Lecture 15 Highlights…
Exam #2 Review
Quantifying how soluble a
compound is ?
• Solubility product constant….
Common Ion Effect
•What if the solvent already
contains a soluble species ?
Exam #2 Review: Exam may
include, but is not limited to….…
Cations, anions and ionic compounds
Naming ionic compounds
Polyatomic ions and their charges
Lattice Energy (Physical Properties)
Chemical Formula (MgCl2) and
calculating the molar mass of
compounds (from Periodic Table
data).
Polyatomic Ions
Ones to know..
acetate
bicarbonate
carbonate
cyanide
dichromate
hydride
hydroxide
phosphate
Sulfate
CH3COOHCO3CO32CNCr2O72HOHPO43SO42-
1
Oxoanions of chlorine
ClO4ClO3ClO2ClO-
perchlorate
chlorate
chlorite
hypochlorite
More oxygen atoms
Need to memorize these!
More For Exam #2…
Chemical Reaction Types
What is Stoichiometry ?
Balancing Equations
Empirical Formula (wt% to formula and
formula to wt%).
Limiting Reactants
Reaction Yield (%): actual yield (g) /
theoretical yield from balanced equation
(g)
Chemical Reaction Types
Combination (A+B → C)
Decomposition (C → A+B)
Single Displacement (A+BC → AC +B)
Double Displacement (AB+CD→ AD +CB)
Combustion (C + O2 → CO2 + Heat)
Redox (Oxidation / Reduction)
• Electrons are exchanged
2
More (2) For Exam #2…
Solutions: number of moles = n = grams of
solute /molar mass
Solution Concentrations (M,m, %)
number of moles of solute in a specific volume
of solution = n = V x M
Dilution Problems: V x M = V x M
Dissociation of Ionic Compounds into solvated
ions. (Electrolytes → number of moles in
solution)
No “solubility” rules
More (3) For Exam #2…
Colligative properties:
BPt elevation
FPt depression
Osmotic Pressure
Plus much, much more !!!!!
☺
Colligative Solution Properties
Properties that depend on the number of
moles of solute present in solution….
1 mole NaCl
1 mol Na+ + 1 mol Cl-= 2
moles total in solution !
One mole of glucose sugar (C6H12O6)
dissolves into one mole of glucose !
Seawater has a total ion concentration
of 1.135M (1.14m) from its salts.
3
Calculating Freezing Point Depression
∆Tf = Tfo - Tf = Kf.m, where:
•
•
•
•
•
∆Tf = Change in Freezing Point
Tf = Observed Freezing Point
Tfo = Pure Solvent Freezing Point
Kf = 1.86°C/m - water freezing point depression constant.
m = Molality = moles of Solute / kg of Solvent
Example: What is the freezing point of a 1.14m
seawater solution ?
∆Tf = (1.86 °C / m).(1.14m) = 2.12°C
Tf = Tfo - ∆Tf
Tf = 0°C - 2.12 °C = -2.12 °C
Calculating Boiling Point Elevation
∆Tb = Tb - Tbo = Kb.m, where:
•
•
•
•
•
∆Tb = Change in Boiling Point
Tb = Observed Boiling Point of solution
Tbo = Pure Solvent Boiling Point
Kb = 0.52°C/m - water boiling point elevation constant.
m = Molality = moles of Solute / kg of Solvent
For chloride ion in seawater:
0.558M or 0.565m - Almost the same!
Example: What is the boiling point for a 1.14m
seawater solution ?
∆Tb =
(0.52 °C / m).(1.14m) = 0.60°C
Tb = ∆Tb + Tbo
Tb = 0.60°C + 100°C = 100.6 °C
Calculating Osmotic Pressure (π)
π
= MRT where:
•M = Molarity (mol/L)
•R = 0.082 atm.L / mol.K
•T = Temperature (K)
Units of
π: atm
Note similarity to
PV=nRT
π
across a membrane
Example: Calculate
separating pure water from seawater at 25ºC.
4
Chemistry
The Science in Context
Chapter 16
Solubility Product Constant or
“Solubility Product”
For electrolytes and strong acids / bases, the
compounds totally dissociate into ions in solution.
NaCl (aq) → Na+ (aq) + Cl- (aq)
(Complete Dissociation to Ions)
For poorly soluble compounds, only a small
fraction dissolves and yields ionic
species in solution.
Example: Water seeping through
caves…..It tries to dissolve
calcium carbonate
Solubility Product Constant or
“Solubility Product” (Ksp)
Example: Calcium carbonate in water…
CaCO3(solid) ↔ Ca2+ (aq) + CO32- (aq)
Ksp = [Ca2+ ] x [CO32-] where
[ ]= concentration in moles / L
a.k.a. “molar concentration” or just
“solubility”
Ksp = 5.0 x 10-9 (Low !)
5
“Solubility Product” (Ksp)
More Generally:
Ionic Compound Dissolves: MmXx ↔ mM+ + xX“equilibrium”
Definition of Ksp = [M]m . [X]x
Ksp values are tabulated for many
compounds ….. Appendix Table A5.3
CaF2 in water, Ksp = [Ca2+] x [F-]2 = 3.9 x 10-11
Appendix Table A5.3
More on “Solubility Product” (Ksp)
Caution, Beware, This Gets Tricky Here !!
Dissolving CaF2 in water ↔ Ca2+(aq) + 2F- (aq)
What is the solubility of CaF2 ?
Solubility = [CaF2 ] in mol / L (This is the fraction
that dissolves)
If you start with 1 mol of CaF2 you can get one mole
of Ca2+ and 2 moles of F- in solution if it all dissolves
(it doesn’t). So, [s] ↔ [s] + [2s]
6
More on “Solubility Product” (Ksp)
If Ksp = [Ca2+] . [F-]2 = 3.9 x 10-11
And, [Ca2+] = s and [F-] = 2s
Then: s . [2s]2 = 4s3 = 3.9 x 10-11
So, s = cube root of 3.9 x 10-11 =
2.1 x 10-4 mol / L = M
2.1 x 10-4 M = [CaF2] = It’s “solubility”
“Solubility Product” (Ksp) Continued
Definition of Ksp = [M]m . [X]x
Where: MmXx ↔ mM+ + xXAnd: s = number of moles of MmXx that
dissolve, then
Ksp = (s.m)m
.
(s.x)x
See Text: page 825
What Real Use is Ksp ?
Find the solubility of barium sulfate in water.
The Ksp is 1.0 x 10-10.
1 mole BaSO4 (s)
[s]
1 mole Ba2+ (aq) + 1 mole SO42- (aq)
[s]
+ [s]
Ksp = [Ba2+][SO42-] = 1.0 x 10-10
[Ba2+] = [SO42-] = s mole/l
[s]2 = 1.0 x 10-10
[s] = 1.0 x 10-5 mole/l = “solubility” of BaSO4 (s)
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What Real Use is Knowing Ksp ?
Find the solubility of lead chloride in water.
The Ksp is 1.6 x 10-5.
PbCl2 (s)
Pb2+ (aq) + 2Cl- (aq)
Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5
[Pb2+] = s mole/l ; [Cl-] = 2s mole/l
Ksp = s(2s)2 = 4s3 = 1.6 x 10-5
s3 = 4 x 10-6 ; s = 1.6 x 10-2
[s] = 1.6 x 10-2 mole/l = “Solubility” of PbCl2(s)
Common Ion Effect in Solution
“The solubility of a slightly soluble
salt is decreased by the presence of
a second solute that furnishes a
common ion”
Common Ion Effect in Solution
LeChatelier’s Principle
CaF2 ↔ Ca++(aq) + 2F- (aq)
If you add extra F- (or Ca++) to the
aqueous solution, the “equilibrium”
is shifted to the left !
CaF2
Ca++(aq) + Extra 2F- (aq)
*** For example, barium sulfate is less soluble in sea
water than fresh water (Text Page 826-27)
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Ksp and the “Common Ion Effect”
Find the solubility of lead chloride in 0.10 M NaCl.
The Ksp is 1.6 x 10-5; Let [s] = solubility of PbCl2
Cl- is a “common ion”, but assume that it’s absent
for now…..
Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5
Assuming No Common Ion: [Pb2+] = s mole/L
[Cl-] = 2s mole/L
Ksp = s(2s)2 = 4s3 = 1.6 x 10-5
[s] = 1.6 x 10-2 mole/L )
Ksp and the “Common Ion Effect”
Find the solubility of lead chloride in 0.10 M NaCl.
The Ksp is 1.6 x 10-5; Let [s] = solubility of PbCl2
Now, assume that Cl- is a “common ion”
Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5
[Pb2+] = s mole/l
[Cl-] = 0.1 + 2s mole/l ≈ 0.1 mole/l
Ksp = s(0.1)2 = 1.6 x 10-5
[s] = 1.6 x 10-3 mole/L
(vs. 1.6 x 10-2 mole/L )
Almost Finished with Ksp …..
For AB ↔ A+ + B“equilibrium”
Definition of Ksp = [A] . [B]
This assumes that the concentration of the
“solute” has not changed much during the solution
process where the original compound disassociates
into ions. Not a bad assumption ….Mostly
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Nearly Finished with Ksp …..
For example, at “equilibrium”:
One mole of barium sulfate
(BaSO4 , 233.4 grams)
yields only 9.5 x 10-6 grams of Ba+2
ion and 9.5 x 10-6 grams of SO4+2 ion
in aqueous solution = 0.000019 g
Final Mass of BaSO4 = 233.4 – 0.000019
= 233.4 more or less.
Original Solute
+ . .
Dissolved Ions
Residual Solute
Almost Finished – Continued
For AB ↔ A+ + BIn reality Ksp = [A] . [B]
[AB]
Since [AB] doesn't change much and its
very high, chemists kind of ignore it but
it’s part of the Ksp values reported in
books.
However ….
For AB ↔ A+ + BKequilibrium =
[A] . [B]
[AB]
If, say, 50% of the [AB] solute “dissociates”
and 50% remains, then you need to use the
expression above to solve for K, the so-called
“equilibrium constant”.
Original Solute
Residual Solute
+
+
Dissolved Ions
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