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Chemistry 1B Summer School 2016: Part Two
Topic 9 – Phase diagrams and Solubility
1
Topic 10 – Transition Metals and Coordination Chemistry
24
Topic 11 – Kinetics
43
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2012 (John Wiley)
ISBN: 978 1 74246 707 8
Topic 9 - Phase diagrams and Solubility
Phase diagrams
The phase (solid, liquid or gas) of a substance depends on temperature, T and pressure, P. A
phase diagram describes the conditions of T & P under which each phase is stable.
Question: Indicate the phase change indicated by the following term. The first is given as an
example.
S→L
Melting
Condensation
Freezing
Sublimation
Vapourisation
Deposition
Key Terms:
• Triple Point: The conditions under which three phases (solid, liquid and gas) are at equilibrium.
• Critical Point: The highest temperature at which it is possible to liquefy a gas.
• Normal Boiling Point: The boiling point at 1 atmosphere pressure.
• Normal Melting Point: The melting point at 1 atmosphere pressure.
Carbon dioxide
Water
P
P
critical point
solid
critical point
liquid
solid
liquid
1 atm
triple point
1 atm
- 78 oC
gas
gas
0 oC
T
1
100 oC
T
2.5
vapour pressure (atm)
[WS 27] Vapour Pressure
A liquid boils when the pressure of the vapour
escaping from the liquid is equal to the pressure
exerted on the liquid by its surroundings.
The pressure of the vapour escaping is called
the vapour pressure.
2
1.5
1
0.5
0
0
The graph opposite shows how the vapour
50
100
temperature ( ˚C)
pressure of water changes with temperature.
Questions:
1. The normal boiling point of water is 100 ˚C. What is the vapour pressure of water at this
temperature?
2. The safety valve on commercial pressure cookers is set to allow a maximum pressure of 2 atm.
What is the boiling point of water at this pressure?
3. The air pressure at the highest point in Australia is Mount Kosciuszko (2,228 m) is around 0.75
atm. What is the maximum temperature of a cup of tea on top on the mountain?
4. The graph below shows the low temperature part of the vapour pressure diagram for water.
vapourpressure(atm)
0.05
(a) How could we make water boil at 25 ˚C?
0.04
0.03
0.02
(b) The pressure on the surface of Mars is about
0.01
0.006 atm. If the temperature on Mars in summer
0
0
10
20
30
reaches 20 ˚C, what phase would water be in?
temperature(°C)
2
Supercritical Fluids
•
Critical point is the T & P at which the boundary between liquid & vapour disappears.
•
If sealed tube of CO2 liquid + vapour heated, liquid-vapour phase boundary
disappears at 31 °C and 73 atm:
•
Supercritical fluids: substance above critical temperature (Tc).
•
Can behave as solvents dissolving a wide range of substances; also new industrial reaction
medium.
•
Environmentally friendly:
CO2 is harmless solvent
•
Easily removed (open valve to release pressure).
•
Removal
of
caffeine
from
coffee
beans
by
supercritical
removes caffeine from beans but leaves flavour and aroma components.
•
Also used in dry-cleaning & as solvent for chemical industry.
[WS 27] Phase Diagram of CO2
A phase diagram allows us to predict the phase
(solid, liquid or gas) of a substance that is stable
at any given temperature and pressure. The
phase diagram opposite is for CO2.
The phases are divided up by lines, which
represent the temperatures and pressures at
which the substance is in equilibrium between
the phases.
3
CO2:
Questions:
1. Add and label the following points on the diagram and hence predict which phase CO2 will be in
under the stated conditions
(a)
-80 ˚C and 1 atm pressure
(b)
25 ˚C and 1 atm pressure
(c)
25 ˚C and 10 atm pressure.
2. Draw a line connecting points (a) and (b). Describe in words what happens to CO2 as:
(i)
it is warmed from -80 ˚C to -25 ˚C at 1 atm pressure
(ii)
it is cooled from 25 ˚C to -80 ˚C at 1 atm pressure
3. Draw a line connecting points (b) and (c). Describe in words what happens to CO2 as:
(i)
the pressure is increased from 1 atm to 10 atm at 25 ˚C
(ii)
the pressure is reduced from 10 atm to 1 atm at 25 ˚C
4. Carbon dioxide sequestration has been suggested as a way of storing CO2 from burning fossil
fuels so that it does not enter the atmosphere. It involves pumping liquid CO2 into empty oil fields
such as the 2 km deep Otway Basin in western Victoria. What is the minimum pressure required to
liquefy CO2?
4
5. Use the following data to draw a phase diagram for hydrogen. Label the solid, liquid and gas
regions of the graph. NB the pressure scale is logarithmic.
Normal melting point 14 K
Normal boiling point
20.3 K
Triple point
14 K, 0.07 atm
Critical point
33 K, 13 atm
Hydrogen
P / atm
10
1
0.1
5
10
15
20
5
25
30
35
T/K
Solubility
What determines solubility?
ClCl
-
Na+ ClClNa+
+
Na
Cl-
Cl
-
+
Na
H O
Cl Cl-
+
+
Na Cl ClNa+
H
H
Cl-
H
H
O
+
H2O Na
H2O
OH2
OH2
OH2
O
H
H
+
HOH
HOH
Cl
O H
-
HOH
HOH
HOH
Solubility Equilibria
Solubility - 'how much dissolves?'
In solubility SI data book, the solubility is quoted as the grams of solute that dissolves in 100 mL
(100 g) of water.
A note on some units:
1. mass%
= (grams of solute/grams of solution) x 100
e.g. 5.0 g of sugar in a cup (250 mL) = 2 %
2. molarity
= moles solute/litre of solution
e.g. [sugar] = (50 g/342 g mol-1 x 0.25 L) = 0.058 M
3. ppm
= mass% x 10 000
this is 1 mg of a solute in 1 L of solution
Saturated solution ⇒ where the maximum amount of solid is dissolved,
Some solid is undissolved so equilibrium between solution and some undissolved solid.
solute + solvent ⇌ solution
6
Solubility Equilibrium Constant, Ksp
Ksp is the solubility equilibrium constant
(Note: Kso may also be used and is equivalent)
Ksp only involves molecular solutes - solubility and solubility equilibrium constant are the same. It
only involves the concentration of the resulting product as the reactants are pure solids. As all
solids are slightly soluble there will always be some ions in solution.
Ksp
= [products] / [reactants]
= [sugar(aq)] / [sugar(s)] = [sugar(aq)] / 1 (by definition [solid] = 1)
e.g.
for sucrose at 25 ˚C Ksp ≈ 6 M
(solubilty in water is 2000 gL-1, mw = 342.30)
Solubility Product, Ksp
For ionic solutes - solubility is not equal to solubility equilibrium constant. The solubility product
constant is the equilibrium constant for a slightly soluble (or nearly insoluble) ionic compound:
MX(s) ⇌ M+ (aq) + X- (aq)
Ksp = [M+][ X-]
Solubility is how much MX dissolves, in this case = [M+] = [X-]; equilibrium constant is the solubility
product.
At equilibrium the product of ion concentrations is a constant at each temperature. If the
concentration of one ion increases the concentration of the other must decrease to keep equilibrium
product constant.
Example:
How much AgCl dissolves in water at 298 K?
Ksp (AgCl) = 1.6 x 10
-10
M
2
+
every x moles of AgCl produces x moles of Ag and x moles of Cl
7
-
+
-
AgCl (s) ⇌ Ag (aq) + Cl (aq
x
+
x
x
-
Ksp = [Ag ][Cl ] = 1.6 x 10-10 M2 = [x][x] = x2
∴
Question:
x = 1.3 x 10-5 M (solubility in water)
What is the solubility of AgI at 298 K? (Ksp (AgI) = 8 x 10-17 M2)
[WS 31]
Table 1. The results of adding solid Mg(OH)2 to a litre of water at 25 ˚C:
Total amount of Mg(OH)2
added
(g)
(moles)
0.000963
1.65 × 10
-5
0.004815
8.26 × 10
-5
-4
Concentration of ions in
solution
2+
[Mg (aq)]
[OH (aq)] /
/M
M
-5
-5
1.65 × 10
3.30 × 10
8.26 × 10
-5
-4
1.65 × 10
Mass of
Mg(OH)2
that does not
dissolve (g)
2+
[Mg (aq)][O
2
H (aq)]
0.00000
1.80 × 10
-14
-4
0.00000
2.25 × 10
-12
-4
-11
0.009590
0.009630
0.009700
0.010000
0.015000
0.015000
1.64 × 10
-4
1.65 × 10
-4
1.66 × 10
-4
1.71 × 10
-4
2.57 × 10
-4
2.57 × 10
1.64 × 10
-4
1.65 × 10
-4
1.65 × 10
-4
1.65 × 10
-4
1.65 × 10
-4
1.65 × 10
3.29 × 10
-4
3.30 × 10
-4
3.30 × 10
-4
3.30 × 10
-4
3.30 × 10
-4
3.30 × 10
0.00000
0.00000
0.00007
0.00037
0.00537
0.00537
1.78 × 10
-11
1.80 × 10
-11
1.80 × 10
-11
1.80 × 10
-11
1.80 × 10
-11
1.80 × 10
0.020000
0.015000
3.43 × 10
-4
2.57 × 10
-4
1.65 × 10
-4
1.65 × 10
-4
3.30 × 10
-4
3.30 × 10
-4
0.01037
0.00537
1.80 × 10
-11
1.80 × 10
-11
Questions:
1.
What is the maximum number of moles of Mg(OH)2 that can be dissolved in a litre of water?
2.
What is the maximum value of the expression [Mg2+(aq)][OH-(aq)]2?
8
3.
Is it possible for the value of the expression [Mg2+(aq)][OH-(aq)]2 to be less than the value
your answer to Q2?
4.
What will happen if the concentrations of Mg2+(aq) and OH-(aq) are such that the expression
[Mg2+(aq)][OH-(aq)]2 is:
(a)
less than your answer to Q3?
(b)
more than your answer to Q3?
Solubility Product of 2:1 Salts
Solubility product equilibrium constants differ between 1:1 and 2:1 electrolytes
Example:
For PbCl2 (1:2 salt)
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)
for x moles dissolved
x
Ksp = [Pb2+] [Cl-]2 = [x] [2x] 2
Ksp (PbCl2)
∴
Example:
2x
units of Ksp are M3
= 1.6 x 10-5 M3 = x.(2x)2 = 4x3
x = 1.6 x 10-2 M
Alternative type of problem: a saturated solution of PbCl2 at 298 K has
3.2 x 10-2 M Cl-; what is Ksp of PbCl2?
Ksp = (1.6 x 10-2)(3.2 x 10-2)2 = 1.6 x 10-5 M3
Question:
What is the solubility of Ag2S? Ksp (Ag2S) = 6 x 10-50 M3
9
[WS 31] The solubility product
A solution is described as saturated if as much solid has dissolved as is possible so there is an
equilibrium:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
If the reaction is at equilibrium, the expression [Mg2+(aq)][OH-(aq)]2 is the equilibrium constant for
the reaction. The concentration of the solid is constant so does not appear in the expression.
Clearly, this is only true if solid is present: if there is no solid present, the solution is not saturated
and the reaction is not at equilibrium.
The equilibrium constant is known as the ‘solubility product’ and given the symbol Ksp:
Ksp = [Mg2+(aq)][OH-(aq)]2
Questions:
1. If x moles of Mg(OH)2 dissolve in 1.00 L of water, what will [Mg2+(aq)] and [OH-(aq)] be in terms
of x?
[Mg2+(aq)] =
[OH-(aq)] =
2. Ksp of Mg(OH)2 is 1.8 × 10-11 at 25 ˚C, using your answer to Q1, work out [Mg2+(aq)] and [OH-(aq)].
3. Using the same approach, write down the formula for Ksp in terms of x for salts of the following
stoichiometry:
(a)
MX
(b)
M 2X
(c)
MX3
4. Work out the molar solubility of the following silver salts at 25 ˚C.
(a)
AgCl; Ksp = 1.8 × 10-10
(b)
Ag2CrO4; Ksp = 2.6 × 10-12 (Hint: dissolves to give Ag+(aq) and CrO42-(aq))
Relative Solubility
Note It is important to take care when comparing the relative solubility using Ksp values.
Ksp = [cation][anion], if x is the solubility in moles/L then x = (Ksp)1/2:
AgI
Ksp = 1.5 x 10–16
x = (Ksp)1/2 = 1.2 x 10–8 M
CuI
Ksp = 5.0 x 10–12
x = (Ksp)1/2 = 2.2 x 10–6 M
CaSO4
Ksp = 6.1 x 10–5
x = (Ksp)1/2 = 7.8 x 10–3 M
10
Ksp of CuS is larger, but more Bi2S3 dissolves:
CuS
Ksp = 8.5 x 10–45
Ksp= x2
x = 9.2 x 10–23 M
Ag2S
Ksp = 1.6 x 10–49
Ksp = (4x3)
x = 3.4 x 10–17 M
Bi2S3
Ksp = 1.1 x 10–73
Ksp = (4x2 *27x3)
x = 1.0 x 10–15 M
Ion Product, Q
To determine whether an equilibrium system will go in the forward or reverse direction requires that
we evaluate the reaction quotient, Q. This is the product of ionic concentrations before solubility
equilibrium is established.
For the reaction:
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq); QC = [Pb2+]i[Cl-1]i
where i = initial concentration (same expression as equilibrium constant), c – in terms of
concentration.
If Qc > Ksp then a precipitate forms
If Qc = Ksp then the solution is saturated
If Qc < Ksp then more solute can dissolve
Example: 1. Does a precipitate of AgCl(s) form if equal volumes of 1 mM HCl and 1 mM AgNO3
solutions are mixed?
+
AgCl (s) ⇌ Ag (aq) + Cl-(aq)
+
-4
[Ag ] = 5 x 10 M (after mixing)
-4
[Cl-] = 5 x 10 M (after mixing)
Q = [Ag+][ Cl- ]
-7
2
= 2.5 x 10 M > Ksp = 1.6 x 10
-10
M
2
∴ AgCl(s) precipitates, i.e. reaction moves to left toward equilibrium.
11
-5
2. What about equal volumes of 2.0 x 10 M solutions?
-5
[Ag+] = 1.0 x 10 M (after mixing)
-5
[ Cl- ] = 1.0 x 10 M (after mixing)
Q = [Ag+][ Cl- ]
= 1.0 x 10
-10
2
M < Ksol = 1.6 x 10
-10
M
2
∴ AgCl(s) does not precipitate.
Question: Lead iodide is an almost insoluble salt with a dense golden yellow colour. (It is used in
ornamental work requiring a gold-like colour). The solubility product equilibrium constant is 7.1x10-9
mol3/L3. Calculate the solubility of PbI2 (assume excess solid).
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq); Ksp=7.1x10-9
Initial:
0
0
mol/L
Change:
+x
+2x
mol/L
Equilibrium:
x
2x
mol/L
Solve for x and work out how many mol/L of PbI2 will dissolve.
12
[WS 31] As we saw earlier, the solubility product gives the maximum values of the ion
concentrations that are allowed. The concentrations can certainly be less but their product cannot
be more!
If their concentrations such that their product is less than Ksp, then more solid can dissolve.
If [Mg2+(aq)][OH-(aq)]2 < Ksp then more solid can dissolve
If their concentrations are such that their product is more than Ksp then the concentrations must
reduce: precipitation must occur.
If [Mg 2+(aq)][OH-(aq)]2 > Ksp then precipitation must occur
The value of the product can thus be used to predict whether dissolution or precipitation can occur.
Because of its importance, it is called the ‘ionic product‘ and given the symbol Qsp:
Qsp = [Mg2+(aq)][OH-(aq)]2
If Qsp < Ksp then dissolution will occur. If Qsp > Ksp then precipitation will occur.
Questions:
1.
500.0 mL of 0.12 M NaOH solution with 500.0 mL of 0.10 M Mg(NO3)2 are mixed.
(a)
Assuming that no reaction occurs, what would [Mg2+(aq)] and [OH-(aq)] be after mixing?
(b)
Write down the value of the ionic product, Qsp.
(c)
Does a precipitate form?
13
Common Ion Effect
The initial presence of an ion in solution may suppress the solubility of a solid if the solid also
contains the same ion – the equilibrium constant remains the same.
Example:
How much AgCl dissolves in 0.1 M HCl?
AgCl (s) ⇌ Ag+(aq) + Cl-(aq)
x
(0.1+ x) ≈ 0.1
Ksp = [Ag+][Cl-] = 1.6 x 10-10 M2 = [x][0.1 +x] ≈ 0.1x
∴ x = 1.6 x 10-9 M
(This is << than 10-5 M in H2O)
Question: What is the solubility of AgBr in 0.05 M sodium bromide solution at 298 K? (Ksp (AgBr) =
5 x 10-13 M2)
14
[WS 32] Le Châtalier’s principle states: If a change is imposed on a system, the position of
equilibrium will shift in a direction that tends to reduce that change
If the concentration of a reactant is increased, the equilibrium responds by producing more
products. If the concentration of a product is increased, the equilibrium responds by producing
more reactant.
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
PbCl2 is not very soluble in water. The picture shows a test tube containing a
saturated solution of lead chloride in contact with a precipitate of solid.
The effect on this solubility of adding Pb2+(aq) or Cl-(aq) ions from another source is
called the common ion effect.
Questions:
1.
Write down the solubility product expression, Ksp, for lead chloride.
2.
Sodium chloride dissolves completely to give Na+(aq) and Cl-(aq) ions. If sodium chloride is
added to the saturated solution, what would be the effect on the solubility of lead(II) chloride?
(Hint: consider how the equilibrium written above would shift, according to Le Châtelier’s
principle, when these ions are added).
3.
If sodium chloride is added so that [Cl-(aq)] = 0.5 M, rearrange your Ksp expression to give
[Pb2+(aq)].
15
4.
If Ksp = 1.7 × 10-5 at 25 oC, what is [Pb2+(aq)] in 0.5 M NaCl?
5.
What is the effect of adding extra PbCl2(s) to the test tube? (Be careful!)
6.
What is the solubility of PbCl2(s) chloride (in mol L–1) in 1.00 M magnesium chloride solution?
(Hint: what is [Cl-(aq)] for a 1.00 M magnesium chloride solution?)
Solubilities of Hydroxides
Example: 1. What is solubility of Zn(OH)2 in water at 298 K? Ksp = 1.0 x 10-15 M3
Zn(OH)2 ⇌
x dissolves
Zn2+ + 2 OHx
2x
Ksp = (x)(2x)2 = 4x3 = 1.0 x 10-15 M3
∴ x = 6.3 x 10-6 M
2. What is the pH of a saturated solution of Zn(OH)2 ?
[OH-] = 2x = 2 (6.3 x 10-6 M) = 1.2 x 10-5 M
∴ pOH = 4.9
pH = 14.0 - pOH = 9.1
Hydroxides are generally insoluble - except Group 1 (alkali metals) Ca(OH)2 and Ba(OH)2.
Divalent hydroxides are moderately insoluble; trivalent hydroxides are exceedingly insoluble.
16
Example:
What is solubility of Fe(OH)3 in water at 298 K? Ksp = 1.0 x 10-38 M4
Fe(OH)3 ⇌
Fe3+ + 3 OH-
x dissolves
x
3x
K = [Fe3+ [OH-]3 = [x] [3x]3 = 27 x
∴ x = 1.4 x 10
-10
4
M
pH and Hydroxide Solubility
Sometimes it is necessary to account for other reactions aqueous ions might undergo.
•
For example, if the anion is the conjugate base of a weak acid, it will react with H3O+.
Terefore the solubility will be affected by pH.
Example:
1. What are solubilities of Zn(OH)2 and Fe(OH)3 at pH 7?
Zn(OH)2: Ksp = 1.0 x 10-15 M3
2+
-7 2
= [Zn ][OH-]2 = [x ] [10 ]
∴ [Zn2+] = [x] = 0.1 M
Fe(OH)3: Ksp = 1.0 x 10
3+
-38
∴ [Fe ] = [x] = 1.0 x 10
4
-7 3
M = [x] [10 ]
-17
M
1. At what pH is Fe(OH)3 soluble to 0.1 M?
Ksp = 1.0 x 10
-38
3
∴ x = 1.0 x 10
∴ x = 4.6 x 10
4
M = [0.1] [x]
-37
-13
M
3
3
M
∴ pOH = 12.3 and pH = 1.7
pH dependence of solubilities can be calculated and plotted - pH control can be used to separate
metal ions.
17
Application: Al(OH)3 is used to clarify drinking water - how much Al (in ppm) remains in water at
pH 6.6?
3+
3
Ksp = [Al ][OH-] = 3 x 10
at pH 6.6, [OH-] = 10
3+
-7.4
-34
M
4
-8
= 4 x 10 M
-34
-8 3
∴ [Al ] = 3 x 10 / (4 x 10 ) = 5 x 10
and
5 x 10
-12
-1
-12
3
M
mol L-1 x 27 g mol x 10 mg L
-7
-1
-1
= 1 x 10 mg L (ppm)
Example: Consider the following equilibrium – is calcium oxalate likely to be more soluble in acidic
solutions?:
CaC2O4 (s) ⇌Ca2+ (aq) + C2O42- (aq)
Because the oxalate ion is conjugate to a weak acid (HC2O4), it will react with H3O+.
C2O42- (aq) + H3O+ (aq) ⇌HC2O4- (s) + H2O (l)
C2(OOH)2 à C2(OOH)OO-
C2(OOH)OO à C2(OOH)2
(pKa = 1.23)
`
(pKa = 4.12)
According to Le Chatelier’s principle, as C2O42- ion is removed by the reaction with H3O+, more
calcium oxalate dissolves.
Therefore, calcium oxalate should be more soluble in acidic solution (low pH) than in pure
water.
Case Study 1: Kidney Stones and Calcium Oxalate
•
The presence of a common ion reduces the solubility of the salt. In some cases considerably. eg Ca and Kidney Stones
•
Each salt contributes the same cation Ca2+. The effect of the calcium ion provided by the
calcium chloride is to make calcium oxalate less soluble than it would be in pure water
•
CaC2O4(s) ⇌Ca2+(aq) + C2O42-(aq)
•
Ksp of Ca(C2O4) = 1.96 x 108 M2 = [Ca2+][C2O42–]
18
Addition of Ca2+ pushes the equilibrium to the left causing calcium oxalate to precipitate from
solution.
Originally it was thought that consumption of too much calcium could promote the development of
calcium kidney stones.
However, current evidence suggests that the consumption of low-calcium diets is actually
associated with a higher overall risk for the development of kidney stones.
This is perhaps related to the role of calcium in binding ingested oxalate in the gastrointestinal tract.
As the amount of calcium intake decreases, the amount of oxalate available for absorption into the
bloodstream increases; this oxalate is then excreted in greater amounts into the urine by the
kidneys.
In the urine, oxalate is a very strong promoter of calcium oxalate precipitation, about 15 times
stronger than calcium.
The formation of renal stones is related to the urine pH (range 4.5 – 8.0). Patients being treated for
renal calculi are frequently given diets or medications to change the pH of the urine so that kidney
stones will not form.
Calcium phosphate, calcium carbonate, and magnesium phosphate stones develop in alkaline
urine; when this occurs, the urine is kept acidic.
Uric acid, cystine, and calcium oxalate stones precipitate in acidic urine; in this situation, the urine
should be kept alkaline or less acidic than normal
19
[WS32] Metal hydroxides dissolve to give metal ions and hydroxide ions. For example,
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)
The position of the equilibrium (i.e. the solubility) is very sensitive to pH since this controls [OH-(aq)].
All forms of life depend on iron and the concentration of iron in the oceans and elsewhere is one of
the primary factors limiting the growth rates of the most basic life forms. One reason for the low
availability of iron(III) is the insolubility of Fe(OH)3 which has a Ksp of only 1 x 10--39.
Questions:
1.
Write down the expression for the solubility product, Ksp, for Fe(OH)3.
2.
The pH of the oceans is currently 8.179. Use this to work out [OH-(aq)].
3.
If x moles of Fe(OH)3 dissolve in 1.00 L of water, [Fe3+(aq)] = x mol L-1. Use your answers to
Q1 and Q2 to work out x.
4.
If the amount of CO2 in the atmospheres increases, the pH of the oceans will decrease due
to the equilibrium below. What will happen to [Fe3+(aq)]?
CO2(g) + H2O(l) ⇌ HCO3- + H3O+(aq)
20
Thermodynamics of Solubility
Temperature dependence
The solubility of most substances increases with T, some show little change (e.g. NaCl), for a few
solubility decreases with increase in T.
solute + solvent ⇌ solution ± heat
if dissolution
exothermic,
K decreases with T
endothermic, K increases with T
Heat of solution – this is a small difference between two large quantities (lattice energy and
hydration energy):
∆Ghyd = ∆Hhyd - ∆Hlat - T∆Shyd
Enthalpy of Hydration
The smaller the ionic radius, the more energy is released by surrounding an ion with water
molecules. The more highly charged the ion, the more energy is released.
–ΔHhyd - Always favours dissolution
Lattice Energy
Energy released when crystal is formed from free ions
∆Hlatt ∝ ca
∆Hlatt ∝ cation charge x anion charge/cation radius + anion radius
∆Hlatt ∝ q2/r
ΔHlatt – Always opposes dissolution
Entropy of Hydration
Solid à Dissolved ions
ΔShyd – Almost always favours dissolution
+
Na (g)
-
+ Cl (g)
ΔHhyd = - 783 kJ mol -1
-LE = 786 kJ mol -1
+
NaCl (s)
Na (aq) + Cl (aq)
ΔHsol = + 3 kJ mol -1
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Blackman Fig 10.13
Case Study 2: Cave Formation
Limestone - mainly calcium carbonate, CaCO3
CaCO3 dissolves in CO2-containing ground-water:
CO2(g) ⇌ CO2(aq)
CaCO3(s) + CO2(aq) + H2O(l) ⇌ Ca2+(aq) + 2HCO3–(aq)
Dissolution of the limestone produces limestone caves.
Stalactites & Stalagmites
The reverse of these processes occurs as water saturated with calcium hydrogencarbonate drips
from the ceiling of the cave.
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Loss of CO2 to the atmosphere:
CO2(aq)
⇌
CO2(g)
drives the precipitation of calcium carbonate:
2 HCO3-(aq) + Ca2+(aq)
⇌ CaCO3(s) + H2O(l) + CO2(aq)
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