• In this sample titration, we are trying to
determine the concentration of 20.00
mL of HCl. In the titration we will be
neutralizing the HCl with 0.150 M NaOH.
• Step 1: The NaOH, the titrant, is placed in
the buret. The titrant is the solution of
known concentration that is added from
• Step 2: The HCl is placed in the
Erlenmeyer flask along with approximately
20.00 mL of distilled water and 2-3 drops
of phenolphthalein indicator. Since the
solution in the flask is acidic,
phenolphthalein is colourless.
• Step 3: NaOH is added to the HCl in the
flask. When the NaOH comes in contact
with the solution in the flask, it turns pink
and then the pink colour quickly
disappears. This is because the OH- from
the NaOH interact with the
phenolphthalein to change the
phenolphthalein from colourless to pink.
• The solution becomes clear again as the
hydronium ions from the hydrochloric acid
neutralize the added hydroxide ions. As
more NaOH is added, it takes longer for
the pink colour to disappear.
• As it starts taking longer for the pink colour
to disappear, the sodium hydroxide is
added a drop at a time.
• The equivalence point of the titration is
reached when equal numbers of moles of
hydronium and hydroxide ions have been
• When this happens in this titration, the pH
of the solution in the flask is 7.0 and the
phenolphthalein indicator is colourless.
• This would be a good time to stop,
however the indicator is still colourless, so
must keep going.
• Step 4: Add as little excess NaOH as
possible. We want to add a single drop of
NaOH to the colourless solution in the
flask and have the solution in the flask turn
pink and stay pink while the contents of
the flask are swirled.
• This permanent colour change in the
indicator is known as the endpoint of the
titration and the titration is over.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 527
Solve the problem
• 1st write the equation for the reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
• 2nd solve for the amount of moles of the titrant used.
NaOH mol = 0.150 mol/L x 0.02567 L= 3.85 x 10-3 mol NaOH
Found in titration experiment
• 3rd using stoichiometry, solve for the
concentration of HCl , knowing it is a 1:1 mole
3.85 x 10-3 mol= 0.192 M
SAMPLE PROBLEM 2
• In an acid-base titration, 17.45 mL of
0.180 M nitric acid, HNO3, were completely
neutralized by 14.76 mL of aluminium
hydroxide, Al(OH)3. Calculate the
concentration of the aluminium hydroxide.
SAMPLE ANSWER 2
The balanced equation for the reaction is:
3HNO3(aq) + Al(OH)3(aq) → Al(NO3)3(aq) + 3H2O(l)
The number of moles of nitric acid used is:
y mol = 0.180 mol/L x 0.01745 L = 3.14 x 10-3 mol HNO3
From the stoichiometry of the reaction, the number of moles of
aluminium hydroxide reacted is:
3.14 x 10-3 mol HNO3 x 1 mol Al(OH)3 = 1.05 x 10-3 mol
3 mol HNO3
Therefore, the concentration of the aluminium hydroxide is:
1.05 x 10-3 mol Al(OH)3 = 0.0711 M