Titration of a strong acid with a strong base
Titrate HCl (aq.) with NaOH (aq.)
volume of NaOH added
At the equivalence point, the number of moles hydroxide added = number of moles H3O+
that is present.
The end point is where the indicator changes color. Choose the indicator appropriately so
that the color change occurs on the vertical portion of the titration curve.
Often we wish to determine the pH at various points during a titration.
If we titrate 50.00 mL of 0.1000 M HCl with 0.1000 M NaOH:
0 mL NaOH added pH = -log(0.1000) = 1.0000
After adding 25.00 mL of the NaOH solution? We must determine how many moles of
NaOH were added. (0.02500 L)*(0.1000 M) = 2.500 x 10-3 mol
H3O+
initial moles
5.000 x 10-3
moles after 2.500 x10-3
-OH
+
2.500 x10-3
0
How do we determine the pH at this point?
2 H2O
After addition of 49.90 ml of NaOH, you have added 0.004990 mol hydroxide.
H3O+
initial moles
-OH
+
5.000 x 10-3
moles after 1.000 x10-5
2 H2O
4.990 x10-3
0
New pH = ? Don’t forget to determine new volume.
After addition of 50.00 mL of NaOH (total), we have added 5.00 x10-3 mol NaOH. This
reacts with all of the acid present. This is the equivalence point.
pH will be due to autoionization of water; [H3O+] = 1.0 x10-7 M. pH = 7.0
After addition of 60.00 mL NaOH, we’ve added 10 mL excess NaOH.
How to determine pH now?
pH =
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The titration of a weak acid with a strong base (or vice versa) is a more common
scenario.
Titrate 50.0 mL of 0.100 M formic acid with 0.1000 M NaOH
Is the [H3O+] = 0.100 M? NO, formic acid is a weak acid. Determine pH of initial
solution using an ice table.
O
O
+
I
C
E
H2O
OH
H
O
H
0.100
0
-x
x
0.100-x
x
x2
Ka = 1.8 x10-4 =
Or about
0.100-x
-
+ H3O+
0
x
x
x2
0.100
check assumption..
x= 0.00420 M
What is the pH?
Add 10.00 mL of the NaOH solution. This is 1.000 x10-3 moles of hydroxide. In the
initial solution, we have 5.00 x 10-3 mol formic acid. This acid-base reaction goes to
completion.
O
+
H
initial mole
moles after
OH
O
-OH
H
O
-
+ H2O
After this reaction, we have a weak acid and its conjugate base. This is a buffer-type
situation. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log
[A-]
[HA]
pH = 3.74 -.6000= 3.14
After 25.00 mL of NaOH is added:
O
+
OH
H
moles after
O
H
5.000 x 10-3
initial mole
O
-OH
+ H2O
0
2.500 x10-3
2.500 x 10-3
-
0
2.500 x 10-3
Still have HA and A- present. Use H.H. eqn.
Half of the acid has been titrated. This is the cleverly named half equivalence point.
After 50.00 mL of NaOH is added, 5.00 x 10-3 moles of hydroxide have been added
(equals moles of acid originally present). This is the equivalence point.
O
+
H
5.000 x 10-3
initial mole
moles after
OH
0
O
-OH
O
H
5.000 x10-3
0
-
+ H2O
0
5.000 x 10-3
In this case, the pH of solution at the equivalence point is NOT 7 because formate is a
weak base. It will react to some extent with the water. Use an ICE table, but first must
calculate the concentration of formate.
O
H
O
O
-
+
H2O
H
OH
+ -OH
I
C
E
Past the equivalence point, you are just adding NaOH to the solution. The pH can be
determined simply by knowing how many moles of hydroxide are present and the total
volume.