Calculations with Acids and Bases IB Chemistry Power Points Topic 18 Acids and Bases www.pedagogics.ca
May 11, 2015
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.
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+
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:
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-
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:
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:
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
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
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
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
[ ][ ]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