The titration of 50.0mL of 0.100M HC2H3O2 (Ka=1.8 x 10-5) with 0.100M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added. Use these results to plot the titration curve.
pH
Vol. NaOH
0.0mL
12.5mL
25.0mL
37.5mL
49.9mL
50.0mL
50.1mL
75.0ml
0.0ml of 0.100M NaOH to 50.0ml of 0.100M HC2H3O2
Part 1- No Base, only weak acid
Set up the ice box using the molarity of the weak acid.
(Note the double arrow, use molarity)
HAc
<==>
H+
+ Ac-
I
0.100M
---
---
C
-X
+X
+X
E
0.100M-X
+X
+X
Ka=
[H+][Ac-]
----------------------------------
[HAc]
Plug in the equilibrium values (E)
1.8x10-5=
[X][X]
---------------------------------
[0.100M-X]
If Ka x the molarity is >10-3, then X in the denominator will be too small and can be disregarded.
1.8x10-5=
[X][X]
---------------------------------
[0.100M]
Solve for X
X=[H+]=0.00134M
pH=-log[H+]=-log 0.00134M
pH=2.87
Part 2- Adding base to the weak acid
12.5ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles HAc=50.0mL x 0.100M=5.00mmol
Moles OH-=12.5mL x 0.100M=1.25mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
1.25mmol
------
C
-1.25mmol
-1.25mmol
+1.25mmol
E
3.75mmol
0
1.25mmol
Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.
pH=pKa+log
[Base]
[Acid]
pKa=-log Ka=-log 1.8x10-5
pKa=4.74
You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.
pH=4.74+log
1.25mmol
3.75mmol
pH=4.26
25.0ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles HAc=50.0mL x 0.100M=5.00mmol
Moles OH-=25.0mL x 0.100M=2.50mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
2.50mmol
------
C
-2.50mmol
-2.50mmol
+2.50mmol
E
2.50mmol
0
2.50mmol
This is the halfway point of the titration. When you consume 1/2 the acid, which produces the same number of moles of the base.
pH=pKa+log
[Base]
[Acid]
The ratio is of acid to base is 1:1.
log1=0
Therefore, pH=pKa
pH= pKa=4.74
37.5ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles HAc=50.0mL x 0.100M=5.00mmol
Moles OH-=37.5mL x 0.100M=3.75mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
3.75mmol
------
C
-3.75mmol
-3.75mmol
+3.75mmol
E
1.25mmol
0
3.75mmol
Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.
pH=pKa+log
[Base]
[Acid]
pKa=-log Ka=-log 1.8x10-5
pKa=4.74
You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.
pH=4.74+log
3.75mmol
1.25mmol
pH=5.22
49.9ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles HAc=50.0mL x 0.100M=5.00mmol
Moles OH-=49.9mL x 0.100M=4.99mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
4.99mmol
------
C
-4.99mmol
-4.99mmol
+4.99mmol
E
0.01mmol
0
4.99mmol
Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.
pH=pKa+log
[Base]
[Acid]
pKa=-log Ka=-log 1.8x10-5
pKa=4.74
You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.
pH=4.74+log
4.99mmol
0.01mmol
pH=7.44
Part 3- The equivalence point.
Where the moles of acid= moles of base
50.0ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles H+=50.0mL x 0.150M=7.50mmol
Moles OH-=50.0mL x 0.150M=7.50mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
5.00mmol
------
C
-5.00mmol
-5.00mmol
+5.00mmol
E
0
0
5.00 mmol
The acetate ion will reestablish equilibrium by reacting with the water. The reaction reverses. (Note the double arrow, use Molarity)
Ac-
+H2O
<==> HAc
+ OH-
Find the concentration of the acetate ion using the total volume.
(add the ml of acid and mL of the base)
Molarity=
mol
=
mmol
liters
mL
[Ac-]=
5.00mmol
50.0ml + 50.0ml
[Ac-]=0.05M
Set up the ice box using the molarity of the acetate ion.
Ac- +
H2O
<==> HAc
+ OH-
I
0.05M
------
C
-X
+X
+X
E
0.05M-X
+X
+X
Write the equilibrium expression.
(Note=K is now Kb..there is OH-)
Kb=
[HAc][OH-]
----------------------------------
[Ac-]
Solve for Kb
Kb=
Kw
----------------------------------
Ka
Kb=
10-14
=5.56 x 10-10
----------------------------------
1.8 x 10-5
Plug values into the Kb expression
Kb=
[HAc][OH-]
----------------------------------
[Ac-]
5.56 x 10-10=
[X][X]
----------------------------------
[0.05-X]
If Kb x the molarity is >10-3, then X in the denominator will be too small and can be disregarded.
5.56 x 10-10=
[X][X]
----------------------------------
[0.05]
Solve for X
X=[OH-]=5.27 x 10-6
Solve for pOH
pOH=-log[OH-]=-log 5.27 x 10-6M
pOH=5.28
Solve for pH
pH=14-5.28=8.72
Part 4- Beyond the equivalence point. Excess base.
50.1ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles H+=50.0mL x 0.100M=5.00mmol
Moles OH-=50.1mL x 0.100M=5.01mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
5.01mmol
------
C
-5.00mmol
-5.00mmol
+5.00mmol
E
0
0.01mmol
5.00mmol
There are 2 bases present the weak base Ac- and the strong base OH-. Ignore the Ac-.
Find the concentration of the hydroxide ion using the total volume. (add the ml of acid and mL of the base)
Molarity=
mol
=
mmol
liters
mL
[OH-]=
0.01mmol
=0.000 1M
50.1mL +50.0mL
pOH=-log[OH-]
pOH=-log[0.000 1M]=4.00
pH=14-4.00=10.00
75.0ml of 0.100M NaOH to 50.0mL of 0.100M HC2H3O2
Moles H+=50.0mL x 0.100M=5.00mmol
Moles OH-=75.0mL x 0.100M=7.50mmol
Set up the ice box using the mmol.
(**Note only a one way arrow, use moles)
HAc +
OH-
è Ac-
+H2O
I
5.00mmol
7.50mmol
------
C
-5.00mmol
-5.00mmol
+5.00mmol
E
0
2.50mmol
5.00mmol
There are 2 bases present the weak base Ac- and the strong base OH-. Ignore the Ac-.
Find the concentration of the hydroxide ion using the total volume. (add the ml of acid and mL of the base)
