Safety: Ensure tip of burette is not chipped or broken and that valve turns easily and
smoothly. Wear safety goggles, apron and gloves. Keep phenolphthalein solution
away from flames as it is flammable. Handle acids and bases carefully as they are
corrosive. Rinse skin with lots of cool water if splashed with chemical and report
any spills to the teacher immediately.
Source: Nelson Chemistry 11 (2010)
ACID BASE TITRATION
Purpose: To determine the concentration of solution of sodium hydroxide by acid-base titration and analyse
the shape of a titration curve.
. Pre-Lab:
/ 11 I
/8 K/U
/5C
1) What colour is the indicator phenolphthalein in an acidic solution and in a basic
solution? (2 K/U)
Phenolphthalein is colourless in acidic solutions and pinkish in basic solutions.
2) Why does the clear colour, which forms at the point where the hydrochloric acid
comes into contact with the solution containing phenolphthalein and base in the
receiving flask, disappear more slowly near the endpoint? (2I)
At the start of the titration, the NaOH solution in the flask will be pinkish and the
pH is quite high. As HCl drops are added, the indicator colour will change
momentarily to colourless as the pH is lower in the area around the drop(s). As the
endpoint nears, the clear colour disappears more slowly as the solution has a lower
pH than when the titration started.
3) What ions are taking part in the neutralization reaction? (2 I)
The ions taking part in the neutralization reaction are OH_ and H+.
All ions are aqueous.
Na+ and Cl_ are spectator ions and are not involved in the reaction.
4) In terms of the ions present, what has happened at the endpoint of the titration? (1 I
Bases dissociate to produce OH_ and acids ionize to produce H+ which creates
water H2O.
5) Why is it a good idea to carry out titrations in triplicate? (1 I)
It is a good idea to carry out titrations in triplicate to ensure that the data collected
are reliable. Data from the 3 trials are then averaged and used to determine the
concentration of the solution being analysed.
6) Would the addition of several millilitres of distilled water to the Erlenmeyer
receiving flask during a titration affect the results of the titration? Explain your
answer. (3 I)
Adding several mL of distilled water during the titration would not affect the results
of the titration as the pH reading is not affected by the concentration of the base.
7) Explain what the shape of the titration curve (a plot of pH vs. volume of acid added)
would look like for this experiment. Sketch the curve.(2 I)
The curve would resemble a backward “S”.
volume of acid added (mL)
Source: http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
8) If 27.31 mL of 0.2115 mol/L sodium hydroxide is able to neutralize 37.45 mL of
hydrochloric acid, what is the concentration of the acid? (3 K/U)
VNaOH = 27.31 mL = 0.02731 L
cNaOH= 0.2115 mol/L
VHCl = 37.45 mL = 0.03745 L
cHCl = ?
c = n/V
NaOH + HCl  NaCl + H2O
cNaOH = nNaOH/ VNaOH
nNaOH = cNaOH x VNaOH
= 0.2115 x 0.02731
= 0.005776
= 5.776 x 10-3 mol
nHCl = 5.776 x 10-3 mol x 1 mol HCl/1mol NaOH (coefficients from balanced eqn)
= 5.776 x 10-3 mol
cHCl = nHCl/ VHCl
= 5.776 x 10-3 mol/ 0.03745 L
= 5.776 x 10-3 mol / 3.745 x 10-2
= 1.547 x 10-1 mol/L
The concentration of the acid is 1.547 x 10-1 mol/L.
9) What volume of 0.117 mol/L hydrochloric acid is needed to neutralize 28.67 mL of
0.137 mol/L potassium hydroxide? (3 K/U)
VHCl = ?
cHCl = 0.117 mol/L
VKOH = 28.67mL = 0.02867 L
cKOH = 0.137 mol/L
Find the moles of KOH:
c = n/V
n=cxV
= 0.137 mol/L x 0.02867 L
= 0.0039277 mol
= 3.9277 x 10-3 mol
Balanced equation for the reaction:
HCl + KOH  H2O + KCl
Find the moles of HCl:
n = 3.9277 x 10-3 mol (KOH) x 1 mol HCl/ 1 mol KOH
= 3.9277 x 10-3 mol
Find the volume of HCl:
c = n/V
V = n/c
= 3.9277 x 10-3 mol / 0.117 mol/L
= 3.9277 x 10-3 / 1.17 x 10-1
= 3.3570 x 10-2
= 0.03357 L
= 33.57 mL
The volume of HCl needed to neutralize the KOH is 33.57 mL.
10) Draw a flow chart for this experiment and prepare your data table. (5 C)
Data table is below. Drawing a flow chart on the computer takes me an inordinate
amount of time and I have already spend 2.5 hours on the pre lab questions.
Observations:
Titrations with indicator:
Measurements
Concentration of HCl (mol/L)
Volume of NaOH (mL)
Initial reading of acid on buret (mL)
Final reading of acid on buret (mL)
Total volume of HCl added (mL)
Average Volume of HCl added (mL)
Trial #1
Trial #2
Trial #3
0.117
0.117
0.117
25
25
25
0.10
12.52
25.10
12.52
24.98
37.62
12.42
12.46
12.52
12.42 +12.46+12.52/3 =12.4667
=0.01247 L
Calculations: (show all calculations and formulae clearly)
1. Determine the number of moles of hydrochloric acid used. (2 A)
n = c x V = 0.117 mol/L x 0.01247 L = 0.0014589 mol = 1.4589 x 10-3 mol
1.4589 x 10-3 moles of HCl were used.
2. Determine the number of moles of sodium hydroxide used. (1 A)
NaOH + HCl  NaCl + H2O The balanced equation shows that the coefficients
of NaOH and HCl are both 1
nNaOH = 1.4589 x 10-3 moles of HCl x 1 mol NaOH/1 mol HCl
= 1.4589 x 10-3 mol
There were 1.4589 x 10-3 mol of NaOH used.
3. Determine the concentration of the unknown sodium hydroxide solution. (2 A)
V=25ml =0.025 L
c = n/V = 1.4589 x 10-3 mol/0.025 L
= 1.4589 x 10-3/ 2.5 x 10-2
= 0.58356 x 10-1
= 0.0583 mol/L
The concentration of NaOH is 0.0583 mol/L.
Marking Scheme for Lab:
10 marks Communication based on communication rubric and proper lab format
15 marks Application (3 data table, 5 analysis , 2 error analysis, 2 conclusion, 3 lab performance)
TOTAL
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