Transcript Slide 1

AQUEOUS EQUILIBRIA
 Salts
 Hydrolysis
 The Common-Ion Effect
 Buffer Solutions
 Acid-Base Titrations
 Solubility
Brown et al., Chapter 15, 569 - 606
CHEM171 – Lecture Series One : 2012/01
SALTS AND HYDROLYSIS
From the Brønsted Lowry Theory, an acid is a proton donor and a
base is a proton acceptor.
A salt arises from the reaction of an acid with a base.
e.g., the reaction between HBr and Mg(OH)2 leads to MgBr2.
2HBr(aq) + Mg(OH)2(aq)  MgBr2(aq) + 2H2O(l)
When a solid sample of a salt (the solute) is placed in water
(the solvent) the salt dissolves to give a solution.
NaCl(s)
solute
+
(aq)
solvent

Na+(aq) + Cl-(aq)
solution
SALTS dissociate FULLY in water to form ionic solutions and
are thus STRONG ELECTROLYTES.
CHEM171 – Lecture Series One : 2012/02
Weak electrolytes do not dissociate fully into ions when in
solution.
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
EXAMPLE
What percentage of molecules have dissociated in a 0.100 mol
dm-3 solution of acetic acid ?
Ka(CH3COOH) = 1.75 x 10-5
SOLUTION
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
Initial (M)
Change (M)
Equili. (M)
0.100
-x
(0.100 – x)
0
+x
x
CHEM171 – Lecture Series One : 2012/03
0
+x
x
Assume that x  0.100
x2 = 1.75 × 10-6  x = 1.32 × 10-3
% dissociated = (1.32 × 10-3/0.1) × 100% = 1.32%
SALTS CAN REACT WITH WATER TO BE EITHER
ACIDIC, BASIC OR NEUTRAL.
CHEM171 – Lecture Series One : 2012/04
SALTS OF STRONG ACIDS AND STRONG BASES
NaOH + HCl  NaCl + H2O
NaCl
H2O

Na+
+
Cl-
Na+ + H2O  no reaction
and
Cl- + H2O  no reaction
a solution of NaCl is neutral.
CHEM171 – Lecture Series One : 2012/05
SALTS OF WEAK ACIDS AND STRONG BASES
CH3COOH + NaOH  CH3COONa + H2O
CH3COONa
H2O

CH3COO¯ + Na+
Na+ does not react with water.
CH3COO undergoes hydrolysis
CH3COO + H2O  CH3COOH + OH
 pH of sodium acetate solution > 7 (basic).
CHEM171 – Lecture Series One : 2012/06
SALTS OF STRONG ACIDS AND WEAK BASES
HCl + NH3  NH4Cl
NH4Cl
H2O

NH4+
+
Cl-
NH4+ + H2O  NH3 + H3O+
 pH of NH4Cl solution < 7 (acidic)
CHEM171 – Lecture Series One : 2012/07
SALTS OF WEAK ACIDS AND WEAK BASES
Solution pH depends on relative strengths of weak acid
and weak base.
HNO2 + NH3 
NH4NO2
H2O
 NH4+ + NO2-
NH4+ + H2O  NH3 +
NO2- + H2O 
Kb < Ka
NH4NO2
H3O+
Ka = 5.6 × 1010
HNO2 + OH-
Kb = 2.2 × 1011
 solution acidic
acidic if Ka (cation) > Kb (anion)
neutral if Ka (cation) = Kb (anion)
basic if Ka (cation) < Kb (anion)
CHEM171 – Lecture Series One : 2012/08
EXAMPLE
Calculate the pH of a 0.0924 mol dm-3 solution of NH4Cl at
25oC, Ka = 5.70 × 10-10.
SOLUTION
NH4+ (aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Initial (M)
Change (M)
Equili. (M)
0.0924
-x
(0.0924 – x)
0
+x
x
x = 7.26 × 10-6
From,
pH = - log [H3O+],
pH = 5.14
CHEM171 – Lecture Series One : 2012/09
0
+x
x
EXAMPLE
Calculate the pH of a 0.154 mol dm-3 solution of sodium
acetate, Kb = 5.70 × 10-10.
SOLUTION
CH3COO-(aq) + H2O(l) ⇌ CH3COO-(aq) + OH-(aq)
Initial (M)
0.154
Change (M)
-x
Equili. (M) (0.154 – x)
0
+x
x
x = 9.38 × 10-6
pOH = 5.03, therefore pH = 8.97
CHEM171 – Lecture Series One : 2012/10
0
+x
x
COMMON ION EFFECT
CH3COOH + H2O ⇌ H3O+ + CH3COOWhat happens if sodium acetate is added?
Can be explained by Le Chatelier’s Principle.
When a reactant containing a given ion is added to an
equilibrium mixture that already contains that ion, the
position of equilibrium shifts away from forming more of it.
CHEM171 – Lecture Series One : 2012/11
EXAMPLE
What is the pH of a solution made by adding 0.30 mol
acetic acid (HC2H3O2) and 0.15 mol of sodium acetate
(NaC2H3O2) to enough water to make 1.0 dm3 of solution?
Ka = 1.8 x 10-5
SOLUTION
CH3COOH
Initial
Change
Equilibrium
0.30 M
-x M
(0.30 – x) M
⇌
H+
0
+x M
xM
+
CH3COO0.15 M
+x M
(0.15 + x) M
CHEM171 – Lecture Series One : 2012/12

x = 3.6 × 10-5 M = [H+]
pH = -log(3.6 × 10-5) = 4.44
EXERCISE FOR THE IDLE MIND
Practice Exercises: p 571, p 572 and Exercise 15.10 p 602
CHEM171 – Lecture Series One : 2012/13
BUFFERS
A buffer solution is a solution whose pH remains essentially
constant despite the addition of a small amount of either acid
or base.
Must have an acidic component to react with added OH- and a
basic component to react with added H3O+ ion.
Typical buffer is acetic acid (CH3COOH) and sodium acetate
(CH3COONa)
CH3COOH + H2O 
acid 1
CH3COO- + H3O+
base 1
CHEM171 – Lecture Series One : 2012/14
Some common buffer solutions are:
NaHCO3
+
Brønsted acid
NaH2PO4
Brønsted acid
Na2CO3
conjugate base
+
Na2HPO4
conjugate base
CHEM171 – Lecture Series One : 2012/15
pH < 7.00
Two components – a weak acid and the salt of a weak acid.
WEAK ACID
CH3COOH
acetic acid
+
+
SALT OF A WEAK ACID
CH3COO- Na+
sodium acetate
The weak acid is there to react with any OH- ions which might
be introduced into the system.
CH3COOH(aq) + OH-(aq) ⇌ CH3COO-(aq) + H2O(l)
Result - to convert a strong base (OH-) into a weak base
(CH3COO-). Although the pH will increase due to more base being
present, the increase will be relatively small as the acetate ion is a
weak base
(Kb = 5.7 × 10-10).
CHEM171 – Lecture Series One : 2012/16
The salt of the weak acid is there to react with any H3O+ ions
which might be introduced into the system.
CH3COO-(aq) + H3O+(aq) ⇌ CH3COOH(aq) + H2O(l)
Result - a strong acid (H3O+) has been converted into a
weak acid (Ka (CH3COOH) = 1.75 ×10-5). The pH of the
solution will decrease but only by a small amount.
This buffer solution can be described by the following
equilibrium:
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
CHEM171 – Lecture Series One : 2012/17
EXAMPLE
Calculate the pH of a solution which contains 0.1046 mol
dm-3 of CH3COOH and 0.1247 mol dm-3 of sodium acetate.
SOLUTION
We first determine the pH using a method described before.
CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
Initial
Change
Equilibrium
0.1046 M
-x M
(0.1046 – x) M
0
+x M
xM
0.1247 M
+x M
(0.1247 + x) M
CHEM171 – Lecture Series One : 2012/18

x = 1.47 × 10-5 M = [H+]
pH = -log(1.47 × 10-5) = 4.83
Can use another method to calculate the pH of a buffer.
Employ the equation:
Called the Henderson Hasslbach Equation.
CH3COOH is the proton donor and CH3COO-, the proton acceptor.
CHEM171 – Lecture Series One : 2012/19
[CH3COOH] = 0.1046 mol dm-3
[CH3COO-] = 0.1247 mol dm-3
Ka = 1.75 × 10-5,  pKa = -log[Ka]
 pH = 4.76 + log(0.1247/0.1046) = 4.84
CHEM171 – Lecture Series One : 2012/20
EXAMPLE
Calculate the pH of the solution prepared from adding
50.00 mL of 0.1011 mol dm-3 NaOH to 100.0 mL of a
0.1122 mol dm-3 CH3CH2COOH solution.
(pKa CH3CH2COOH = 4.88).
SOLUTION
Before (mol)
0.01122
0.005055
0
CH3CH2COOH + OH- → CH3CH2COO- + H2O
After (mol)
0.006165
0
0.005055
CH3CH2COOH + H2O → CH3CH2COO- + H3O+
Initial (M)
Change (M)
Equilibrium
0.0411
-x
(0.0411 – x)
0.0337
+x
(0.0337 + x)
CHEM171 – Lecture Series One : 2012/21
+x
x
x = 1.61 × 10-5 mol dm-3
pH = -log(1.61 × 10-5) = 4.79
OR
 pH = 4.88 + log(0.0337/0.0411) = 4.79
CHEM171 – Lecture Series One : 2012/22
pH > 7.00
Weak base and the salt of a weak base.
WEAK BASE
NH3
ammonia
+
+
+
SALT OF A WEAK BASE
NH4Cl
ammonium chloride
The weak base will take care of any added H3O+.
NH3(aq) + H3O+(aq) ⇌ NH4+ + H2O(l)
The salt of a weak base will react with any added OH-
NH4+ + OH-(aq) ⇌ NH3(aq) + H2O(l)
and the strong base gives way to a weak base (NH3).
CHEM171 – Lecture Series One : 2012/23
The equilibria which determines the final pH of the buffer will be
NH4+ + H2O(l) ⇌ NH3(aq) + H3O+(aq)
If instead we had written the equilibrium as
NH3(aq) + H2O(l) ⇌ NH4+ + OH-(aq)
then
CHEM171 – Lecture Series One : 2012/24
EXAMPLE
Calculate the pH of a solution prepared from 100.0 mL
of 0.02041 mol NH3 and 100.0 mL of 0.03267 mol
NH4Cl (Ka (NH4+) = 5.71 × 10-10).
SOLUTION
NH4+ + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Initial (mol)
0.03267
Change (mol)
-x
Equili(mol)
(0.03267 – x)
Equili(M)
(0.1634 – x)
0.02041
+x
(0.02041 + x)
(0.1021 + x)
CHEM171 – Lecture Series One : 2012/25
+x
x
x
x = 9.14 × 10-10 mol dm-3
pH = -log(9.14 × 10-10 ) = 9.04
OR
 pH = 9.24 + log(0.1021/0.1634) = 9.04
EXERCISE FOR THE IDLE MIND
Practice Exercise p 575and Exercise15.14, 15.16 p 603
CHEM171 – Lecture Series One : 2012/26
What happens to the pH of a buffer solution on
adding base or acid?
When an acid is added to a buffer solution, it is consumed by the
weak base present. When a base is added to a buffer solution, it
reacts fully with the weak acid present.
EXAMPLE
Consider the addition of 0.20 cm3 of a 0.50 mol dm-3 HNO3
solution to a buffer solution consisting of 50 cm3 0.045 mol dm-3
aqueous CH3CO2H (pKa = 4.77) and 50 cm3 0.045 mol dm-3
mol dm-3 Na[CH3CO2]. Assume that the addition of HNO3 causes
minimal increase in the total volume and it remains at 100 cm3.
CHEM171 – Lecture Series One : 2012/27
Before (mol)
0.00225
0.00225
CH3COO-(aq) + H+ (aq)  CH3COOH(aq)
After (mol)
0.00215
0.00235
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
Initial (mol)
0.00235
Change (mol)
-x
Equili(mol)
(0.00235 – x)
Equili(M)
(0.0235 – x)
0.00215
+x
(0.00215 + x)
(0.0215 + x)
CHEM171 – Lecture Series One : 2012/28
+x
x
x
x = 1.86 × 10-5 mol dm-3
pH = -log(1.86 × 10-5) = 4.73
OR
 pH = 4.77 + log(0.0215/0.0235) = 4.73
CHEM171 – Lecture Series One : 2012/29
EXERCISE FOR THE IDLE MIND
1. A buffer is made by adding 0.300 mol HC2H3O2 and
0.300 mol of NaC2H3O2 to enough water to make 1.0 dm3
of solution (pH of the buffer is 4.74).
(a) Calculate pH after 0.020 mol NaOH is added.
(b) Calculate pH after 0.020 mol HCl is added to original
solution. (Ka = 1.8 x 10-5)
2. Calculate the pH of a buffer solution of 0.50 mol dm-3 HF
and 0.45 mol dm-3 F- before and after addition of 0.01 mol
NaOH to 1.0 dm3 of the buffer. (Ka of HF = 6.8 x 10-4)
CHEM171 – Lecture Series One : 2012/30
MAKING UP A BUFFER SOLUTION OF SPECIFIED pH
EXAMPLE (Sample Exercise 15.4 p 575)
How many moles of NH4Cl must be added to 2.0 dm3 of 0.1
M NH3 to form a buffer whose pH is 9.00? (Assume that the
addition of NH4Cl does not change the volume of the
solution: Kb = 1.8 × 10-5)
SOLUTION
NH3(aq) + H2O(l) ⇌ NH4+ + OH-(aq)
pOH = 14.00 – 9.00 = 5.00
[OH-] = 1.0 × 10-5 M
CHEM171 – Lecture Series One : 2012/31
Amount of NH4Cl must be added = (0.18 mol dm-3) × (2.0 dm3)
= 0.36 mol
EXERCISE FOR THE IDLE MIND
Practice Exercise p 576 and Exercise15.18, 15.22 p 603
CHEM171 – Lecture Series One : 2012/32
ACID-BASE TITRATIONS
Titration - one solution of known concentration is used to
determine the concentration of another solution.

Indicators mark the endpoint of the titration by changing
colour.
HIn(aq) + H2O(l) ⇌ H3O+(aq) + In-(aq)

acid colour

base colour
Various indicators that cover a wide range of pH values.
CHEM171 – Lecture Series One : 2012/33
COLOURS AND APPROXIMATE pH RANGE OF SOME
COMMON ACID-BASE INDICATORS
pH
CHEM171 – Lecture Series One : 2012/34
TITRATION CURVES
Strong Acid/Strong Base
Weak Acid/Strong Base
CHEM171 – Lecture Series One : 2012/35
SOLUBILITY
PRECIPITATION REACTIONS
What happens when we add a solution of a salt to a solution
of another salt?
If we add the two following salts to each other
KNO3(aq) + NaCl(aq)  ?
Some students might write
KNO3(aq) + NaCl(aq)  KCl(aq) + NaNO3(aq)
If we write it out more fully
Initial: K+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
Final: K+(aq) + Cl-(aq) + Na+(aq) + NO3-(aq)
CHEM171 – Lecture Series One : 2012/36
A reaction will only happen between salts in
solution if ions are removed from solution.
One of the ways ions can be removed from solution is through
formation of a precipitate.
• Salts containing Group 1 metal ions, NH4+, nitrates and
acetates are soluble.
• Chlorides, bromides and iodides are soluble except those of
Ag+, Pb2+ and Hg22+.
• Sulfates are soluble except those of Pb2+, Hg22+, Sr2+ and
Ba2+. Ag2SO4 and CaSO4 are only slightly soluble.
• Carbonates, phosphates and sulfites are insoluble (except
those of group 1 and NH4+).
• Sulfides (S2-) are insoluble except those of Groups 1 and 2
and NH4+.
CHEM171 – Lecture Series One : 2012/37
If, therefore, we do the following reaction
BaCl2(aq) + Na2SO4(aq)  ?
This time a reaction will take place as when the Ba2+ ions
meet the SO42- ions in solution they will form a precipitate
and be removed from solution.
BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
When will a precipitate form?
If Q = [Ba2+][SO42-] is greater than Ksp
If Q  Ksp, precipitation occurs until Q = Ksp
Q = Ksp, equilibrium exists (saturated solution)
Q  Ksp, solid dissolves until Q = Ksp
CHEM171 – Lecture Series One : 2012/38
SPARINGLY SOLUBLE SALTS
Consider the following reaction:
BaSO4(s) + (aq) ⇌ Ba2+(aq) + SO42-(aq)
The equilibrium constant for this type of equation is known as
the solubility product, Ksp.
Ksp = [Ba2+][SO42- ]
For BaSO4, Ksp = 1.1 × 10-10. If the value of Ksp is very
small (as for BaSO4), then that means the ionic product
([Ba2+] × [SO42- ] in this case), is very small which in turn
means there are very few ions in solution. Salts which have a
Ksp value (i.e. those which are not fully soluble in water) are
said to be sparingly-soluble.
CHEM171– Lecture Series One : 2012/39
EXAMPLE (Practice Exercise p 586)
A saturated solution of Mg(OH)2 in contact with undissolved
solid is prepared at 25°C. The pH of the solution is found to
be 10.17. Assuming that Mg(OH)2 dissociates completely in
water and that there are no other simultaneous equilibria
involving the Mg2+ or OH- ions in the solution , calculate the
Ksp for this compound.
SOLUTION
The equilibrium equation and the expression for Ksp are
Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq)
Ksp = [Mg2+][OH-]2
CHEM171 – Lecture Series One : 2012/40
pOH = 14.00 – 10.17 = 3.83
[OH-] = 1.48 × 10-4 M
Now [Mg2+]/[OH-] = ½,  [Mg2+] = ½[OH-] = 7.4 × 10-5 M
Ksp = [Mg2+][OH-]2
= (7.4 × 10-5)(1.48 × 10-4)2 = 1.6 × 10-12
CHEM171 – Lecture Series One : 2012/41
MOLAR SOLUBILITIES
The molar solubility of a salt in water is calculated from the
Ksp value.
EXAMPLE
Calculate the molar solubility of AgCl at 25°C, given the
relevant Ksp value is 1.82 × 10-10.
SOLUTION
For every 1 mole of AgCl that dissolves, 1 mole of Ag+ and
1 mole of Cl- results.
Therefore, [AgCl] = [Ag+] = [Cl-]
If we can calculate the molar concentration of either ion then
we will have solved the problem.
CHEM171 – Lecture Series One : 2012/42
Ksp = [Ag+] [Cl-] = 1.82 × 10-10
But [Ag+] = [Cl-] and therefore,
[Ag+] = (1.82 × 10-10)½ = 1.35 ×10-5 mol dm-3
[AgCl] = 1.35 × 10-5 mol dm-3
EXAMPLE (Practice Exercise p 587)
The Ksp for LaF3 is 2.0 × 10-19. What is the solubility of LaF3
in water in moles per litre.
SOLUTION
LaF3(s) ⇌ La3+(aq) + 3F-(aq)
Let [La3+] = x,  [F-] = 3x
CHEM171 – Lecture Series One : 2012/43
Ksp = [La3+][F-]3 = (x)(3x)3 = 2.0 × 10-19
27x4 = 2.0 × 10-19  x4 = 7.41 × 10-21  x = 9.3 × 10-6 M
[LaF3] = 9.3 × 10-6 M
EXERCISE FOR THE IDLE MIND
Exercise15.37, 15.38 p 604
CHEM171 – Lecture Series One : 2012/44
COMMON-ION EFFECT
Consider a case where we have both NaCl and AgCl
dissolved in the same body of water. Let’s imagine that the
dissolution of NaCl, being freely soluble, results in a
chloride ion concentration of 0.100 mol dm-3. How will this
affect the solubility of AgCl?
Ksp = [Ag+] [Cl-] = 1.82 ×10-10
The chloride ion concentration has two inputs – the total
concentration (0.100 mol dm-3) arises from the NaCl plus
the Cl- ions from AgCl.
Let us set the molar solubility of [AgCl] as S mol dm-3 then
CHEM171 – Lecture Series One : 2012/45
[AgCl] = [Ag+] = S mol dm-3
[Cl-] = (0.100 + S ) mol dm-3
Now we can substitute these concentrations into the Ksp
expression.
Ksp = [Ag+] [Cl-] = (S)(0.100 + S) = (S)(0.100)
Since 0.100 >> S
0.100 S = 1.82 × 10-10
S = 1.82 × 10-9
 [Ag+] = [AgCl] = 1.82 × 10-9 mol dm-3
CHEM171 – Lecture Series One : 2012/46
EXAMPLE
Calculate the molar solubility of PbI2 (Ksp = 7.9 × 10-9) in a
solution containing 0.214 mol dm-3 iodide ions.
SOLUTION
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
[PbI2] = [Pb2+] = S mol dm-3
[I-] = (0.214 + 2S ) mol dm-3
Now we can substitute these concentrations into the Ksp
expression.
Ksp = [Pb2+] [I-]2 = (S)(0.214 + S)2 = (S)(0.214)2
since 0.214 >> S
CHEM171 – Lecture Series One : 2012/47
(0.214)2S = 7.9 × 10-9
S = 1.73 × 10-7 mol dm-3
 [Pb2+] = [PbI2] = 1.73 × 10-7 mol dm-3
THE EFFECT OF pH ON MOLAR SOLUBILITIES
The solubility of any substance whose anion is basic is affected
by the pH of the solution.
Consider the following:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
If we decrease the pH of the solution, say by adding an acidic
buffer, then what will happen to the solubility of Mg(OH2)?
CHEM171 – Lecture Series One : 2012/48
By adding an acid we increase [H+] and consequently
decrease [OH-] as Kw must be maintained.
If we decrease the amount of OH-(aq) in the system the
position of equilibrium will shift to the right.
If the position of equilibrium moves to the right then the
concentration of Mg2+ increases.
As [Mg(OH)2] = [Mg2+], the molar solubility of magnesium
hydroxide increases as the pH is lowered.
THE MOLAR SOLUBILITY OF SPARINGLY-SOLUBLE
HYDROXIDES INCREASES AS THE pH IS DECREASED.
CHEM171 – Lecture Series One : 2012/49
EXAMPLE
Calculate the solubility of Zn(OH)2 (Ksp = 3.0 × 10-16) in
(a) pure water (b) a buffer at pH = 7.00 and (c) a buffer
at pH = 11.00
SOLUTION
Zn(OH)2(s) ⇌ Zn2+(aq) + 2OH-(aq)
(a)
Let [Zn2+] = x, [OH-] = 2x
[Zn2+] [OH-]2 = (x)(2x)2 = 3.0 × 10-16
x = 4.2 × 10-6
[Zn(OH)2] = [Zn2+] = 4.2 × 10-6 mol dm-3
CHEM171 – Lecture Series One : 2012/50
(b)
At pH = 7.00, [OH-] = 1.00 × 10-7
[Zn2+] [1.00 × 10-7]2 = 3.0 × 10-16
[Zn2+] = 3.0 × 10-2 mol dm-3
 [Zn(OH)2] = [Zn2+] = 3.0 × 10-2 mol dm-3
(c)
At pH = 11.00, [OH-] = 1.0 × 10-3
[Zn2+] [1.00 × 10-3]2 = 3.0 × 10-16
[Zn2+] = 3.0 × 10-10 mol dm-3
 [Zn(OH)2] = [Zn2+] = 3.0 × 10-10 mol dm-3
CHEM171 – Lecture Series One : 2012/51