TITRATION CURVES - KEY POINTS
Weak Acid with Strong Base
A: Initial pH of acid = pH of weak acid of given concentration.
From A pH rises steeply at first as strong base is added.
B: Region B = buffer region. pH changes little (centred around pKa of weak
acid) – 1 pH unit either side of pKa value. Occurs at ‘½ neutralisation value’
of added base.
At this point [HA] = [A-]
C: Equivalence point / Stoichiometric point
n(acid) = n(base). Occurs at midpoint of steepest part of curve. pH at this
point = pH of salt of weak acid, half the concentration of the original acid.
e.g. if initial acid is 0.100 mol L-1 HA (titrated with NaOH), then at
equivalence point, pH is that of a 0.0500 mol L-1 NaA solution.
--
--
A + H2O
Kb =
HA + OH
[ HA][OH  ]
[ A ]
[OH-] = Kb [A-]
[OH-]2 = Kb [A-]
K b = K w / Ka
[H3O+] = Kw / [OH-]
pH = -log[H3O+]
D: End of titration pH ; Volume of added base has diluted the solution. From
the stoichiometric point onwards the solution has no way of reacting with
with added base and it is merely diluted.
e.g. If 10.0 mL of NaOH is needed to reach equivalence point and a
further 10.0 mL is added after this, the final volume is 30.0 mL.
(conc NaOH = 0.100 molL-1). pH is that of 0.0333 mol L-1 NaOH
solution.
Example: Consider a titration of 10.0 mL of 0.100 mol L-1 CH3COOH with
0.100 mol L-1 NaOH. PKa (ethanoic acid ) = 4.7. Follow and complete the
calculations below and mark each point (A to E) on the graph on the next
page.
A: Initial pH: Ka =
[ H 3O  ][CH 3COO  ]
[CH 3 COOH ]
[H3O] = 2.00 X 10-5 X 0.100
LRY 2013
[H3O] = Ka [CH3COOH]
= 1.414 X 10-3
pH = 2.8
Titration Curve
C: At equivalence point –
PH = that of 0.0500 mol L-1
solution of CH3COONa.
[OH-] = Kb[CH3COO-]
=
=
[H3O+] =
pH =
pH
B: At ½ way to equivalence
point
pH = pKa = 4.7
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0
D: At 20 mL of added NaOH, pH
is that of 0.0333 mol L-1 NaOH.
[OH-] = 0.0333 mol L-1
[H3O+] =
pH =
2
4
6
8
10 12 14 16 18 20
vol NaOH (mL)
E: At 15 mL added base – the unreacted base = 15 – 10 mL = 5 mL
Therefore n(unreacted NaOH) = 0.100 x 5 x 10-3 mol
=
Total volume = 25 mL. Therefore [NaOH] =
=
mol L-1
[OH-] =
[H3O+] =
pH =
Choice of indicator:
The best indicator for this titration is one which changes colour in the steepest
part of the curve. The pH at the end point is 8.7 and since an indicator
changes colour over a range of 2 pH units near its pKa value, we need an
indicator whose pKa value lies in that region of the curve. What would be
suitable?
LRY 2013