Calculation of pH in AcidBase Titrations

1
Calculation of pH in Acid-Base Titrations
Pavan Balabathula

• pH is determined by the acid dissociation
  reaction of BH22
• BH22 ? BH H Ka1Kw/Kb2
  
• F-x x x
• x2 / (0.033 3-x) 1.010-5 ? x
  5.7210-4
• ? pH3.24

Titrations of Diprotic Bases
A titration is a technique where a solution of
known concentration is used to determine the
concentration of an unknown solution. Typically,
the titrant (the known solution) is added from a
buret to a known quantity of the analyte (the
unknown solution) until the reaction is complete.

Titration of Weak Acid with Strong Base
Titration of 50.00 mL of 0.020 00 M MES with
0.100 0 M NaOH MES 2-(N-morpholino)ethanesulfonic
acid, pKa 6.27 HA OH- ? A- H2O Volume(Vb)
of base needed to reach equivalence point Vb(mL)
(0.100 0M) (50.00 mL) (0.020 00M) Vb 10.00 mL
Buffer region
Excess OH-
Titration of 10.00 mL of 0.100 M base (B) with
0.100 M HCl. Base is dibasic, with pKb1 4.00
and pKb1 9.00 The titration curve is
reasonably sharp breaks at both equivalence
points, corresponding to the reactions
B H ? BH
BH H ? BH22 The volume at the first
equivalence point is 10 mL Ve (mL)
(0.100 M) (10.00mL) (0.100 0 M)
? Ve 10.00 mL
When you carry out a simple acid-base titration,
you use an indicator to tell you when you have
the acid and alkali mixed in exactly the right
proportions to "neutralize" each other. When the
indicator changes color, this is often described
as the end point of the titration. In an ideal
world, the color change would happen when you mix
the two solutions together in exactly equal
proportions. That particular mixture is known as
the equivalence point.
Beyond the second Eq. Point
Va gt20.0 mL, pH of the solution can be
calculated from the volume of strong acid added
to the solution Let Va 25.0 mL, means there is
an excess of 5.00 mL of 0.100 M HCl in a total
volume of 10.0025.0035.00 mL H (0.100 M)
(5.00/35.00) 1.4310-2 M ? pH 1.85
Before Base is Added
Solution of 0.020 00 M HA with pKa 6.27 So
simply weak acid problem HA ? H A-
Ka 10-6.27 F-x x
x x2 / (0.020 00-x) Ka ? x
1.0310-4 ? pH
3.99
The volume at the second equivalence point must
be 2Ve as the second reaction requires the same
number of moles of HCl as the first reaction pH
calculations are similar to those for
corresponding points in the titration of a
monobasic compound
Titration of Strong Base with Strong Acid
First write chemical reaction between titrant and
analyte Let 50.00 mL of 0.020 00 M KOH titrated
with 0.100 0 M HBr H(aq) OH-(aq) ? H2O(l)
K 1/Kw 1014 Any amount of H added will
consume a stoichiometric amount of OH- (Ve (mL)
) (0.100 0 M) (50.00 mL) (0.020 00
M) mmol of HBr at equivalence point
mmol of OH- being titrated
Ve 10.00 mL
First Buffer region
Second buffer region
Excess H
Before the Equivalence Point
Titration of Weak Base with Strong Acid

• Adding OH- creates a mixture of HA and A-
• This mixture is a buffer whose pH cna be
  calculated using Henderson-Hasselbalch equation
  from quotient A- / HA
• pH pKa log (A- / HA)
• Let 3.00 mL of OH- added
• HA
  OH- ? A- H2O
• Relative initial quantities (HA1) 1
  3/10
• Relative final quantities
  7/10 3/10
• ? pH pKa log (3/10)/ (7/10) 5.90
• When Vb ½ Ve , pH pKa

Reverse the titration of a weak acid with a
strong base B H ? BH 1. Before acid is
added , the solution contains just the weak base,
B, in water pH is determined by Kb of the
reaction B H2O ? BH OH- F-x
x x 2. Between initial
point and equivalence point, there is a mixture
of B and BH , buffer pH pKa (for BH) log
(B/BH )
Before the Equivalence Point

• Let 3.00 mL HBr is added from buret, then
• Total moles of OH- in the solution
• 50.0010-3 L 0.0200 M 1.0010-3 moles of
  OH-
• Number of moles of H added
• 3.00 10 -3 L 0.1000 M 0.3010-3 moles of
  H
• Thus remains
• (1.0010-3 ) (0.3010-3 ) 0.7010-3 moles of
  OH-
• Total volume of the solution is
• 50.0010-3 L 3.00 10 -3 L 53.00 10 -3
  L
• OH- (0.7010-3 moles)/(53.00 10 -3 L )
  0.0132 M
• H Kw / OH- (1.010-14
  )/(0.0132)7.5810-13 M
  
• pH -log H 12.12

3. At the equivalence, B has been converted into
BH , a week acid. The pH is calculated by
considering the acid dissociation reaction of
BH BH ? B H
KaKw/Kb F'-x x
x The formal conc. Of BH,F', is not the original
formal concentration of B, because some dilution
has occurred. The solution contains BH at the
equivalence point, so it is acidic. The pH at the
equivalence point must be below 7. 4. After the
equivalence point , the excess strong acid
determines the pH. We neglect the contribution of
weak acid, BH
Point A Before acid is added, solution contains
just weak base B B H2O
? BH OH- 0.100-X
X X X2
/(0.100-X) 1.0010-4 ? X 3.1110-3
H Kw / x ? pH 11.49
At the Equivalence Point
There is exactly NaOH to consume HA
HA OH- ? A-
H2O Relative initial quantities 1
1 Relative final quantities
1 A- H2O
? HA OH- KbKw/Ka
F-x
x x The formal concentration of A- in
no longer 0.020 00 M , which was the initial
concentration of HA. The A- has been diluted by
NaOH from the buret.
Point B At any point between A(the initial
point) and C (the first equivalence point), we
have a buffer containing B and BH Point B is
halfway to the equivalence point ? B BH
pH is calculated from Henderson-Hasselbalch
equation for the weak acid, BH, whose acid
dissociation constant is Ka2 (for BH22 )
Kw/Kb1 10-10.00 pH pKa2 log B / BH
10.00 log 1 10.00 To calculate B / BH
at any point in the buffer region, just find what
fraction of the way from point A to point C the
titration has progressed.
At the Equivalence Point
H2O ? H OH- , let at equilibrium H OH-
x Kw x2 ? x 1.0010-7 M
pH 7.00
initial volume of HA
Strong base

• F (0.020 00 M) 50.00
  0.0167 M
• 50.0010.00
• Initial conc. of Dilution
  factor
• HA
• x2 / (F-x) KbKw/Ka 1.8610-8 ? x 1.76
  10-5 M
• pH -log H -log (Kw/x) 9.25
• The equivalence point pH will always above 7
  for the titration of weak acid, because the acid
  is converted into its conjugate base at the
  equivalence point

After the Equivalence Point
Adding excess HBr to the solution The
concentration of excess H , say 10.50 mL H
(0.100 0M) 0.50
0.0132 M
50.0010.50 Initial conc.
of Dilution factor
H pH
-log H 3.08
Total volume of solution
Weak base
Volume of excess H
For example, if Va 1.5mL, then,
(B / BH) (8.5/1.5) because 10.00mL are
required to reach the equivalence point and we
added just 1.5mL The pH at Va 1.5mL is
pH10.00log(8.5/1.5)10.75
Total volume of solution
After the Equivalence Point
Point C At first equivalence point , B has
converted in to BH, the intermidiate form of the
diprotic acid, BH22 pH ? ½ (pKa1/ pKa2) Point
D At any point between C and E, there is a
buffer containing BH(base) and BH22 (acid)
When Va 15 mL, BH BH22 pH pKa1log
(BH/BH22 )5.00 log 15.00
Excess OH-
Excess H

• Adding NaOH to a solution of A- , means pH is
  determined by the excess OH-
• Let Vb 10.10 mL
• OH- (0.100 0 M) 0.10
  1.6610-4 M
• 50.0010.00
• Initial conc. of
  Dilution factor
• OH-
• pH - log (Kw / OH- )
  10.22

Volume of excess H
http//www.ausetute.com.au/titrcurv.html
Reference
Total volume of solution
Quantitative Chemical Analysis (Seventh
Edition) Daniel C. Harris

• Point E Second equivalence point
• Formal conc. Of BH22 is
• F (0.100 M) (10.0/30.0) 0.033 3 M

http//www.chem.ubc.ca/courseware/pH/section14/con
tent.html
Acknowledgements
Dr. Daniel Tofan CHE 805