1. AQUEOUS EQUILIBRIA Salts Hydrolysis The Common-Ion Effect
Buffer Solutions
Acid-Base Titrations
Solubility
Brown et al., Chapter 15,
CHEM171 – Lecture Series One : 2012/01

2. 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) (aq)  Na+(aq) + Cl-(aq)
solute solvent solution
SALTS dissociate FULLY in water to form ionic solutions and are thus STRONG ELECTROLYTES.
CHEM171 – Lecture Series One : 2012/02

3. 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 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) x x x
Equili. (M) (0.100 – x) x x
CHEM171 – Lecture Series One : 2012/03

4. SALTS CAN REACT WITH WATER TO BE EITHER ACIDIC, BASIC OR NEUTRAL.
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

5. SALTS OF STRONG ACIDS AND STRONG BASES
NaOH + HCl  NaCl + H2O
H2O
NaCl  Na Cl-
Na+ + H2O  no reaction
and
Cl- + H2O  no reaction
a solution of NaCl is neutral.
CHEM171 – Lecture Series One : 2012/05

6. SALTS OF WEAK ACIDS AND STRONG BASES
CH3COOH + NaOH  CH3COONa + H2O
H2O
CH3COONa  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

7. SALTS OF STRONG ACIDS AND WEAK BASES
HCl + NH3  NH4Cl
H2O
NH4Cl  NH Cl-
NH H2O  NH3 + H3O+
 pH of NH4Cl solution < 7 (acidic)
CHEM171 – Lecture Series One : 2012/07

8. SALTS OF WEAK ACIDS AND WEAK BASES
Solution pH depends on relative strengths of weak acid and weak base.
HNO2 + NH3  NH4NO2
H2O
NH4NO2  NH NO2-
NH H2O  NH H3O+ Ka = 5.6 × 1010
NO H2O  HNO OH- Kb = 2.2 × 1011
Kb < Ka  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

9. EXAMPLE
Calculate the pH of a mol dm-3 solution of NH4Cl at 25oC, Ka = ×
SOLUTION
NH4+ (aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Initial (M)
Change (M) x x x
Equili. (M) ( – x) x x
x = 7.26 × 10-6
From, pH = - log [H3O+], pH = 5.14
CHEM171 – Lecture Series One : 2012/09

10. EXAMPLE
Calculate the pH of a mol dm-3 solution of sodium acetate, Kb = ×
SOLUTION
CH3COO-(aq) + H2O(l) ⇌ CH3COO-(aq) + OH-(aq)
Initial (M)
Change (M) x x x
Equili. (M) (0.154 – x) x x
x = 9.38 × 10-6
pOH = 5.03, therefore pH = 8.97
CHEM171 – Lecture Series One : 2012/10

11. COMMON ION EFFECT CH3COOH + H2O ⇌ H3O+ + CH3COO-
What 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

12. 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 ⇌ H CH3COO-
Initial M M
Change x M x M x M
Equilibrium (0.30 – x) M x M ( x) M
CHEM171 – Lecture Series One : 2012/12

13. EXERCISE FOR THE IDLE MIND
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 p 602
CHEM171 – Lecture Series One : 2012/13

14. 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  CH3COO H3O+
acid base 1
CHEM171 – Lecture Series One : 2012/14

15. NaHCO3 + Na2CO3 NaH2PO4 + Na2HPO4 Some common buffer solutions are:
Brønsted acid conjugate base
NaH2PO Na2HPO4
Brønsted acid conjugate base
CHEM171 – Lecture Series One : 2012/15

16. pH < 7.00 Two components – a weak acid and the salt of a weak acid.
WEAK ACID + SALT OF A WEAK ACID
CH3COOH CH3COO- Na+
acetic acid 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

17. CH3COO-(aq) + H3O+(aq) ⇌ CH3COOH(aq) + H2O(l)
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

18. EXAMPLE
Calculate the pH of a solution which contains mol dm-3 of CH3COOH and mol dm-3 of sodium acetate.
SOLUTION
We first determine the pH using a method described before.
CH3COOH(aq) ⇌ H+(aq) CH3COO‑(aq)
Initial M M
Change x M x M x M
Equilibrium ( – x) M x M ( x) M
CHEM171 – Lecture Series One : 2012/18

19. Can use another method to calculate the pH of a buffer.
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

20. [CH3COOH] = 0.1046 mol dm-3 [CH3COO‑] = 0.1247 mol dm-3
Ka = 1.75 × 10-5,  pKa = -log[Ka]
 pH = log(0.1247/0.1046) = 4.84
CHEM171 – Lecture Series One : 2012/20

21. EXAMPLE
Calculate the pH of the solution prepared from adding mL of mol dm-3 NaOH to mL of a mol dm-3 CH3CH2COOH solution.
(pKa CH3CH2COOH = 4.88).
SOLUTION
Before (mol)
CH3CH2COOH + OH- → CH3CH2COO- + H2O
After (mol)
CH3CH2COOH + H2O → CH3CH2COO- + H3O+
Initial (M)
Change (M) x x x Equilibrium ( – x) ( x) x
CHEM171 – Lecture Series One : 2012/21

22. x = 1.61 × 10-5 mol dm-3 pH = -log(1.61 × 10-5) = 4.79 OR
CHEM171 – Lecture Series One : 2012/22

23. pH > 7.00 Weak base and the salt of a weak base.
WEAK BASE SALT OF A WEAK BASE
NH NH4Cl
ammonia ammonium chloride
The weak base will take care of any added H3O+.
NH3(aq) + H3O+(aq) ⇌ NH H2O(l)
The salt of a weak base will react with any added OH-
NH OH-(aq) ⇌ NH3(aq) + H2O(l)
and the strong base gives way to a weak base (NH3).
CHEM171 – Lecture Series One : 2012/23

24. The equilibria which determines the final pH of the buffer will be
NH H2O(l) ⇌ NH3(aq) + H3O+(aq)
If instead we had written the equilibrium as
NH3(aq) + H2O(l) ⇌ NH OH-(aq)
then
CHEM171 – Lecture Series One : 2012/24

25. EXAMPLE
Calculate the pH of a solution prepared from mL of mol NH3 and mL of mol NH4Cl (Ka (NH4+) = 5.71 × 10-10).
SOLUTION
NH H2O(l) ⇌ NH3(aq) + H3O+(aq)
Initial (mol)
Change (mol) x x x Equili(mol) ( – x) ( x) x
Equili(M) ( – x) ( x) x
CHEM171 – Lecture Series One : 2012/25

26. EXERCISE FOR THE IDLE MIND
x = 9.14 × mol dm-3
pH = -log(9.14 × ) = 9.04 OR
 pH = log(0.1021/0.1634) = 9.04
EXERCISE FOR THE IDLE MIND
Practice Exercise p 575and Exercise15.14, p 603
CHEM171 – Lecture Series One : 2012/26

27. 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 cm mol dm-3 aqueous CH3CO2H (pKa = 4.77) and 50 cm 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

28. CH3COO-(aq) + H+ (aq)  CH3COOH(aq)
Before (mol)
CH3COO-(aq) + H+ (aq)  CH3COOH(aq)
After (mol)
CH3COOH(aq) + H2O(l) ⇌ CH3COO‑(aq) + H3O+(aq)
Initial (mol)
Change (mol) x x x Equili(mol) ( – x) ( x) x
Equili(M) ( – x) ( x) x
CHEM171 – Lecture Series One : 2012/28

29. x = 1.86 × 10-5 mol dm-3 pH = -log(1.86 × 10-5) = 4.73 OR
CHEM171 – Lecture Series One : 2012/29

30. EXERCISE FOR THE IDLE MIND
1. A buffer is made by adding mol HC2H3O2 and mol of NaC2H3O2 to enough water to make 1.0 dm3 of solution (pH of the buffer is 4.74).
(a) Calculate pH after mol NaOH is added.
(b) Calculate pH after 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

31. 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) ⇌ NH OH-(aq)
pOH = – 9.00 = 5.00
[OH-] = 1.0 × 10-5 M
CHEM171 – Lecture Series One : 2012/31

32. EXERCISE FOR THE IDLE MIND
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, p 603
CHEM171 – Lecture Series One : 2012/32

33. 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

34. COLOURS AND APPROXIMATE pH RANGE OF SOME COMMON ACID-BASE INDICATORS
CHEM171 – Lecture Series One : 2012/34

35. TITRATION CURVES Strong Acid/Strong Base Weak Acid/Strong Base
CHEM171 – Lecture Series One : 2012/35

36. 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

37. Sulfides (S2-) are insoluble except those of Groups 1 and 2 and NH4+.
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

38. If Q = [Ba2+][SO42-] is greater than Ksp
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

39. 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 × 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

40. 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 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) OH-(aq)
Ksp = [Mg2+][OH-]2
CHEM171 – Lecture Series One : 2012/40

41. Now [Mg2+]/[OH-] = ½,  [Mg2+] = ½[OH-] = 7.4 × 10-5 M
pOH = – = 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

42. MOLAR SOLUBILITIES EXAMPLE SOLUTION
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 ×
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

43. EXAMPLE (Practice Exercise p 587)
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 × 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

44. EXERCISE FOR THE IDLE MIND
Ksp = [La3+][F-]3 = (x)(3x)3 = 2.0 × 10-19
27x4 = 2.0 ×  x4 = 7.41 ×  x = 9.3 × 10-6 M
[LaF3] = 9.3 × 10-6 M
EXERCISE FOR THE IDLE MIND
Exercise15.37, p 604
CHEM171 – Lecture Series One : 2012/44

45. Let us set the molar solubility of [AgCl] as S mol dm-3 then
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 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

46. [AgCl] = [Ag+] = S mol dm-3
[Cl-] = ( S ) mol dm-3
Now we can substitute these concentrations into the Ksp expression.
Ksp = [Ag+] [Cl-] = (S)( S) = (S)(0.100)
Since >> 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

47. EXAMPLE
Calculate the molar solubility of PbI2 (Ksp = 7.9 × 10-9) in a solution containing mol dm-3 iodide ions.
SOLUTION
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
[PbI2] = [Pb2+] = S mol dm-3
[I-] = ( S ) mol dm-3
Now we can substitute these concentrations into the Ksp expression.
Ksp = [Pb2+] [I-]2 = (S)( S)2 = (S)(0.214)2
since >> S
CHEM171 – Lecture Series One : 2012/47

48.  [Pb2+] = [PbI2] = 1.73 × 10-7 mol dm-3
(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

49. 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

50. 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

51.  [Zn(OH)2] = [Zn2+] = 3.0 × 10-2 mol dm-3
(b) At pH = 7.00, [OH‑] = 1.00 × 10-7
[Zn2+] [1.00 × 10-7]2 = 3.0 × 10-16
[Zn2+] = × 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+] = × mol dm-3
 [Zn(OH)2] = [Zn2+] = 3.0 × mol dm-3
CHEM171 – Lecture Series One : 2012/51