Experiment 10Class XI · CBSEChemistry 043

To Study the pH Change in the Titration of a Strong Base with a Strong Acid

For twenty millilitres the pH barely moves. Then half a drop takes it from 4 to 10. That vertical jump is the whole reason an indicator works.

SectionpH and pH Change

In the examContent based experiment · 6

01Aim

To study the change in pH during the titration of a strong base with a strong acid, using pH paper or a pH meter, and to plot the titration curve and find the equivalence point from it.

02Requirements

No.	Material	Specification / purpose
1	Strong base	0.1 M NaOH, 20 ml taken in a beaker
2	Strong acid	0.1 M HCl in the burette
3	Burette, 50 ml	To deliver the acid in measured portions
4	pH meter or pH paper	A meter is far better here
5	Beaker, 100 ml, and glass rod	To hold and stir the solution
6	Pipette, 20 ml	To measure the base
7	Graph paper	To plot pH against volume added
8	Phenolphthalein	To compare the visual end point with the curve

03Principle

When a strong acid is added to a strong base, both are completely ionised, and the only reaction that matters is

H⁺+OH⁻⟶H₂OΔH=−57.1kJ mol⁻¹

At the equivalence point the moles of acid added exactly equal the moles of base present. The solution contains only sodium chloride and water, and since neither ion hydrolyses, the pH is exactly 7.

Why the curve is flat and then vertical

pH is a logarithmic quantity. Near the start the solution holds a large reservoir of hydroxide ions, so neutralising a little of it changes [OH⁻] by only a small fraction, and the pH hardly moves.

Close to the equivalence point almost all the hydroxide has gone. Each further drop now removes a large fraction of what remains, so [H⁺] changes by powers of ten and the pH leaps. Between 19.9 ml and 20.1 ml — about four drops — the pH falls from about 10.4 to about 3.6.

Why any indicator works for this titration. The vertical part of the curve spans pH 4 to 10. Both methyl orange, which changes at 3.1 to 4.4, and phenolphthalein, which changes at 8.3 to 10.0, lie inside that jump. For a weak acid against a strong base the jump is much shorter and only phenolphthalein will do.

The three regions of the curve

Region	What is in excess	The pH is set by
Before the equivalence point	Unreacted NaOH	The hydroxide left over
At the equivalence point	Neither	The ionisation of water alone, so pH = 7
After the equivalence point	Added HCl	The excess hydrogen ions

Make the additions drop by drop near the end point. Adding a millilitre at a time there will carry you straight past the jump and the curve will have no vertical portion at all.

04The titration curve

Plate I

pH against the volume of 0.1 M HCl added to 20 ml of 0.1 M NaOH, showing the flat regions and the vertical jump at the equivalence point.

05Procedure

Rinse the burette with the 0.1 M HCl and fill it, removing the air bubble below the stopcock. Note the initial reading.
Pipette 20 ml of 0.1 M NaOH into a 100 ml beaker.
Measure the pH of the base before any acid is added. It should be near 13.
Add the acid in portions of 2 ml, stirring well after each addition.
After each addition measure the pH, either with the meter or by taking out one drop on a glass rod and touching it to pH paper.
When about 18 ml has been added, reduce the portions to 0.5 ml.
From 19.5 ml onwards add drop by drop, measuring the pH after every drop. This is where the whole shape of the curve is decided.
Continue past the equivalence point until about 25 ml has been added, going back to 1 ml portions once the pH has settled below 3.
Record every reading in a table of volume against pH.
Plot pH on the y axis against volume of acid on the x axis. Draw a smooth curve and read off the volume at the steepest point of the vertical portion. That is the equivalence point.

06Observations

No.	Volume of 0.1 M HCl added, ml	pH	Region
1	0.0	13.0	Only NaOH present
2	5.0	12.8	Excess base, pH falls slowly
3	10.0	12.5	Excess base
4	15.0	12.2	Excess base
5	18.0	11.7	Approaching the equivalence point
6	19.5	11.1	The curve begins to steepen
7	19.9	10.4	Entering the vertical portion
8	20.0	7.0	The equivalence point
9	20.1	3.6	Just past it, pH has fallen by 6.8 units
10	20.5	2.9	Excess acid
11	22.0	2.3	Excess acid
12	25.0	2.0	Excess acid, pH falls slowly again

M₁V₁=M₂V₂⇒0.1×20.0=0.1×V_acid⇒V_acid=20.0ml

The pH fell by only 1.3 units over the first 18 ml, and then by 6.8 units between 19.9 and 20.1 ml.
The equivalence point from the graph is at 20.0 ml, which is exactly what the stoichiometry predicts.
The pH at the equivalence point is 7.0, because the salt formed is sodium chloride and it does not hydrolyse.
Phenolphthalein changed colour at about 19.92 ml and methyl orange at about 20.08 ml. Both lie inside the vertical jump, so either is acceptable here.

07Result

The pH change during the titration of 20 ml of 0.1 M NaOH with 0.1 M HCl was studied and the curve plotted. The pH changes very slowly away from the equivalence point and very sharply near it, falling from 10.4 to 3.6 on the addition of 0.2 ml. The equivalence point is at 20.0 ml and at pH 7.0, as expected for a strong acid against a strong base.

08Precautions

Remove the air bubble from the burette tip before starting.
Stir thoroughly after every addition before measuring the pH.
Add drop by drop near the equivalence point; that is the only part of the curve that matters.
Rinse the pH electrode with distilled water between readings and do not let it touch the bottom of the beaker.
Calibrate the meter with buffers of pH 4, 7 and 10.
Both solutions must be accurately 0.1 M.
Plot pH on the y axis and volume on the x axis, with a scale open enough to show the jump.

09Viva voce

Q1What is the equivalence point?

ANSThe point at which the acid added is chemically equivalent to the base present, so that neither is in excess.

Q2What is the difference between the equivalence point and the end point?

ANSThe equivalence point is the true stoichiometric point. The end point is where the indicator changes colour. A well chosen indicator makes the two practically the same.

Q3Why is the pH exactly 7 at the equivalence point here?

ANSThe product is sodium chloride, a salt of a strong acid and a strong base. Neither ion hydrolyses, so the solution is simply water.

Q4Why does the pH change so sharply near the equivalence point?

ANSAlmost all the hydroxide has been used up, so each further drop removes a large fraction of what remains. Since pH is a logarithm, the hydrogen ion concentration changes by powers of ten.

Q5Why can both methyl orange and phenolphthalein be used here?

ANSThe vertical portion of the curve covers pH 4 to 10, and the range of both indicators falls inside it.

Q6What would the curve look like for a weak acid against a strong base?

ANSThe equivalence point would be above 7, around 8.7, and the vertical jump would be shorter, roughly pH 7 to 10. Only phenolphthalein would be suitable.

Q7Why must the solution be stirred after each addition?

ANSThe acid is denser and sinks. Without stirring the electrode reads a local concentration and not the true pH of the mixture.

Q8Where exactly is the equivalence point read from the curve?

ANSAt the midpoint of the vertical portion, which is the point of steepest slope. It can be found exactly by plotting the first derivative.

Class 11 Chemistry Experiment 10 - To Study the pH Change in the Titration of a Strong Base with a Strong Acid