Chapter 17-4
Taylor Walsh
Shiv Patel
Emily Penn
Philip Adejumo
 Be
able to read a titration curve
 Understand how titrations work
 Perform titration calculations
 Equivalence
point- the point at which
stoichiometrically equivalent quantities of
acids and bases have been brought together
http://www.chemguide.co.uk/physical/ac
idbaseeqia/phcurves.html
 Titration-
when a solution containing a
known concentration of base is slowly added
to an acid (or vice versa)

Titration enables us to find the equivalence point
of the acid-base solution
http://www.dartmouth.edu/~chemla
b/techniques/graphics/titration/titr
ation6.gif
A
titration curve is a graph of the pH as a
function of the volume of the added acid or
base
 There are 3 types of titrations with distinct
titration curves:



Strong acid-strong base
Weak acid-strong base
Polyprotic acid-strong base
Final pH
Rapid Rise Portion
Initial pH
Equivalence Point
http://www.files.chem.vt.edu/chemed/titration/graphics/titration-strong-acid-35ml.gif
http://0.tqn.com/d/chemistry/1/0/f/g/s
atitration.JPG
The initial pH
1.
4
The initial pH is a purely acidic
solution
a.
Between the initial pH and
the equivalence point
3
2.
pH slowly rises at first, then more
Rapidly when it gets close to the
Equivalence point

3.
4.
The equivalence point
After the equivalence point
2
1
Ex. .100 M
NaOH added to
50.0 mL of .100
M HCl
1.
2.
First determine how many moles of H+ were
originally present and how many moles of OHwere added
Subtract the two values (moles) to calculate
moles of H+
Ex. Calculate the pH when the following
quantities of 0.100 M NaOH solution have
been added to 50.0 mL of 0.100 M HCl
solution
There are more moles of
H+ than moles of OH-, so
the resulting value will be
moles of H+
Ex. Calculate the pH of 49.0 mL of 0.100 M NaOH solution after 50.0 mL of
0.100 M HCl solution was added
(0.0500L
0.100 𝑚𝑜𝑙 𝐻 +
soln)(
1 𝐿 𝑠𝑜𝑙𝑛
(0.0490 L
)= 5.00 x 10-3 mol H+
0.100 𝑚𝑜𝑙 𝑂𝐻 −
soln)(
)=
1 𝐿 𝑠𝑜𝑙𝑛
4.90 x 10-3 mol OH-
(5.00 x 10-3 mol H+) – (4.90 x 10-3 mol OH- ) = 0.10 x 10-3 mol H+
[H+]
𝑚𝑜𝑙𝑒𝑠 𝐻 +
=
𝑡𝑜𝑡𝑎𝑙 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑛
=
pH= -log (1.0 x 10-3) = 3.00
0.10 × 10−3 𝑚𝑜𝑙
0.10 × 10−3 𝑚𝑜𝑙
=
= 1.0 x 10-3
0.0500 𝐿 + 0.0490 𝐿
0.09900 𝐿
4
1.
The initial pH

2.
pH of the acid
3
2
Between the initial pH and the
1
Equivalence point
3.
The equivalence point
4.
After the equivalence point
 The
solution of the weak acid has a higher
initial pH than a solution of a strong acid of
the same concentration
 The pH change at the rapid-rise portion of
the curve is smaller for the weak acid than
it is for the strong acid
 The pH at the equivalence point is above
7.00 for the weak acid-strong base titration

Equivalence point for strong acid-strong base is
always at 7.00 pH
 Calculate
[HX] and [X-] after reaction
 Use [HX], [X-], and Ka to calculate
 Use [H+] to calculate pH
Ex: Calculate the pH of the solution formed
when 45.0 mL of 0.100 M NaOH is added to
50.0 mL of 0.100 M HC2H3O2, (Ka = 1.8 x 10-5)
(.0500 L soln)(
HC2H3O2
(.0450 L
Before rxn
0.100 𝑚𝑜𝑙 𝐻𝐶2𝐻3𝑂2
1 L soln
0.100 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
soln)(
)
1 𝐿 𝑠𝑜𝑙𝑛
(5.00 x 10−3 mol)
HC2H3O2 (aq)
After rxn
(0.50 x 10−3 𝑚𝑜𝑙)
)= 5.00 x 10-3 mol
= 4.50 x 10-3 mol NaOH
(4.50 x 10−3 mol)
+
OH0.0 mol
0.0 mol
C2H3O2- (aq)
4.50 x 10−3 mol
45.0 mL + 50.0 mL = 0.0950 L
[HC2H3O2] =
[C2H3O2-] =
0.50 × 10−3 𝑚𝑜𝑙
=
0.0950 𝐿
.0053 M
4.50 x 10−3 mol
= .0474 M
0.0950 𝐿
[H+][C2H3O2−]
Ka =
= 1.8 x 10-5
[HC2H3O2]
[H+] = Ka x
[HC2H3O2]
-5) x 0.0053 = 2.0 x 10-6 M
=
(1.8
x
10
0.0474
[C2H3O2−]
pH = -log(2.0 x 10-6) = 5.70
 Calculate
the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M HC2H3O2
with 0.100 M NaOH
Moles=M x L= (0.100 mol/L)(0.0500 L) = 5.00 x 10-3 mol
HC2H3O2
[C2H3O2-]=
5.00 × 10−3 𝑚𝑜𝑙
0.1000 L
= (0.0500 M)
Since C2H3O2- is a weak base:
C2H3O2- (aq) + H2O (l)
Kb=
𝐾𝑤
𝐾𝑎
HC2H3O2 (aq) + OH- (aq)
(1.0x10−14)
=
= 5.6x10-10
(1.8x10−5)
Calculate the pH at the equivalence point in the titration of
50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
(x)(x)
[HC2H3O2−][OH−]
-10
Kb=
=
=
5.6
x
10
(0.0500 – x)
[C2H3O2−]
X = [OH-] = 5.3 x 10-6 M
pOH = 5.28
pH = 8.72
 When
weak acids contain more than one
ionizable H atom (H3PO3)
 Neutralization occurs in a series of steps
H3PO3
HPO3-2
H2PO3http://www.files.chem.vt.edu/RVGS/APChem/lab/Exp
eriments/images/titration_curve.jpg