Titration Curve of a Weak Acid with a Strong Base

A 30 mL sample of 0.35 M of lactic acid is titrated with a 0.25 M NaOH solution. Calculate the pH before NaOH is added. After 20 mL of NaOH is added. After 42 mL. After 45 mL. After 60 mL. After 80 mL. Graph the titration curve.

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... neutralized and 0.005 mol of the conjugate base will be added.

0.0105 mol - 0.005 mol = 0.0055 mol / 0.050 L = 0.11 M HA
0.005 mol / 0.050 L = 0.1 M A-
(please see the attached file)

Now you can use the pH equation:
(please see the attached file)

After 45 mL

When NaOH is added, it neutralizes the acid. You need to know the moles of each:

0.030 L * 0.35 M = 0.0105 mol lactic acid
0.045 L * 0.25 M = 0.01125 mol NaOH

0.01125 mol of lactic acid will be neutralized and 0.01125 mol of the conjugate base will be added. Some NaOH will be left as well.

0.0105 mol - 0.01125 mol = 0 M HA
0.01125 mol / 0.075 L = 0.15 M A-

Now you will need to use ...