This document provides calculations to determine the pH at different points in a titration between 0.3 M HC2H3O2(aq) and 0.3 M NaOH(aq). It also provides similar calculations for a titration between 0.1 M NH3(aq) and 0.1 M HCl(aq). ICE tables and equilibrium expressions are used along with mole calculations to determine pH before, at the equivalence point, and after the equivalence point for both strong-weak acid base titrations. Concentration expressions and additional ICE tables are set up as needed when the reaction reaches the equivalence point and changes direction.

1. TITRATION CALCULATIONS

2. In a titration 20.0 mL of 0.300 M HC2H3O2(aq) with 0.30 M NaOH(aq), what
is the pH of the solution after the following volumes of NaOH(aq) have
been added?
a) 0.00 mL b) 10.0 mL c) equivalence pt
d) 30.0 mL
HA(aq) <===> H+
(aq) + A-
(aq)
a)
I 0.300M 0 0
C -x +x +x
E 0.300-x x x
Ka = [H+
(aq)][C2H3O2
-
(aq)]
[HC2H3O2(aq)]
1.8x10-5
= [x][x]
[0.300-x]  Assumption used
2.32379x10-3
= x
pH = -log [H+
]
pH = -log [2.32x10-3
]
pH = 2.63
.: pH = 2.63
STRONG-WEAK TITRATIONS

3. I 0.006 0.003 0
C -x -x +x
E 0.6-x 0.3-x x
b) nNaOH=CV
nNaOH=(0.3M)(0.010L)
nNaOH= 0.003 mol
nHC2H3O2
=(0.300M)(0.020L)
nHC2H3O2
=0.006mol
STRONG-WEAK TITRATIONS
HA(aq) + OH-
(aq) <===> H2O(l) + A-
(aq)
TIP: Use moles instead of concentration in the ICE table for this type of question
Since all of the strong base
reacts, the value of x is the
number of moles of base
-0.003-0.003 +0.003
0.003 0 0.003
Solution is a buffer (HA and A-
present) so we can use this formula:
[H3O+
] = Ka x nHA
nA-
= 1.8 x 10-5
(0.003)
(0.003)
pH = -log [H+
]
pH = -log [1.8x10-5
]
pH = 4.74
.: pH = 4.74

4. c)
STRONG-WEAK TITRATIONS
HA(aq) + OH-
(aq) <===> H2O(l) + A-
(aq)
I 0.006 0.006 0
C -x -x +x
E 0.6-x 0.6-x x
-0.006-0.006 +0.006
0 0 0.006
Moles are used in
this ICE table
Since this is all that remains,
1)convert to concentration
2)flip the chemical equation
3)set-up a new ICE table
Vtotal = Vacid + Vbase
Vbase= nNaOH
C
Vbase= 0.006mol
0.3M
Vbase= 0.02L
Vtotal = 0.020L + 0.020L
Vtotal = 0.040L
C= n
Vtotal
C= 0.006mol
0.040L
C= 0.15 M

5. c)
STRONG-WEAK TITRATIONS
I 0.15 0 0
C -x +x +x
E 0.15-x x x
A-
(aq) + H2O(l) <===> HA(aq) + OH-
(aq)
Since the equation is flipped, you must use kb instead of ka
Kb = [OH-
(aq)][HA(aq)]
[A-
(aq)]
Kb = (x)(x)
(0.15-x)
Kb = Kw
Ka
Kb = 1.0x10-14
1.8x10-5
Kb = 5.5x10-10 5.5 x 10-10
= (x)(x)
(0.15) Assumption used
9.1287x10-6
M = x
9.1287x10-6 9.1287x10-6
pOH = -log [OH-
]
pOH = -log [9.1287x10-6
]
pOH = 5.03959
pH = 14-5.03959
pH = 8.96
.: pH = 8.96

6. d)
nNaOH=CV
nNaOH=(0.3M)(0.030L)
nNaOH=0.009mol
nHA = 0.006mol
nNaOH – nHA = 0.009mol – 0.006mol
STRONG-WEAK TITRATIONS
HA(aq) + OH-
(aq) <===> H2O(l) + A-
(aq)
Solve for pH from remaining concentration of base (no ICE table needed)
= 0.003mol NaOH remaining
C = nNaOH
Vtotal
C = 0.003mol
0.050L
C = 0.06M
pOH = -log [OH-
]
pOH = -log [0.06]
pOH = 1.2218
pH = 14-1.2218
pH = 12.78
.: pH = 12.8

7. In a titration 20.0 mL of 0.1 M NH3(aq) is titrated with 0.1 M HCl(aq).
What is the pH of the resulting solution after the following volumes of
HCl(aq) have been added?
a) 0.00 mL b) 10.0 mL c) equivalence pt
d) 30.0 mL
NH3(aq) + H2O(l) <===> OH-
(aq) + NH4
+
(aq)
a)
I 0.1M 0 0
C -x +x +x
E 0.1-x x x
Kb = [OH-
(aq)][NH4
+
(aq)]
[NH3(aq)]
1.8x10-5
= [x][x]
[0.1-x] Use assumption
1.34164x10-3
= x
pOH = -log [OH-
]
pOH = -log [1.34x10-3
]
pOH = 2.87
pH = 14 - 2.87
pH = 11.13
.: pH = 11.13
STRONG-WEAK TITRATIONS

8. NH3(aq) + H+
(aq) <===> H2O(1) + NH4
+
(aq)
b) nNH3
=CV
nNH3
=(0.1M)(0.020L)
nNH3
=0.002 mol
nHCl=CV
nHCl=(0.1M)(0.010L)
nHCl=0.001mol
STRONG-WEAK TITRATIONS
I 0.002 0.001 0
C -x -x +x
E 0.002-x 0.001-x x
-0.001-0.001 0.001
0.001 0 0.001
Since all of the strong acid
reacts, the value of x is the
number of moles of acid
Moles are used
in this ICE table
Solution is a buffer (B and HB+
present) so we can use this formula:
[OH1
] = Kb x nB
nHB+
= 1.8 x 10-5
(0.001)
(0.001)
= 1.8 x 10-5
M
pOH = -log [OH-
]
pOH = -log [1.8x10-5
]
pOH = 4.74
.: pH = 9.26
pH = 14 - 4.74
pH = 9.26

9. NH3(aq) + H+
(aq) <===> H2O(1) + NH4
+
(aq)c)
STRONG-WEAK TITRATIONS
I 0.002 0.002 0
C -x -x +x
E 0.002-x 0.001-x x
-0.002-0.002 0.002
0 0 0.002
Moles are used
in this ICE table
Since this is all that remains,
1)convert to concentration
2)flip the chemical equation
3)set-up a new ICE table
Vtotal = Vbase + Vacid
Vacid= nHCl
C
Vacid= 0.006mol
0.3M
Vacid= 0.02L
Vtotal = 0.020L + 0.020L
Vtotal = 0.040L
C= n
Vtotal
C= 0.002mol
0.040L
C= 0.05 M

10. H2O(1) + NH4
+
(aq) <===> NH3(aq) + H+
(aq)c)
STRONG-WEAK TITRATIONS
I 0.05 0 0
C -x +x +x
E 0.05-x x x
5.555x10-10
= [x][x]
[0.05-x] Use assumption
5.555x10-5
= [x][x]
[0.05]
5.27x10-6
= x
pH = -log [H+
]
pH = -log [5.27x10-6
]
pH = 5.28
.: pH = 5.28
Since the equation is flipped, you must use ka instead of kb

11. NH3(aq) + H+
(aq) <===> H2O(1) + NH4
+
(aq)
d)
STRONG-WEAK TITRATIONS
Solve for pH from remaining concentration of acid (no ICE table needed)
nNH3
=CV
nNH3
=(0.1M)(0.020L)
nNH3
=0.002 mol
nHCl=CV
nHCl=(0.1M)(0.030L)
nHCl=0.003mol
nHCl – nNH3
= 0.003mol – 0.002mol
= 0.001mol HCl remaining
C = nHCl
Vtotal
C = 0.001mol
0.050L
C = 0.02M
pH = -log [H+
]
pH = -log [0.02]
pH = 1.69897
.: pH = 1.70