62.3018 g/mole) = 0.073 g MgF2 /L
6 © Prof. Zvi C. Koren 20.07.2010
Problem:
Calculate the solubility product constant, Ksp, of magnesium fluoride.
Solubility = 1.17x10–3 mole/L @250C = [Mg2+]
Solution:1. Write the equilibrium dissolution rxn: MgF2(s) Mg2+ + 2F–
2. Write the Ksp expression: Ksp = [Mg2+][F–]2
Case 2: Solubility Ksp
3. Do the stoichometric “ICE-box” right on the equilibrium rxn.
Let x = solubility of the salt in moles/L = 1.17x10–3 mole/L
x 2x
4. Calculate using Ksp expression:
Ksp = (x)(2x)2 = (1.17x10–3)(2.34x10–3)2 = 6.41x10–9
Experimental
Determination of
Ion Concentrations
Atomic Absorption (AA)
Spectrometer
7 © Prof. Zvi C. Koren 20.07.2010
Ksp in your Kishkes
8 © Prof. Zvi C. Koren 20.07.2010
The Meaning of the Ion-Product in Ksp
Consider MX(s) M+ + X–
Ksp = [M+][X–]
This equation pertains to:
• just before precipitation, or
• after precipitation
Maximum Concentrations BEFORE Precipitation:
[M+] and [X–] are the maximum concentrations that can “live with each
other” (“peaceful coexistence”) without “attacking each other” and
creating a ppt.
Maximum Concentrations AFTER Precipitation:
[M+] and [X–] are the maximum concentrations that can “live with each
other” after a precipitate of MX(s) is formed, no matter how much ppt is
present: Saturated solution.9 © Prof. Zvi C. Koren 20.07.2010
Al2S3(s) 2Al3+ + 3S2-
K = [Al3+]2 [S2-]3sp
Recall:
eq eq
Q = [Al3+]2 [S2-]3no ppt ppt no ppt;
equil.:
max concs.
“ppt” = precipitate
Unsaturatedsolution
Saturatedsolution
Reaction Quotient, Q (Again):Can a Precipitation Reaction Occur?
Q < Ksp Q > Ksp Q = Ksp
(continued)10 © Prof. Zvi C. Koren 20.07.2010
Question:
50.0 mL of 0.34 M Pb(NO3)2 solution is mixed with 25.0 mL of
0.10 M NaCl. Will a precipitate be formed?
Answer:
Overall rxn.: Pb(NO3)2 (aq) + NaCl(aq) PbCl2(s) + NaNO3(aq)2 2
Net ionic rxn.: Pb2+ + 2Cl– PbCl2(s)
?
?
Ksp = [Pb2+] [Cl–]2 = 1.7x10–5 Qsp = [Pb2+][Cl–]2
Calculate concentrations after mixing, but before rxn.:
[Pb2+] = (50.0/75.0)(3.4x10-1 M) = 2.27x10–1 M
[Cl–] = (25.0/75.0)(1.0x10-1 M) = 3.33x10–2 M
Q = (2.27x10-1)(3.33x10-2)2 = 2.5x10–4 > Ksp = 1.7x10–5 ppt!
What are the concentrations of all ions and how many moles of
precipitate are obtained? (see next slide)
eq eq
Can a Precipitation Reaction Occur?
11 © Prof. Zvi C. Koren 20.07.2010
Net ionic rxn.: Pb2+ + 2Cl- PbCl2(s),
Ksp
What are the concentrations of all ions and how many moles of
precipitate are obtained?
0.227 0.0333I
-0.0333 +0.0167-0.0167CE,I 0.210 - 0.0167
C +x +2x -x
E 0.210+x 2x 0.0167-x
1.7x10-5 = = [Pb2+][Cl-]2
= (0.210+x)(2x)2 (0.210)(4x2), x = 4.5x10-3 M
Concentrations of ions:
[Cl-] = 2x = 9.0x10-3 M
[Pb2+] = 0.210+x 0.21 M
Moles of ppt formed:
PbCl2(s): 0.0167-0.0045 (0.0122 mole/L)x(0.075 L) = 9.15x10-4 mole
K >> 1
Check: Ksp = [Pb2+][Cl-]2 1.7x10-5
(continued)
K >> 1:
12 © Prof. Zvi C. Koren 20.07.2010
MX(s) M+ + X–sparingly soluble salt:+ MA(aq) M+ + A- or LX(aq) L+ + X–soluble salt:
What happens to the solubility of a salt when there is a common ion?
according to Le Chatelier’s Principle, the solubility of the sparingly
soluble salt will decrease in the presence of a common ion.
The Common Ion Effect (Again)
13 © Prof. Zvi C. Koren 20.07.2010
Question:
The solubility of AgCl is 0.0019 g/L.
If some AgCl is placed in 500.0 mL of a 0.50 M solution of NaCl, how
many grams of AgCl dissolve?
Answer:
Note the presence of 2 salts, one sparingly soluble and one soluble.
0.50 M
2. Write the equilibrium dissolution rxn: AgCl(s) Ag+ + Cl–
1. Write the total dissolution rxn: NaCl(aq) Na+ + Cl–
4. Do the stoichometries (and ICE-box for sp. sol. salt) right on the rxns.
Let x = solubility of the salt in moles/L
3. Write the Ksp expression for the sparingly soluble salt:
Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)
0.50 M
x x+0.50
(continued)
The Common Ion Effect - Calculations
14 © Prof. Zvi C. Koren 20.07.2010
(continued)
Need to calculate Ksp from solubility:
Given: The solubility of AgCl is 0.0019 g/L.
From before: Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)
x = Solubility = 0.0019 g/L (1 mole/143.321 g) 1.33x10-5 M
[Ag+] = x = 1.33x10-5 M, [Cl-] = x = 1.33x10-5 M.
Ksp = 1.77x10-10
1.77x10-10 = Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)
x = 3.54x10-10 mole/L 5.07x10-8 g AgCl dissolve /L (0.500 L)
2.5x10-8 g AgCl dissolve!
15 © Prof. Zvi C. Koren 20.07.2010
Selective Separation of Ions & Solubility
Ag+
Pb2+
Cu2+
Cl–
AgCl(s)
PbCl2(s)
Cu2+
Δ
AgCl(s)
Pb2+ + CrO42– PbCrO4(s)
16 © Prof. Zvi C. Koren 20.07.2010
Separation of Two Ions by Difference in Solubility: Cl– and CrO42–
Which salt is more insoluble, AgCl or Ag2CrO4?
Ksp: 1.8x10–10 9.0x10–12
Check Solubilities!!! Let x = solubility of a salt in moles/L:
x x
Ksp = [Ag+][Cl-] = x2 x = 1.3x10–5 mol/L
Ag2CrO4(s) 2Ag+ + CrO42– ,
2x x
Ksp = [Ag+]2 [CrO42–] = (2x)2(x) x = 1.3x10–4 mol/L
Surprise, surprise!
(continued)
Selective Separation of Ions & Solubility - Calculations
AgCl(s) Ag+ + Cl– ,
17 © Prof. Zvi C. Koren 20.07.2010
Question:
A solution contains 0.010 M NaCl and 0.0010 M K2CrO4.
Solid AgNO3 is added slowly to the solution.
a) Which precipitates first?
b) What is [Cl-] when the second precipitate begins to form?
Answer:
First calculate [Ag+] required to just begin the precipitation of each salt:
To just begin precipitation of AgCl when [Cl–] = 0.010 M:
Ksp = [Ag+][Cl-] [Ag+] = 1.8x10-8 M.
To just begin precipitation of Ag2CrO4 when [CrO42–] = 0.0010 M:
Ksp = [Ag+]2 [CrO42–] [Ag+] = 9.5x10-5 M.
AgCl precipitates first!
When does Ag2CrO4 begin to precipitate?
When [Ag+] = 9.5x10-5 M Ksp = [Ag+][Cl-] [Cl-] = 1.9x10-6 M
% Cl- remaining in solution = 100x(1.9x10-6 M)/(0.010 M)= 0.019%
Separation of Two Ions by Difference in Solubility: Cl– and CrO42–
Selective Separation of Ions & Solubility (continued)
18 © Prof. Zvi C. Koren 20.07.2010
PbCl2(s) + CrO42– PbCrO4(s) + 2Cl–, K = ?
PbCl2(s) Pb2+ + 2Cl–, K1 = Ksp(PbCl2)
Pb2+ + CrO42– PbCrO4(s), K2 = 1/Ksp(PbCrO4)
?
K = K1•K2 = 9.4x108
Simultaneous Equilibria:
Two or more equilibria rxns occurring at the same time
When a salt solution is added to a salt precipitate … Check Krxn!!!
PbCl2(s) is only sparingly soluble in water (Ksp = 1.7x10–5).
What is its solubility in certain other ionic solutions, e.g., CrO42– ?
PbCl2 is “soluble” in a solution of CrO42–, but it becomes another ppt.
Simultaneous Equilibria
19 © Prof. Zvi C. Koren 20.07.2010
Some sparingly soluble salts are more soluble in water than normally
expected. Why???
Answer: Because the resultant anion can undergo hydrolysis as a base.
For example:
PbS(s) Pb2+ + S2–, Ksp(PbS) = 3.2x10–28
AND
S2– + H2O HS– + OH–, K = Kb(S2–) = Kw/Ka(HS–) =
*(Note: For H2S: Ka1 = 1x10–7, Ka2 = 1x10–19)
K2(H2S)*
1x105
PbS(s) + H2O Pb2+ + HS– + OH–,
K = 3x10–23 >> Ksp
(continued)
PbS is more soluble in water than expected
Solubility and pH
20 © Prof. Zvi C. Koren 20.07.2010
How can we determine the ability of a salt to be dissolved by an acid?(A powerful display of the wonders of chemical principles!!!)
Example: CaCO3(s) + H+? What is the value of K for this rxn?
The Trick: “Build” this rxn from other known salt solubility and acid-
base hydrolysis rxns.:
CaCO3(s) Ca2+ + CO32-, K = Ksp = 3.8x10-9
CO32- + H2O HCO3
- + OH-, K = Kb(CO32-) = Kw/Ka(HCO3
-) = 2.1x10-4
H+ + OH- H2O, K = 1/ Kw = 1.0x1014
CaCO3(s) + H+ Ca2+ + HCO3
-, K = 80.0 > 1 (but read on)
But Also:
HCO3- + H+
H2CO3, K = 1/Ka1 = 1/4.2x10-7 = 2.4x106
But Also:
H2CO3(aq) CO2(g) + H2O, K 105
In general: Must examine all possible subsequent rxns.
CaCO3(s) + 2H+ Ca2+ + CO2(g) + H2O, K 1013
(Note: CO2(g) bubbles out of system, so equil. moves even more to right!)
Solubility and pH – Salt in an Acidic Solution
21 © Prof. Zvi C. Koren 20.07.2010
Question:
The Ksp of Mg(OH)2 is 1.5x10-11.
If solid Mg(NO3)2 is added to a
solution with a pH of 9.00,
at what [Mg2+] does precipitation
begin?
Answer:
What happens to the solubility of
this hydroxide as the pH is changed,
up or down?
Mg(OH)2(s) Mg2+ + 2OH-,
Ksp = [Mg2+][OH-]2
pH = 9.00 [OH-] = 1.0x10-5 M.
[Mg2+] = 0.15 M.
Solubility and pH – Metal Hydroxides
22 © Prof. Zvi C. Koren 20.07.2010
Complex Metal surrounded by two or more groups
(neutral or cation) (anion or molecule )
Complex-Ion Formation and “Formation Constant”:
Ag+ + 2NH3(aq) Ag(NH3)2+,
Rxn in two steps:
Ag+ + NH3(aq) Ag(NH3)+, K1
Ag(NH3)+ + NH3(aq) Ag(NH3)2
+, K2
Kf = K1•K2 = 1.6x107
Dissolving a Precipitate by Complex-Ion Formation:
“ligands”
AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl-. K = ?
If K > Ksp, then bingo!
We know that the solubility of AgCl(s) in water is only slight, Ksp =
1.8x10-10. Can another solvent, e.g., NH3(aq), better disssolve the solid?
Build this rxn from others. That’s the trick!
AgCl(s) Ag+ + Cl-, Ksp
Ag+ + 2NH3(aq) Ag(NH3)2+, Kf
K = Ksp•Kf
= 2.9x10-3. Yesh!
Solubility and Complex Ion Formation
23 © Prof. Zvi C. Koren 20.07.2010
Problem:
How many moles of ammonia must be added to dissolve 0.050 mol
of solid AgCl present in 1.0 L of water?
Answer:
AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl-,
0.050 mol/LI
C -0.050 +0.050 +0.050
E - y-0.10 0.050 0.050
2.9x10-3 = K = [Ag(NH3)2+][Cl-] / [NH3]
2
y-0.10 = [NH3] = 0.93 M
y
-0.10
y = 1.03 mol/L
And
n = (1.03 mol/L) x (1 L) = 1.03 mol NH3 must be added initially.
K = 2.9x10-3.
Solubility and Complex Ion Formation - Calculations
24 © Prof. Zvi C. Koren 20.07.2010