1. Refresh
For each of the following equilibria, write the expression for the equilibrium constant Kc and state its units:
2NO2(g) == N2O4(g)
CH3CH2CO2H(l) + CH3CH2OH(l) == CH3CH2CO2CH2CH3(l) + H2O(l)
H2(g) + I2(g) == 2HI(g)
2SO2(g) + O2(g) == 2SO3(g)
N2(g) + 3H2(g) == 2NH3(g)

2. Answers

3. Calculate the equilibrium constant for the following
Answers
28.65
48.33
1.1*109
3.5

4. More extreme
In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
N2(g) + 3H2(g) == 2NH3(g)
Hydrogen chloride can be oxidised to chlorine by the Deacon process:
4HCl(g) + O2(g) == 2Cl2(g) + 2H2O(g)
0.800 mol of hydrogen chloride was mixed with mol of oxygen in a vessel of volume 10 dm3. At equilibrium it was found that the mixture contained mol of hydrogen chloride. Calculate Kc for the reaction.
A 0.04 sample of SO3 is introduced into a 3.04 litre vessel and allowed to reach
equilibrium. The amount of SO3 present at equilibrium is found to be mole. Calculate the value of Kc for the reaction 2SO3(g) == 2SO2(g) + O2(g).
Answers: 1-> 6.6mol-2dm > 1013mol-1dm > 3.2*10-4 mol dm-3

5. Acids and Bases (HL) - Lesson 7
Ka/pKa and Kb/pKb

6. Lesson 6: Ka/pKa and Kb/pKb
Objectives:
Understand the concepts of Ka and Kb
Understand the concepts of pKa and pKb

7. Weak Acids: Ka and pKa pKa = -log10(Ka)
Weak acids dissociate to form an equilibrium
HA(aq)  H+(aq) + A-(aq)
This has the equilibrium constant (aka acid dissociation constant), Ka
The values for Ka are often very small, so we use ‘pKa’ to make them easier to handle:
pKa = -log10(Ka)

8. Ka and pKa in action In order of decreasing acid strength:
Smaller Ka  weaker acid
Smaller pKa  stronger acid
Acid Ka
pKa
Hydronium ion, H3O+
1.00
0.00
Oxalic acid, HO2CCO2H
5.9x10-2
1.23
Hydrofluoric, HF
7.2x10-4
3.14
Methanoic, CHOOH
1.77x10-4
3.75
Ethanoic, CH3COOH
1.76x10-5
4.75
Phenol, C6H5OH
1.6x10-10
9.80

9. Weak Bases: Kb and pKb pKb = -log10(Kb)
Weak bases dissociate to form an equilibrium
BOH(aq)  B+(aq) + OH-(aq)
This has the equilibrium constant (aka base dissociation constant), Kb
The values for Kb are often very small, so we use ‘pKb’ to make them easier to handle:
pKb = -log10(Kb)

10. Kb and pKb in action In order of decreasing base strength:
Smaller Kb  weaker base
Smaller pKb  stronger base
Base Kb
pKb
Diethylamine
1.3x10-3
2.89
Ethylamine
5.6x10-4
3.25
Methylamine
4.4x10-4
3.36
Ammonia
1.8x10-5
4.74

11. Measuring Ka/pKa and Kb/pKb
pKa and pKb can be determined experimentally
At the point of half neutralisation:
pH = pKa
pOH = pKb
This is a convenient artefact of the mathematics
Follow on the instruction sheet

12. Key Points At the point of half-neutralisation: For a weak acid:
AND pKa = -log10(Ka)
For a weak base:
AND pKb = -log10(Kb)
At the point of half-neutralisation:
pKa = pH AND pKb = pOH