2012 topic 18 1 calculations involving acids and bases

Calculations

with

Acids and Bases

IB Chemistry Power Points

Topic 18

Acids and Baseswww.pedagogics.ca

Kw : the ionic product constant of water2 H2O H3O

+ + OH-

H2O H+ + OH- (simplified)

32

2 2

14 3

[ ][ ] [ ][ ][ ][ ]

[ ] [ ]

1 10 mol dm (at 25 )

w

w

H O OH H OHK H OH

H O H O

K

Kw : depends on temperature

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

30 1.471 x 10-14 6.92

40 2.916 x 10-14 6.77

50 5.476 x 10-14 6.63

100 51.3 x 10-14 6.14

Pure water is always neutral i.e. [H+] = [OH-]

This means the pH value that is “neutral” i.e. [H+] =

[OH-] changes with temperature

Be aware of this. 99% of the time you can assume the Kw value is 1 x 10-14

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.

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?

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?

5.8 6

14

14

14 8.2 5.8

[ ] 10 1.58 10

pH pOH

pOH pH

pOH

OH

53 3

3

[ ][ ]1.738 10

[ ]a

CH COO H OK

CH COOH

54

3

[ ][ ]1.778 10

[ ]b

NH OHK

NH

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+

ammonia: NH3 + H2O NH4+ + OH-

PracticeCalculate the pH of a) a 0.75 M solution of ethanoic acid

Compare to the pH of a 0.75 M HCl solution

PracticeCalculate the pH of a) a 0.75 M solution of ethanoic acid

53 3

3

5

5 3

[ ][ ]1.738 10

[ ]

[ ][ ]1.738 10

[0.75 ]

0.75 1.738 10 0.00361 mol dm

log(0.00361) 2.44

a

CH COO H OK

CH COOH

x x

x

x

pH

Compare to the pH of a 0.75 M HCl solution

PracticeCalculate the pH of b) a 0.75 M solution of ammonia

PracticeCalculate the pH of b) a 0.75 M solution of ammonia

54

3

5

5 3

[ ][ ]1.778 10

[ ]

[ ][ ]1.778 10

[0.75 ]

0.75 1.778 10 0.00365 mol dm

14 14 log(0.00365) 11.6

b

NH OHK

NH

x x

x

x

pH pOH

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:





The equilibrium expression for the reaction of the acid is:

The equilibrium expression for the reaction of the conjugate base is:

3

3

[ ][ ]

[ ]a

CH COO HK

CH COOH

3

3

[ ][ ]

[ ]b

CH COOH OHK

CH COO

Now write an expression for Ka multiplied by Kb


3 3

3 3

[ ][ ] [ ][ ]

[ ] [ ]a b

CH COO H CH COOH OHK K

CH COOH CH COO


3 3

3 3

[ ][ ] [ ][ ]

[ ] [ ]a b

CH COO H CH COOH OHK K

CH COOH CH COO

[ ][ ]a b wK K H OH K

ethanoic acid CH3COOHKa = 1.738 × 10-5 pKa =

methanoic acid HCOOHKa = 1.778 × 10-4 pKa =

Consider 2 weak acids, use your data booklet to find the missing values

now 2 weak basesammonia NH3

Kb = 1.778 × 10-5 pKb =

methylamine CH3NH2

Kb = 4.365 × 10-4 pKb =

ethanoic acid CH3COOHKa = 1.738 × 10-5 pKa =

methanoic acid HCOOHKa = 1.778 × 10-4 pKa =

Consider 2 weak acids, use your data booklet to find the missing values

now 2 weak basesammonia NH3

Kb = 1.778 × 10-5 pKb =

methylamine CH3NH2

Kb = 4.365 × 10-4 pKb =

4.76

3.75

4.75

3.36

conclusions?

Can you relate acid/base strength to

pK values?

Easy math between Ka and pKa?

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 CH3COOHKa = 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 + pK

b = pK

w

Not to be confused withpH + pOH = pK

w

ethanoic acid CH3COOHKa = 1.738 × 10-5 pKa =4.76

ethanoate anion CH3COO-

Ka = 5.75 × 10-10 pKb = 9.24

2012 topic 18 1 calculations involving acids and bases

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