Transcript Kinetics again

Kinetics and
Equilibrium again
Kinetics and the
Rate constant
Equilibrium constants
Kc and Kp
Acids and alkalis
and pH
Curves and
neutralisation
Buffers
Revised: 18-Jul-15
Need to do:
• clarify graphs
• Expand rate-limiting step & mechanism
Remember:
• 5 factors that affect rate:
– Temperature
– Catalyst
– Surface area (S), Pressure (G),
Concentration (L)
• They actually affect:
– Frequency of collision (all)
– Energy of collision (temperature)
– Activation energy / path (catalyst)
• Need to know definition of Ea
• Use Ea in your answers
Measuring rate
• How much
• Unit of time
• Real pattern is
• ∆concentration of A
•
∆time
(similar to speed )
Pressure and
concentration
• Both expressed as [species]
• Read brackets as “concentration of”
• So ∆[NO2] reads “change in
concentration
of NO2”
How to measure [product]
or [reactant]
• Collect gas – volume (say temp. constant)
• Gas reactions – pressure
(say temp. constant)
• Colour change –use colorimeter
• Change in liquid volume (tiny)
(called dilatometry)
• pH using meter/data logger
• Use radioisotope tracer (rare!)
Methods continued...
• Polarisation of light (optical isomers)
• Sample + quench at low temperature, then:
– Halides – use to displace iodine and do a starch
titration
OR use colour change
– Acid/base – neutralise one and/or titrate
• All these need “final” measurement for
calibration before you can work out the rate.
Rate equations
• The rate of reaction is generally
proportional to the concentration of at
least one reactant
• So a reaction like mA + nB  C can be
represented by a rate equation of the
form:
• Rate = k [A]m[B]n
Note:
•
•
•
•
The multiplier has become a power
k is called the rate constant
k is affected by temperature.
k is affected by the presence (or
effectiveness) of a catalyst.
• Not by concentration.
• Note: the initial gradient is the one we
want, after that, concentrations have
changed from the ones we set.
e.g.
[NO] (moles dm-3)
350.00
300.00
[NO] mol dm-3
250.00
200.00
[NO] (moles dm-3)
150.00
100.00
50.00
0.00
0
50
100
Time(s)
150
200
250
e.g.
[NO] (moles dm-3)
120.00
100.00
[NO] mol dm-3
80.00
60.00
[NO] (moles dm-3)
40.00
20.00
0.00
0
50
100
150
200
Time(s)
250
300
350
Reaction rate equation
• General form
• Rate = k[A]m[B]n .....
• The m and n terms are described as
“the order of reaction with respect to...”
order
• Order of 0 means concentration makes
no difference to rate
• Order of 1 means direct proportion
• Order of 2 means square-law proportion
• Yes, fractional orders do occur but not at
A level.
A good first guess....
• It happens that, for many reactions, a
“first guess” for the order of reaction with
respect to one of the reactants is its
stoichiometric ratio:
• E.g. the reaction N2 + 3H2  2NH3 is
probably 1st order with respect to N2 and
3rd order w.r.t. H2.
• And the backwards reaction is 2nd order
w.r.t. NH3
In a concentration-time
graph:
• A zero-order reaction gives a straight line.
• A first-order reaction does not give a
straight line, it’s like a half-life curve.
• Only first-order reactions have constant
half-lives.
• A second-order reaction will give a
steeper curve.
In a rate-concentration
graph:
• A zero-order reaction gives a horizontal
line.
• A first-order reaction gives a straight line.
• A second-order reaction will give a curve.
• This curve will be a straight line only if
you plot rate2 against time.
Typical data
initial [A] initial [B]
Run
Run 1
1
1
initial rate
mol dm-3 s-1
10
Run 2
1
2
20
Run 3
2
1
10
Run 4
3
2
20
So the rate equation is...
•
•
•
•
•
•
Rate = k [A]0[B]1
Or rate = k [B] (when simplified.)
And k is:
From 1st line of table...
10 = k x 1
Therefore k = 10
And again....
initial [A] initial [B]
Run
Run 1
1
1
initial rate
mol dm-3 s-1
10
Run 2
1
2
10
Run 3
2
1
20
Run 4
3
2
30
So the rate equation is...
• Rate = k [A]1[B]0
• Or rate = k [A] (when simplified.)
•
•
•
•
And k is:
From 1st line of table...
10 = k x 1
Therefore k = 10
And again....
initial [A] initial [B]
Run
Run 1
1
1
initial rate
mol dm-3 s-1
10
Run 2
1
2
20
Run 3
2
1
20
Run 4
3
2
60
So the rate equation is...
•
•
•
•
•
•
Rate = k [A]1[B]1
Or rate = k [A][B] (when simplified.)
And k is:
From 1st line of table...
10 = k x 1
Therefore k=10
And....
initial [A]
initial [B]
initial rate
mol dm-3 s-1
Run 1
1
1
10
Run 2
1
2
40
Run 3
2
1
20
Run 4
3
2
120
Run
So the rate equation is...
•
•
•
•
•
•
Rate = k [A]1[B]2
Or rate = k [A][B]2 (when simplified.)
And k is:
From 1st line of table...
10 = k x 1 x 1
Therefore k=10
Or more usually....
initial [A]
mol dm-3
initial [B]
mol dm-3
initial rate
mol dm-3 s-1
Run 1
0.02
0.3
2x10-3
Run 2
0.02
0.6
4x10-3
Run 3
0.04
0.3
4x10-3
Run 4
0.04
0.6
8x10-3
Run
So the rate equation is...
•
•
•
•
•
•
Rate = k [A]1[B]1
Or rate = k [A][B] (when simplified.)
And k is:
From 1st line of table...
2 x 10-3 = k x 0.02 x 0.3
Therefore k = 0.333
What affects k?
•
•
•
•
It is not obvious that k can be changed:
Rate = k[A]m[B]n .....
If temperature goes up, k goes up
Adding a catalyst increases k.
What about catalysts?
• If a catalyst is involved in a reaction it
must have an effect on rate, so it will be
in the rate equation.
• It (probably) won’t be in the balanced
(stoichiometric) equation, though you may
see it drawn above the arrow.
Units of k
• k is independent of
moles and volume for
order 1
• You need to do the
same operations to
the units as you have
done to the numbers
• Or maybe memorise:
Order
units of k
0
mol dm-3 s-1
1
s-1
2
mol-1 dm3 s-1
3
mol-2 dm6 s-1
Typical (easy) exam
question:
The following data were obtained in a series of
experiments on the rate of the reaction between
compounds A and B at a constant temperature.
Experiment
Initial concentration
Initial
of A/mol dm–3
concentration of
B/mol dm–3
Initial rate
mol dm–3 s–1
1
0.12
0.15
0.32 × 10–3
2
0.36
0.15
2.88 × 10–3
3
0.72
0.30
11.52 × 10–3
(i) Deduce the order of reaction with respect to A.
(ii) Deduce the order of reaction with respect to B.
(2 marks)
Mechanism
• Reactions usually have more than one
step:
• ABC
• Or written as:
• A  B followed by B  C
• Often have a rate-determining step
(the slowest)
How to recognise a ratedetermining step:
• Any step after the rate-determining step will not
appear in the rate equation because it can have
little or no effect on the overall rate.
• E.g. if a reaction proceeds in two steps:
• Step 1: A  B then step 2: B + C  D
• If the rate equation is
• Rate = k[A] then step 2 is not rate-determining
• Rate = k[A][C] then step 2 must be.
• .
Equilibria (again)
Remember:
• Equilibrium involves:
– Forward and backward reactions
– Equal rates
– Closed system:- nothing leaves or enters,
temperature and pressure constant
– Constant concentrations, NOT equal
concentrations
• Note – an equilibrium is not actually much
use to us.
Two reactions:
• For a hypothetical equilibrium:
• aW +bX Ý cY + dZ
• The forward and backward reactions have
rate equations that must be equal
• So rate = k1[W]a[X]b = k2[Y]c[Z]d
• We can get an overall equation for the
equilibrium constant:
[Y ]c [ Z ]d .....
Kc 
a
b
[W ] [ X ] .....
Some nice little points
....
• Anything in large excess at the beginning
of the reaction, particularly solvents
(usually water) and solids don’t count in
the calculations and end up included in Kc
• Because if their concentrations do not
change, they are constant
• UNLESS they are part of the reaction, like
in hydrolysis
• Concentrations must be measured at
equilibrium, obviously.
What affects Kc and Kp?
• Temperature (and nothing else)
• Forward reaction exothermic: increased
temp decreases Kp, make more reactant
• Forward reaction endothermic: increased
temp increases Kp, make more product
Think about units
• It is worth “renaming” the unit of mol dm-3 as,
say, “c” or “u”, because you can manipulate the
unit and then convert the answer back again
• E.g. if the equation is
[ A ]2
• Then you could say the
[B ]3
units are c2 / c3
• The c2 cancels with the c3,
giving c(2-3), so the units are c-1
• This is (mol dm-3)-1
• This is mol-1 dm3
Let’s visit an old friend.....
• For
N2 + 3H2 Ý 2NH3
[NH 3(g )]2
Kc 
[N 2(g )][ H 2(g )]3
• And the units will be:
• C2 / C4 or C(2-4) or C-2
• That is mol-2 dm6
If we use Kp....
• We will also be able to use the symbol Kp
for gas equilibria, instead of Kc; we use
partial pressures instead of concentrations.
• For N2 + 3H2 Ý 2NH3
pNH 3(g )2
Kp 
pN 2(g ) pH 2(g )3
• And the units will be:
• p2 / p4 or p(2-4) or p-2
• That is, atm-2 or Pascals-2, depending on
the question
Partial pressures
• To calculate Kp, we need to know the
partial pressure of each species, using:
moles of gas
p
 total pressure
total moles
• Or alternatively,
• Partial pressure = mole fraction x total
pressure
• E.g. if the total pressure of the air is 1 atm
and mole fraction of N2 is 0.78, then pN2(g)
is 0.78 atm.
• E.g:
• (atm – atmospheres – one times standard atmospheric pressure)
• In a sample of gas, there is 3 moles of N2,
4 moles of NO and 3 moles of O2. If the
total pressure is 12 atmospheres, what
are the three partial pressures?
e.g. consider:
• 2NO Ý N2O2
• Given: at equilibrium, the partial pressures
are:
• NO
3.5 atm
•
(atm – atmospheres – one times standard atmospheric pressure)
• N2O2
4 atm
• Assuming that the equilibrium is
[N 2O 2 ]
represented by :
[NO ]2
• What is the value of Kc?
examples
• PbSO4(s) + H2O(l) Ý Pb2+(aq) + SO42-(aq)
• Remember, water is a solvent so we can
ignore it and the solid can also be
ignored, as its concentration doesn’t
change.
• So Kc = [SO42-] [Pb2+]
• And the units will be mol2 dm-6
Or
• E.g. 2
• In the flame of a burning hydrogen
balloon, just after it is ignited, there is 12
moles of H2, 482 moles of H2O and 6
moles of O2. If the total pressure is 22
atmospheres, what are the three partial
pressures?
Kp
• For the reaction:
• N2(g) + 3H2(g) Ý 2NH3(g)
• Calculate Kp for the Haber process, given
that the number of moles of N2(g) = 1.2,
H2(g)=0.4 and NH3(g)=2.3, and the total
pressure = 6 atm.
Surprise!
• Kp doesn’t change for overall pressure
changes.
• You’d think it would, but it doesn’t, you just
calculate it from the pressures given.
• And that’s why Le Chatelier’s principle
works, because if Kp can’t change, then
the equilibrium must respond to change.
A typical exam question
3 (a)
The expression for an equilibrium constant, Kc, for a
homogeneous equilibrium reaction is given below.
Kc = [A]2[B]
[C][D]3
(i) Write an equation for the forward reaction.
(ii) Deduce the units of Kc
(iii) State what can be deduced from the fact that the value
of Kc is larger when the equilibrium is established at a lower
temperature.
Jan 06 (3 marks)
And then the nasty
parts:
Some questions, like this one, give you partial pressures
after equilibrium, so you have to deduce the individual
partial pressures:
(b) A 36.8 g sample of N2O4 was heated in a closed flask of
volume 16.0 dm3. An equilibrium was established at a
constant temperature according to the following equation:
N2O4(g) 2NO2(g)
The equilibrium mixture was found to contain 0.180 mol of N2O4
(i) Calculate the number of moles of N2O4 in the 36.8 g sample.
For N2O4 Mr = 92.0, Mol = 36.8 ÷ 92.0 = 0.400 moles
(ii) Calculate the number of moles of NO2 in the equilibrium mixture.
moles N2O4 reacted = 0.400 – 0.180 = 0.220
moles NO2 formed = 0.440
And the rest:
(iii) Write an expression for Kc and calculate its value under these
conditions.
Expression for Kc:
Calculation:
Kc = [NO2]2 ÷ [N2O4]
Kc = (0.44/16)2 ÷ 0.18/16 = 0.067
(iv) Another 36.8 g sample of N2O4 was heated to the same
temperature as in the original experiment, but in a larger flask.
State the effect, if any, of this change on the position of equilibrium and
on the value of Kc compared with the original experiment.
Effect on the position of equilibrium:
Effect on the value of Kc:
Moves to right/forwards/to NO2
None
(9 marks)
Here’s another for you...
(Courtesy of somebody on Yahoo!Answers)
Is this equation at equilibrium?
• The Kc for this equation = 0.98
2FeBr3 (s)
2FeBr2 (g) + Br2 (g)
• A 3.0 L reaction vessel contains 3.6 moles of iron(ll)
bromide, 1.2 moles of iron(lll) bromide, and 2.1 moles of
bromine gas at this temperature.
Is this system at equilibrium?
“My notes say that when K < 1, the reaction is not at
equilibrium, but this question asks me to show my work,
and I don't know how to? Thanks in advance!”
• Now go back and teach them about the
graphs.
Acids and alkalis
Acids and alkalis
• Arrhenius definition:
– Acids create a supply of H+ ions in solution
• Bronsted & Lowry definition:
– Acids are proton donors
– Bases/Alkalis are proton acceptors
• Useful to know the solubility rules
(next page)
Solubility rules (abridged)
1. Group 1 ions always dissolve
2. Ammonium ions always dissolve
3. Metal oxides, carbonates and
hydroxides don’t dissolve, except as
rules 1 & 2
•
And some others that don’t affect us now....
Acid, base or alkali?
H+
OH-
NH3
NH4+
SO42-
I-
CH3COOH
ZnO
H3O+
H2O
NaOH
CH3COO-
• Note: water can be an acid or a base, losing
a proton to make OH-(aq) or picking-up a
proton to make H3O+(aq). It is amphoteric.
Oh, no, it’s another
equilibrium!
• H2O(l) H+(aq) + OH-(aq)
• (Or 2H2O(l) H3O+(aq) + OH-(aq))
• “Autoionisation of water”
• (You may see H3O+(aq) in some texts, called
hydronium or hydroxonium ion)
• The equilibrium lies very much to the left
under normal circumstances.
• Write an expression for the equilibrium
constant for this reaction
•
•
•
•
Kw = [H+(aq)][OH-(aq)]
Note new symbol Kw
Note [H2O] missing..... Why?
Note also, [H+(aq)] = [OH-(aq)] because
pure water must be neutral.
• Memorise – the equilibrium constant of
water (at 298K) is
•
Kw = 1 x 10-14
• So if [H+(aq)][OH-(aq)] = 1 x 10-14
• Then both are equal to 1 x 10-7
• Because 10-7 x 10-7 = 10-14
• An increase/excess of H+ will mean
reduction/decrease in OH• Because Kw can’t change for anything
except temperature
• So acid and base activity have opposite
effects, increasing or decreasing the
concentrations at equilibrium
• Two new notations; pH and pOH.
Logs (very briefly!) (Don’t
copy)
• If I want to multiply 1,000 by 10,000, I can
do it three ways:
This
index is This
• Get calculator and get answer a log index is
also a log
• Long multiplication
• Know 1,000 by 10,000 is 103 x 104
This
index is
• Add the 3 and the 4 to get 107
also a log
7
• Make 10 to be 10,000,000
• Wow, I just used logs without knowing!
pH
• Defined as
• pH = -log10[H+]
• Pure water: [H+] = 10-7 so pH=7
• 0.1M solution of HCl: [H+] = 0.1M
so pH = -log10(0.1) = -log10(10-1) so pH=1
• 0.1M solution of NaOH: [OH-] = 0.1,
• so [H+] = 1 x 10-14 / 0.1 = 1 x 10-13
• So pH = 13
Try these:
•
1.
2.
3.
4.
Assume all are strong acids or bases
and have been fully dissociated in
water.
0.05 moles/dm3 solution of HCl
0.02 moles/dm3 solution of H2SO4
0.15 moles/dm3 solution of NaOH
0.25 moles/dm3 solution of KOH
And from pH to
concentration:
• [H+] = 10-pH
• That is, you make the pH negative, then
raise 10 to that power.
• (you may need to force your calculator to give the
answer in scientific notation or do that yourself)
So, for practice:
• What is the hydrogen ion concentration of
the following, again assuming they refer
to strong acids or alkalis?
• A solution of pH 1.5
• A solution of pH 4.2
• A solution of pH 8.2 (and what is the
concentration of OH- in this solution?)
• A solution of pH 11.5 (and what is the
concentration of OH- in this solution?)
• Notation for any acid:
• HA, which can dissociate to make H+ and
a conjugate base A-
Weak Acids and Ka
• So much for strong acids, which can be
assumed to have totally dissociated
• Weak acids are more of an equilibrium:
• HA H+ + A• Write an expression for the dissociation
constant for this equation.
• Ka = [H+][A-]
[HA]
• Units, calculated as before, give mol dm-3
• Values of Ka much less than 1 are
generally weak acids, those with Ka much
greater than 1 are strong.
• E.g. HNO3............ 4.0 x 101
• CH3COOH........... 1.7 x 10-5
pKa
• Now we can quantify how much
dissociation is happening, we can derive
an expression for pKa:
• pKa = -log10(Ka)
• Smaller/lower values for pKa imply a
stronger acid than a higher value, as for
pH.
• E.g. pKa for CH3COOH is 4.75
• For HNO2 is 3.37
• For HCN is 9.31
Approximating Ka
• If the acid is very weak and does not
dissociate well, so the denominator can be
assumed to be the same as the original [HA]
• Also, [H+] must be the same as [A-]
• So we get
HAtot is just the total
+
2
• Ka ≈ [H ]
acid you dissolved
in the water at start
•
[HAtot]
• (same units, mol dm-3)
• This is the “weak acid approximation”
pH from Ka
• Use weak-acid approximation:
• Ka ≈ [H+]2
•
[HAtot]
• Therefore
Memorise!

[H ]  Ka [HA]tot
• Then derive pH using usual equation:
• pH = -log10[H+]

[H ]  Ka [HA]tot
e.g.
• What is the pH of a 0.1 mol dm-3
solution of methanoic acid?
(Ka = 3.6 x 10-4 mol dm-3)
• [HAtot] = 0.1
• So

[H ]  3.6 x 10 x 0.10
-4
•And so
•[H+] = 6.0 x 10-3 mol dm-3 and pH = 2.22
Practice (2-step)
• What is the pH of a 0.15M solution of
methanoic acid?
(Ka = 3.6 x 10-4 mol dm-3)
• What is the pH of a 0.1M solution of formic
acid?
(Ka = 1.8 x 10-4 mol dm-3)
• What is the pH of a 0.1M solution of citric(III)
acid (HC6H5O72-)?
(Ka = 4.1 x 10-7 mol dm-3)
• What is the pH of a 0.4M solution of
hypochlorous acid?
(Ka = 3.5 x 10-8 mol dm-3)
A different practice
• What is the Ka of an unknown acid, if a
0.1M solution has a pH = 2.5?
• What is the Ka of an unknown acid, if a
0.16M solution has pH = 3.1?
• What is the Ka of an unknown acid, if a
0.5M solution has pH = 2.8?
• What is the Ka of an unknown acid, if a
0.05M solution has pH = 4.1?
And another different
practice
• What is the concentration of an unknown
acid, given Ka = 4.3 x 10-5 & pH = 2.5?
• What is the concentration of an unknown
acid, given Ka = 1.9 x 10-4 & pH = 2.0?
• What is the concentration of an unknown
acid, given Ka = 2.8 x 10-3 & pH = 4.1?
• What is the concentration of an unknown
acid, given Ka = 3.0 x 10-2 & pH = 1.5?
•
Some problems
•
•
•
•
•
Calculate the pH of these solutions:
A. 0.20 M HCN
B. 1.00 M NaC2H3O2
C. 6.00 M HBr
D. 0.40 M Ba(OH)2
– (From Yahoo!Answers)
Titration and equivalence
• Equivalence is the point at which the
amounts of acid and base are equal.
• In strong/weak titrations, the equivalence
point, which we have previously thought
of as neutralisation, is not actually
neutral.
The curves
Strong acid,
strong base
Weak acid,
strong base
Weak acid,
weak base
Strong acid,
weak base
Practice
14
pH
7
1
Volume of added base
Titration and equivalence
• Try titrating the following with universal
indicator or a pH meter/logger:
• 25 ml 0.1M HCl and 25 ml 0.1M NaOH
• 25 ml 0.1M HCl and 25 ml 0.1M
ammonium hydroxide
• 25 ml 0.1M ethanoic acid and 25 ml 0.1M
ammonium hydroxide
• What differences are there in the pH
curves and why?
• So in a strong base/weak acid interaction,
the base is dominant at equivalence
because of the presence of the conjugate
anion – e.g. in
CH3COOH + NaOH, the presence of
CH3COO- anion (a Bronsted-Lowry base)
means that the solution is on the alkaline
side of neutral, even at equivalence.
• The same applies to strong acids and
weak bases:
HCl + NH4OH leaves some NH4+ in solution
at equivalence and this is a proton donor
and B-L acid
Skip titration
Titrations
VacidCacid ValkaliCalkali

nacid
nalkali
• Are calculated as in module 1:
• Equation as last year (above)
• E.g. 28.4 cm3 of 0.22 mol dm-3 Ba(OH)2
neutralises 25 cm3 solution of HCl.
What is the concentration of the acid?
• May need:
• 2HCl + Ba(OH)2  BaCl2 + 2H2O
• Can also get moles of known; find
equivalence; get moles of unknown.
Solution
• 2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + H2O(l)
• Moles Ba(OH)2 = 0.00625 mol
• Stoichiometry: 2 moles HCl needs one
Ba(OH)2
• Need 0.0125 moles in 25 cm3 so acid
concentration is 0.5 moles dm-3
Buffers
• Looking at the graphs, there are areas
where the pH value stays almost
constant, as more acid or alkali is added
• If we can force this situation, then we
have a buffer solution
• Buffers are: solutions that resist changes
in their pH caused by the addition of a
small amount of acid or base.
The principle
• A buffer solution contains either a
weak acid and a salt of its conjugate
base, or a weak alkali and a salt of its
conjugate acid
• (for information)
– Acid buffers are as far as the syllabus goes,
not basic buffers
– Blood is buffered at pH 7.4 by carbonate,
phosphate and protein systems; a blood pH
below 7.0 or above 7.8 quickly leads to death.
How it works
• Make a fairly concentrated solution of a
weak acid, HA
• This dissociates to give small amounts of
H+(aq) and A-(aq)
• HA(aq) H+(aq) + A-(aq)
• Add some NaA, giving lots of A- in solution
• NaA Na+ + A• This forces the equilibrium to the
left, creating more undissociated acid.
• E.g
• CH3COOH(aq) CH3COO-(aq) + H+(aq)
(With CH3COONa)
• If an acid is added, the large amount of Awill form more dissolved HA, since the
original acid is weak
• If a base is added, the free H+ is used-up
but the large amount of HA re-creates it by
dissociation. (base probably contains
metal ion, e.g. Na, Mg)
E.g. if you made a buffer solution of 1M
methanoic acid, CHOOH, with 0.5M of
sodium methanoate added:
• Consider the concentrations for 1 mole of acid
and 0.5 moles of salt in solution:
acid
1.00
H+
0
HCOO0.5
change in conc.
-y
+y
+y
at equilibrium
1-y
y
0.5 + y
initial
• At equilibrium, you get
• Ka = y (0.5+y)
(and Ka is usually given)
(1.0-y)
Calculating buffer pH
• Because the salt and acid are in excess,
compared to the H+ concentration we can
approximate:
 [H ][A - ]  [H ][salt]
 
Ka  
[acid]
 [HA] 
Ka [acid]

So H 
Memorise!
[salt]
• Then you can use the usual pH = -log10[H+]
• This assumes that the concentrations are
relatively large.
An odd thing
• Notice, in the equation:
 [H ][A - ]  [H ][salt]
 
Ka  
[acid]
 [HA] 
• If you add more water, then [salt] and [acid] are
reduced by dilution.
• In equal proportions
• So the effects cancel out
• So dilution does not affect the pH of a buffer.
• Also it’s the absolute amounts of substance that
matter, not the concentration, after all, you could put
vol=1 dm3, then the amounts would equate with the
concentrations.
Calculating buffer pH
• Alternatively, we can derive the expression:
 [HA(aq)]
  pH
pKa  log10   [A (aq)] 
• So buffer pH:
 [HA(aq)]

pH  pKa - log10   [A (aq)] 
[ACID]
OR pH  pKa - log10
[SALT ]
Memorise!
• Notice, as written, this is the ratio of
concentrations, not absolute values.
e.g.
• For a 1-litre buffer solution that is 0.3 mol dm-3
with respect to ethanoic acid (pKa = 4.76) and
0.2 mol dm-3 w.r.t. sodium ethanoate
• The initial pH is:
[ACID]
pH  pKa - log10
[BASE]
• = 4.76 – log10(0.3/0.2) = 4.58
And adding acid....
• When you add acid, the concentrations change,
one up, one down.....
• E.g. adding 50ml of 1 mol dm-3 hydrochloric
acid:
[ACID]
(0.3  0.05)
pH  pKa - log10
 4.76 - log10
 4.57
[BASE]
(0.2  0.05)
• Or, for an alkali, e.g. adding 50ml of 1 mol dm-3
of sodium hydroxide:
[ACID]
(0.3  0.05)
pH  pKa - log10
 4.76 - log10
 4.60
[BASE]
(0.2  0.05)
• Note, the pH changes logarithmically, so
the increase and decrease are not always
the same.
• Either way, the pH has only changed by
0.01 or 0.02!
Try:
• What is the pH of a buffer solution which
is 0.4 mol dm-3 with respect to ethanoic
acid and 0.2 mol dm-3 w.r.t. sodium
ethanoate? pKa = 4.76
• pH = 4.76 – log10(0.4/0.2)
• = 4.76 – log10(2)
• = 4.76 – 0.301 = 4.459 or 4.46
• And how will the pH change when 20ml of
2 mol dm-3 hydrochloric acid is added to a
litre of this buffer?
[ACID]
(0.4  0.04)
pH  pKa - log10
 4.76 - log10
 4.32
[BASE]
(0.2  0.04)
• So, pH changes from 4.46 to 4.32
• Not a big change, but notice, the base
concentration has reduced significantly in
this operation. If we had used less of the
buffer, it could have been all used up.
An odd exam question:
• Does a mixture of carbonic acid, H2CO3,
a weak acid, and sodium bicarbonate,
NaHCO3, in water constitute a buffer? If
no, explain why not.
If yes, explain why and use chemical
equations to show what happens when
either OH- or H3O+ is added to the
solution.
Another one....
• Not an amazingly well-worded question, but....
• A 1.0L sample of an aqueous solution is made to contain 0.200
mol CH3COOH and 0.100 mol CH3COONa.
A small quantity, 1.0 mL, of 12M HCl is added to the system.
a) Write a system of three (3) equations that outlines this
chemical situation.
b) Calculate the initial concentration of each of the chemicals
introduced into the system.
c) If the Ka of acetic acid is 1.8 x 10^-5, calculate the hydronium
ion concentration at equilibrium.
d) Hence calculate the pH of the solution before adding HCl
e) Calculate the pH of the system after the addition of HCl.
And another
• (from Yahoo Answers):
• Chemistry base strengths please help!?
• The equilibrium constants (Ka) for HCN and HF
in H2O at 25°C are 6.2 × 10–10 and 7.2 × 10–4,
respectively.
The relative order of base strengths is:
A) F– > H2O > CN–
B) H2O > F– > CN–
C) CN– > F– > H2O
D) F– > CN– > H2O
E) none of these