Course Name: Volumetric and Gravimetric
Analytical Chemistry
Course Code: 4022133-3
Lecture 7 contents
Blank titration
Polyprotic acid
Acid-Base Titrations
• Strong acid with strong base
• Weak acid with strong base
Titration of strong acid with strong base
• The chemical reaction between titrant and analyte is:
H+ + OH- H2O
• We have three region in the titration curve that represents
three different kinds of calculations:
1- Before reaching the eq. point:
the pH is determined by excess H+ in the solution
H+ + OH- H2O + H+
much little excess
2- At the eq. Point:
the pH is determined by the dissociation of water
H2O H+ + OH-
3- After the eq. Point:
the pH is determined by the excess OH- in the solution
H2O + OH- H2O + OH-
excess
Titration of strong acid with strong base
Example (1):
consider the titration of 50.0 mL of 0.100 M HCl using
a titrant of 0.200 M NaOH.
Step1:Calculate the volume of NaOH needed to reach the
equivalence point
moles ofClH= moles ofaOH
N
Ma.Va = Mb.Vb
‘a’ indicates the acid, HCl, and ‘b’ indicates the base, NaOH
Veq =Vb =VaMa = (0.100M
)(50.00ml
) = 25.00ml
Mb 0.200M
Titration of strong acid with strong base
Step 2: Before adding the titrant, the pH is determined by
the analyte, which in this case is a strong acid.
At the start of the titration the solution is 0.100 M HCl,
because HCl is a strong acid, means that the pH is
pH = − log[ H ] = − log[ HCl ] = − log( 0 .100 ) = 1 .00
+
Titration of strong acid with strong base
Step 3: Calculate pH values before the equivalence
point by determining the concentration of
unreacted analyte.
Before the equivalence point, HCl is present in excess and
the pH is determined by the concentration of unreacted
HCl.
Titration of strong acid with strong base
After adding 10.0 mL of NaOH the concentration of
excess HCl is
the pH increases to 1.30
Titration of strong acid with strong base
Step 4: calculate the pH at the equivalence point for
the titration of a strong acid with a strong base.
At the equivalence point the moles of HCl and the
moles of NaOH are equal. Since neither the acid nor
the base is in excess, the pH is determined for the
produced salt (salt of strong acid with strong base).
+ − + 2
K = [ H O ][OH ] = [ H 3 O ] = 1.00 X 10 − 14
w 3
[ H 3 O + ] = 1.00 X 1 0 − 7
the pH at the equivalence point is 7.00.
Titration of strong acid with strong base
Step 5: Calculate pH values after the equivalence point by
determining the concentration of excess titrant.
For volumes of NaOH greater than the equivalence
point, the pH is determined by the concentration of
excess OH–.
Titration of strong acid with strong base
After adding 30.0 mL of NaOH the concentration of
excess NaOH is
pOH = -Log [OH-] = -Log0.0125 = 1.90
pH =14-pOH = 12.10
Table1:Titration of 50.0 ml of HCl with 0.200M NaOH
Titration curve for the titration of 50.0 mL of 0.100 M HCl
with 0.200 M NaOH.
The red points correspond to the data in Table 1 . The blue
line shows the complete titration curve.
Example
Calculate the pH during the titration of 50.00 mL of 0.0500 M
NaOH with 0.1000 M HCl at 25°C after the addition of the
following volumes of reagent:
(a) 24.50 mL, (b) 25.00 mL, (c) 25.50 mL.
(a)
Example
Titration of weak acid with strong base
Example (1):
Consider the titration of 50.0 mL of 0.100 M acetic acid
CH3COOH using a titrant of 0.200 M NaOH. (Ka =
1.75x10-5 )
Step1:Calculate the volume of NaOH needed to reach the
equivalence point
Moles of CH3COOH = moles of NaOH
Ma x Va = Mb x Vb
‘a’ indicates the acid, CH3COOH, and ‘b’ indicates the base, NaOH
Veq =Vb = VaMa (0.100M
= )(50.00ml
) = 25.00ml
Mb 0.200M
Titration of weak acid with strong base
Step 2: Before starting titration (before adding the titrant),
the pH is determined by the analyte, which in this case is a
weak acid.
-Before adding NaOH the pH is that for a solution of 0.100 M
acetic acid.
-Because acetic acid is a weak acid,
CH3CCOH (aq) + H2O(l) H3O+(aq) + CH3COO-(aq)
Titration of weak acid with strong base
[H3O+] =[H+] = ka[acid]
pH= -Log [H+] = - Log 1.75 x10-5 x 0.1 =2.88
At the beginning of the titration the pH is 2.88.
Titration of weak acid with strong base
Step 3: Before the equivalence point, the pH is determined
by a buffer containing the analyte and its conjugate form.
Before the equivalence point the concentration of unreacted
acetic acid is
and the concentration of acetate is
Titration of weak acid with strong base
After adding 10.0 mL of NaOH the concentrations of
CH3COOH and CH3COO– are (buffer solution)
They form buffer
solution
Titration of weak acid with strong base
pH = pKa + log [salt]
[acid]
How to calculate the
pH of a buffer solution
Titration of weak acid with strong base
Step 4: The pH at the equivalence point is determined by the
analyte΄s conjugate form, which in this case is a weak base
-At the equivalence point the moles of acetic acid initially
present and the moles of NaOH added are identical.
- Because their reaction effectively proceeds to
completion, the predominate ion in solution is CH3COO–,
which is a weak base.
-To calculate the pH we first determine the concentration of
CH3COO–
Titration of weak acid with strong base
we calculate the pH of the produced salt (salt of week
acid and strong base)
[H+]= Kw.Ka
Csalt
[H+]= 10-14 X 1.75 x10 -5 /0.0667 = 1.62 x 10-9
pH = -Log [H+] = -Log 1.62 x 10-9
pH at the equivalence point is 8.79.
Titration of weak acid with strong base
Step 5: Calculate pH values after the equivalence point
by determining the concentration of excess titrant.
-After the equivalence point, the titrant is in excess and the
titration mixture is a dilute solution of NaOH.
-We can calculate the pH using the same strategy as in the
titration of a strong acid with a strong base.
Titration of weak acid with strong base
after adding 30.0 mL of NaOH the concentration of
OH– is
poH = -Log [OH-] = -Log0.0125 = 1.90
pH =14-pOH = 12.10
Table2: Titration of 50.0 mL of 0.100 M
Acetic Acid with 0.200 M NaOH
Titration curve for the titration of 50.0 mL of 0.100 M
CH3COOH with 0.200 M NaOH. The red points
correspond to the data inTable 2. The blue line shows
the complete titration curve.
Homework
Calculate the pH at 0, 10, 25, 50, 60 and 80 ml titrant, in the
titration of 50 ml of 0.1M CH3COOH with 0.1M NaOH.
