weak acid weak base weak acid strong base
strong acid weak base strong acid strong base
4.4 (a) Acid-base conjugate pairs
acid:- substance consisting of molecules or ions which
donates protons, (a proton donor).
base :- substance consisting of molecules or ions which
accepts protons, (a proton acceptor).
HCl(aq) + H20 (l) <--------> H3O+
(aq) + Cl-(aq)
H3O+ called hydronium ion.
acid 1 base
1
acid 2
base 2
conjugate acid conjugate base
HCl and Cl- formed from each other by the
loss or gain of a proton is a conjugate acid-base pair.
Weak acids donate donate a proton to H20 in
an equilibrium reaction.
CH3COOH(aq) + H20(1) <----->
H30+(1) + CH3COO-(1)
HCl and CH3COOH, can only donate one proton
each so are called monoprotic acids and each only have a single conjugate
base so are called monobasic acids. Water, which accepts the proton,
is a base. In the reverse reaction, hydronium ion is an acid because
it donates a proton to acetate ion and acetate ion is a base because it
accepts a proton. An acid which donates more than one proton is polyprotic.
An acid which has two conjugate bases is dibasic e.g. H2SO4
Water can also act as an acid by donating a proton to
a weak base.
H2O(aq) + NH3(aq) <----->
NH4+(aq) + OH-(aq)
acid 1 base 1
conjugate
acid 2 conjugate base 2
Acid - Base reactions can occur in the absence of water
for example nitric acid below acts as a base.
H2SO4(l) + HNO3(l) <----->
HSO4-(l) + H2NO3+(l)
4.4 (b) pH
pH = -log[H+(aq)]
where [H+] = the hydrogen ion concentration at equilibrium
What is the pH of a solution in which [H+]
is 0.025M i.e. 2.50*10-2moldm-3?
pH = -log[H+]
pH = -log2.50*10-2
pH = -(-1.60) = 1.60
What is [H+] of a solution of HCl of pH 2.30?
pH = -log[H+]
2.30 = -log[H+]
log[H+] = -2.30
[H+] = 10 -2.30 = 5.01*10-3 moldm-3
4.4 (c) Strong and weak acids and bases
A weak acid is an acid which is only slightly ionised
and so forms an equilibrium mixture.
HA(aq) <===> H+(aq) +A-(aq)
A strong acid, hydrochloric acid, (HCl), for example,
is completely completely ionised;
HCl(aq) ------> H+(aq) + Cl-(aq)
A strong acid will have a extremely large value for Ka,
whereas a weak acid will have a small value.
A concentrated acid has a high initial value for [HA]
and a dilute acid has a low initial [HA].
pH depends on [H+] which depends on concentration
and acid strength.
4.4 (d) Ka and Kw
In a pure sample of water, some of the molecules will
spontaneously transfer a proton from one water molecule to another. The
equilibrium reaction is written as:
2 H20 (1) <----> H3O+(aq)
+ OH-(aq)
Kw= [H3O+(aq)][ OH-(aq)]
= 1.0 X 10-14 mol2dm-6
Kw is known as the ionisation constant or ionic product
of water. It is a measure of how many water molecules ionise spontaneously.
Measurements have shown that this equilibrium is far to the left, with
only a small concentration of the ions.
For the general weak acid HA we can write the following
equilibrium:
HA(aq) + H2O(l) <-----> H3O+(aq)
+ A-(aq)
The equilibrium constant (acid dissociation constant)
Ka is
Ka= [ H3O+][ A-] / [HA]
units moldm-3
In each case the concentration of water is incorporated into Kc so
Kw =Kc[H2O] and Ka=Kc[H2O]
Task 4.4d What is Ka for a 0.1M solution of a weak
acid HA of pH 5?
What is the pH of a 0.2M solution of sodium hydroxide given that Kw=1.0 X 10-14 mol2dm-6?
4.4 (e) pKa and pKw
These quantities are normally very small inconvenient numbers. E.g. Ka
for ethanoic acid is about 10-5 mol dm-3 and Kw
is about 10-14 mol2dm-6. pKa
and pKw give more convenient numbers in the same way that pH values
are easier than hydrogen ion concentrations.
pKa = -logKa
pKw = -logKw
4.4 (f) pH of solutions of strong acids and bases
The hydrogen ion concentration and hence the pH of a strong acid can be
worked out directly from the concentration of the acid.
For 0.2M HCl, conc of HCl = 0.2 mol dm-3, one mole of HCl contains
one mole of H+ so the concentraion of hydrogen ions is 0.2 mol dm-3.
pH = -log 0.2 = 0.7
For 0.2M H2SO4, conc of H2SO4 = 0.2
mol dm-3, one mole of H2SO4 contains two moles
of H+ so the concentration of hydrogen ions is 0.4 mol dm-3.
pH = -log 0.4 = 0.4.
The hydrogen ion concentration and hence the pH of a strong base can be worked
out directly from the concentration of the base and the value for Kw.
For 0.1 M NaOH, conc of NaOH = 0.1 mol dm-3, as one mol of NaOH
contains one mole of OH- then conc of OH- = 0.1 mol dm-3.
Kw= 10-14 mol2 dm-6 = [H+][OH-]
= [H+]*0.1mol dm-3.
[H+] = 10-14 mol2 dm-6 / 0.1mol dm-3
[H+] = 10-13 mol dm-3
pH = -log [H+] = -log 10-13
pH = 13
4.4 (g) Calculations of pH and Ka
for weak acids
Find Ka if the pH of a solution of acid HA is 5.40 at 25oC and
the initial [HA] =0.100 moldm-3
pH = -log[H+] so [H+] = antilog -pH = antilog -5.40
= 3.98*10-6 mol dm-3
initial conc
0.100
0.00
0.00
HA <---->
H+ +
A-
equilb conc 0.100 - 3.98*10-6
3.98*10-6 3.98*10-6
assume 0.100 - 3.98*10-6 = 0.100
Ka = [H+][A-]/[HA] = (3.98*10-6)2
/0.100 mol dm-3
Ka = 1.58*10-10 mol dm-3
Find the pH of a solution of 0.100 mol dm-3
HA if Ka = 2.50*10-5 mol dm-3
let the equilibrium concentration of H+ be x
initial conc
0.100
0.00 0.00
HA <----> H+ + A-
equilb conc 0.100 - x x x
assume 0.100-x = 0.100
Ka = [H+][A-]/[HA]
so 2.50*10-5 mol dm-3 = x2/0.100
mol dm-3
x = [H+] = 1.58*10-3 mol dm-3
pH = -log [H+] = -log 1.58*10-3 =2.80
4.4 (h) Principles of acid-base titrations
An acid/base titration is a procedure used in quantitative
chemical analysis, in order to determine the concentration of either an
acid or a base. Generally, an alkaline solution of unknown concentration,
and of known volume, is added to a conical flask, by means of a 25.0 cm3 pipette.
An acid of known concentration is then added to the conical flask using
a burette, until the equivalence point is reached, i.e. when the stoichiometric
amount of acid has been added to the base, this is when all the alkali
has been neutralised and there is no excess acid or alkali present in the
solution. Normally, a visual indicator is used in order to help determine
the equivalence point by noticing the exact point at which the colour of
the solution changes. The point when the colour changes is the end point
of the titration. If the indicator is chosen correctly the end point and
the equivalence point are the same. A pH meter, or conductimetric method, can also
be used to determine the equivalence point in an acid/base titration.
4.4 (i) pH curves for titrations
weak acid weak
base
weak acid strong base
strong acid weak
base
strong acid strong base
4.4 (j) Determining Ka from titration curves
Plot the graph of pH of 0.1M weak acid e.g. ethanoic acid, against volume of
0.1M strong
alkali added as below.
For ethanoic acid Ka = [H+(aq)][CH3COO-(aq)]/[CH3COOH(aq)]
[H+(aq)] = Ka[CH3COOH(aq)]/[CH3COO-(aq)]
When half of the acid has reacted [CH3COOH(aq)] = [CH3COO-(aq)]
so
[CH3COOH(aq)]/[CH3COO-(aq)] =
1 and
[H+(aq)] = Ka
On the graph above this happens when 12.5 cm3 of alkali have been
added.
From the graph the pH at this point is 4.8.
Ka = [H+(aq)] = antilog -pH = antilog -4.8 = 1.58*10-5 mol
dm-3
4.4 (k) Choice of indicator from pKIn
values
Indicators are generally weak acids, HIn, in an equilibrium
with water. The colour depends on the position of the equilibrium
which in turn depends on the total [H30+], that is
on the pH of the solution to which the indicator is added.
HIn(aq) + H2O(l) -----> H30+(aq)
+ In-(aq)
colour 1
colour 2
The pH at which the colour changes depends on the acid
dissociation constant Ka, for the indicator known as the indicator constant
KIn.
KIn = [H3O+][In-]
/ [HIn]
pKIn = -logKIn
The colour of the indicator changes from one colour
to another at its end point. At the end point [HIn] = [In-]
Therefore, at the end point for an indicator (this is
when half the indicator is in acid form and half is in the
form of its conjugate base.):
KIn = [H3O+]
therefore -log KIn = -log[H3O+]
and pKIn = pH
To choose an indicator the pKIn value must
be in the right pH range for the titration being done. e.g. methyl orange
pKIn = 5.1 for pH range 4.2 - 6.3 (pKIn + or - one pH
unit)
Strong acid and strong base - any indicator.
Strong acid and weak base - low pH range 3.1- 4.4 e.g.
methyl orange.
Weak acid and strong base - high pH range 8.3 - 10.0
e.g. phenolphthalein.
Weak acid and weak base - narrow pH range, very hard
choice.
In each case pKIn must be matched to pH at
equivalence point of titration.
end point- mid way between 2 colours of indicator (a
property of the indicator)
equivalence point when the stoichiometric amounts of
acid and alkali have been added. End point and equivalence point must coincide
for an effective titration.
4.4 (l) Buffer solutions
A buffer is a solution that absorbs small amounts of
added acid or base while changing little in pH. A common buffer mixture
contains a weak base and its salt or a weak acid and its salt e.g. a mixture
of ethanoic acid and sodium ethanoate is a good buffer. Two equilibria
exist in this buffer.
CH3COOH(aq) <-----> CH3COO-(aq)
+ H+(aq)
If a small amount of H+(aq) is added it reacts
with CH3COO- which shifts the position of equilibrium
to the left thus keeping [H+], and hence pH, almost constant.
If a small amount of OH-(aq) is added it reacts with H+
which
shifts the position of equilibrium to the right producing more H+ thus
keeping the pH almost constant. [CH3COO-(aq)]
is large and kept constant because the equilibrium below lies far to the
right.
CH3COONa(aq) <-----> CH3COO-(aq)
+ Na+(aq)
Buffered solutions do change in pH upon addition of H+
or OH- ions. However, the change is much less than that would
occur if no buffer were present. Solutions which are very strongly
acidic or very strongly basic are rarely deliberately buffered by addition
of a chemical buffer because it already takes a relatively large amount
of strong acid or strong base to change their pH significantly.
Calculate the pH of a buffer made by dissolving 0.200 mol of ethanoic acid and 0.200mol of sodium ethanoate to give 1 dm3 of buffer solution given Ka[ethanoic aid] = 1.7*10-5moldm-3
initial conc
0.200
0.200
0
CH3COOH(aq) <----> CH3COO-(aq)
+ H+(aq)
equilib conc
0.200 - x
0.200 + x
x
assume 0.200 - x and 0.200 + x = 0.200
pH = pKa -log[HA] /[A-] = -logKa - log(0.200/0.200)
pH = -log 1.7*10-5- log(0.200/0.200) = 4.77
4.4 (m) Enthalpy of neutralisation and the
strength of acids and bases
Standard Molar Enthalpy of Neutralisation (/\Hn,m)
is the enthalpy change per mole of water formed in the neutralisation
of an acid by an alkali, (298K and 1 atm).
For the neutralisation of a strong acid, (HCl)and a strong
alkali, (NaOH), (/\Hn,m) =-57.3kJ mol-1 because
the ionic reaction below is common to all neutralisations of strong acids
with strong alkalis;
H+(aq) + OH-(aq) -----> H2O(l)
; /\Hn,m = -57.3 kJ mol-1
Weak acids and bases give a value for (/\Hn,m)
more positive that -57kJmol-1.
CH3COOH(aq) + NaOH(aq) ----->
CH3COO-(aq) + H2O(l)
/\Hn,m = -55.2 kJ mol-1
CH3COOH(aq) + NH4OH(aq)
-----> CH3COONH4(aq) + H2O(l)
/\Hn,m = -50.4 kJ mol-1
The energy missing goes into fully dissociating the weak
acid or base.