1. IB Chemistry Power Points
Topic 18
Acids and Bases
www.pedagogics.ca
Calculations
with
Acids and Bases

2. Kw : the ionic product constant of water
2 H2O  H3O+ + OH-
H2O  H+ + OH- (simplified)
[ H 3O ][OH ] [ H ][OH ]
Kw 2
[ H ][OH ]
[ H 2O ] [ H 2O ]
14 3 
K w 1 10 mol dm (at 25 )

3. Kw : depends on temperature
Pure water is always neutral i.e.
[H+] = [OH-]
T (°C) Kw (mol2 dm-6) pH
0 0.114 x 10-14 7.47
10 0.293 x 10-14 7.27
20 0.681 x 10-14 7.08
25 1.008 x 10-14 7.00 This means the pH value that
30 1.471 x 10-14 6.92 is “neutral” i.e. [H+] = [OH-]
changes with temperature
40 2.916 x 10-14 6.77
50 5.476 x 10-14 6.63
100 51.3 x 10-14 6.14
Be aware of this. 99% of the
time you can assume the Kw
value is 1 x 10-14

4. The Kw value for pure water at 60o C is 9.614 × 10-14. Calculate the
concentration of hydroxide ions and the pH of pure water at this temperature.

5. The Kw value for pure water at 60o C is 9.614 × 10-14. Calculate the
concentration of hydroxide ions and the pH of pure water at this temperature.

6. about pOH
pOH = -log [OH-]
• just like pH, pOH is a measure of concentration
• pH + pOH = pKw
• this means that at 25o pH + pOH = 14
• unless otherwise stated (or a temperature different than 25o is
given), this relationship applies to all pH problems
For example: what is the [OH-] for a solution with pH 8.2 measured at 25o C?

7. about pOH
pOH = -log [OH-]
• just like pH, pOH is a measure of concentration
• pH + pOH = pKw
• this means that at 25o pH + pOH = 14
• unless otherwise stated (or a temperature different than 25o is
given), this relationship applies to all pH problems
For example: what is the [OH-] for a solution with pH 8.2 measured at 25o C?
pH pOH 14
pOH 14 pH
pOH 14 8.2 5.8
5.8 6
[OH ] 10 1.58 10

8. Weak acids and weak bases – dissociation reactions with water
Unlike strong acids and bases, weak acids and bases do not dissociate
100%. This means the concentration acid or base is NOT the same as the
equilibrium concentrations of hydronium or hydroxide ions
ethanoic acid: CH3COOH + H2O  CH3COO- + H3O+
[CH3COO ][ H 3O ] 5
Ka 1.738 10
[CH 3COOH ]
ammonia: NH3 + H2O  NH4+ + OH-
[ NH 4 ][OH ] 5
Kb 1.778 10
[ NH3 ]

9. Practice
Calculate the pH of
a) a 0.75 M solution of ethanoic acid
Compare to the pH of a 0.75 M HCl solution

10. Practice
Calculate the pH of
a) a 0.75 M solution of ethanoic acid
[CH 3COO ][ H 3O ] 5
Ka 1.738 10
[CH 3COOH ]
[ x][ x] 5
1.738 10
[0.75 x]
x 0.75 1.738 10 5 0.00361 mol dm 3
pH log(0.00361) 2.44
Compare to the pH of a 0.75 M HCl solution

11. Practice
Calculate the pH of
b) a 0.75 M solution of ammonia

12. Practice
Calculate the pH of
b) a 0.75 M solution of ammonia
[ NH 4 ][OH ] 5
Kb 1.778 10
[ NH 3 ]
[ x][ x] 5
1.778 10
[0.75 x]
5 3
x 0.75 1.778 10 0.00365 mol dm
pH 14 pOH 14 log(0.00365) 11.6

13. Consider the following simplified equation for the reaction of ethanoic acid in water
CH3COOH  CH3COO- + H+

14. Consider the following simplified equation for the reaction of ethanoic acid in water
CH3COOH  CH3COO- + H+
Now write the equation for the reaction of the conjugate base with water:

15. Consider the following simplified equation for the reaction of ethanoic acid in water
CH3COOH  CH3COO- + H+
Now write the equation for the reaction of the conjugate base with water:
CH3COO- + H2O  CH3COOH + OH-

16. Consider the following simplified equation for the reaction of ethanoic acid in water
CH3COOH  CH3COO- + H+
Now write the equation for the reaction of the conjugate base with water:
CH3COO- + H2O  CH3COOH + OH-
The equilibrium expression for the reaction of the acid is:
The equilibrium expression for the reaction of the conjugate base is:

17. Consider the following simplified equation for the reaction of ethanoic acid in water
CH3COOH  CH3COO- + H+
Now write the equation for the reaction of the conjugate base with water:
CH3COO- + H2O  CH3COOH + OH-
The equilibrium expression for the reaction of the acid is:
[CH 3COO ][ H ]
Ka
[CH 3COOH ]
The equilibrium expression for the reaction of the conjugate base is:
[CH3COOH ][OH ]
Kb
[CH3COO ]

18. Now write an expression for Ka multiplied by Kb

19. Now write an expression for Ka multiplied by Kb
[CH 3COO ][ H ] [CH 3COOH ][OH ]
K a Kb
[CH 3COOH ] [CH 3COO ]

20. Now write an expression for Ka multiplied by Kb
[CH 3COO ][ H ] [CH 3COOH ][OH ]
K a Kb
[CH 3COOH ] [CH 3COO ]

21. Now write an expression for Ka multiplied by Kb
[CH 3COO ][ H ] [CH 3COOH ][OH ]
K a Kb
[CH 3COOH ] [CH 3COO ]
K a Kb [ H ][OH ] K w

22. Consider 2 weak acids, use your data booklet
to find the missing values
ethanoic acid CH3COOH
Ka = 1.738 × 10-5 pKa =
methanoic acid HCOOH
Ka = 1.778 × 10-4 pKa =
now 2 weak bases
ammonia NH3
Kb = 1.778 × 10-5 pKb =
methylamine CH3NH2
Kb = 4.365 × 10-4 pKb =

23. Consider 2 weak acids, use your data booklet
to find the missing values
ethanoic acid CH3COOH
Ka = 1.738 × 10-5 pKa = 4.76 conclusions?
methanoic acid HCOOH Can you relate acid/base
Ka = 1.778 × 10-4 pKa =
3.75 strength to pK values?
now 2 weak bases Easy math between Ka
ammonia NH3 and pKa?
Kb = 1.778 × 10-5 pKb =
4.75
methylamine CH3NH2
Kb = 4.365 × 10-4 pKb =
3.36

24. More neat stuff.
CH3COO- is the conjugate base of the weak acid CH3COOH
(Ka value 1.738 × 10-5).
CH3COO- is therefore a weak base with a Kb value of 5.75 ×
10-10.
What do you notice about the pK values for this acid -
conjugate base pair?
ethanoic acid CH3COOH
Ka = 1.738 × 10-5 pKa =4.76
ethanoate anion CH3COO-
Ka = 5.75 × 10-10 pKb = 9.24

25. ethanoic acid CH3COOH
Ka = 1.738 × 10-5 pKa =4.76
ethanoate anion CH3COO-
Ka = 5.75 × 10-10 pKb = 9.24
for acid/conjugate base pairs & base/conjugate acid pairs
pKa + pKb = pKw
Not to be confused with
pH + pOH = pKw