2012 topic 18 1 calculations involving acids and bases
-
Upload
david-young -
Category
Documents
-
view
1.734 -
download
0
Transcript of 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.
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