AS level chemistry practical (GCE)

Chemistry 101 Laboratory
Experiment 10: Acid/ Base Titrations
Pre-Laboratory
Pre-lab Questions
1. How would your calculated molarity of NaOH be affected by each of the following errors? Elaborate.

a.) The buret is not rinsed with solution before filling.

b.) Some solution splashes out of the flask during titration.

c.) You go past the end point in the titration.
2. How would your calculated value for the equivalent weight of the acid be affected by each of the following errors? Elaborate.
a.) Some acid is spilled out of the flask after it is weighed.

b.) Some solution splashes out of the flask during titration.

c.) You go past the end point of the titration.
3. What is the effect of allowing a solution of NaOH to remain in contact with the air in the laboratory? Be specific as to the two, main effects and their sources.
Laboratory
The reaction of an acid and a base in water involves the reaction of hydroxide and hydrogen ions to form water (neutralization) and the formation of the resulting salt from the conjugate base and acid. The latter may or may not be soluable.

In general, the reaction is given by the following reaction for simple acids and bases:
HA + HOX <-> H2O + A- + X+
In this example, just one hydrogen and one hydroxl ion were available for the reaction. The ionization of the acid therefore produces one "equivalent" (one mole per mole) of hydrogen ions and ionization of the base produces one equivalent of hydroxyl ions. ( It is important to remember that, in water, the hydrogen ion and hydroxyl ion are the strongest acid and base, respectively. For this reason, ionization of any acid or base in water leads to the production of soley these acidic and basic ions.)

It is possible that one mole of the acid or base will produce more than one mole (one equivalent mole) of hydrogen ions or hydroxyl ions. An example of this is the diprotic acid sulfuric H2SO4.

In this instance, a total of 2 moles of hydrogen ions are produced when the acid is completely ionized to the sulfate.
When a titration is carried out, an acid or base of unknown concentration is allowed to neutralize a base or acid known to produce a known number of moles of hydroxyl or hydrogen ions.

As the acid or base is added to the solution of a known number of moles, the acids and bases react to neutralize one another.

At the point where the number of moles of acid equals the number of moles of base, we have reached what is called equivalence.

This is also known as the equivalence point and is shown either by a indicating dye or pH meter.

At the equivalence point we know the number of moles of acid or base reacted (from the fact that we know the number of moles of base or acid originally present from the sample of known amount) and the number of liters of the unknown which were added to neutralize the solution.

From the definition of molarity we have the following relationship:
Equivalent molarity = (# moles originally )/ ( liters of unknown added)
Notice that this differs slightly from our usual definition of molarity.

The reason is that we have not taken into account the fact that our unknown acid or base may have more than one equivalent per mole.

In today's experiment, you will deal with a monoprotic acid and a simple hydroxyide (NaOH) so that the concentration you obtain will be the true molarity of the base.
One of the important concepts in any aspect of chemistry in which quantitative answers are required is that of the primary standard.

This is an agreed upon set of conditions, chemical compound, or anything else which is agreed to represent a measure to which other things are compared.

One such standard concerns hydrogen ion concentration and is the use of potassium hydrogen phthalate (KHC8H4O4, molecular weight 204.224) as a standard for the concentration of hydrogen ions in a solution.

One mole of this salt produces one mole of hydrogen ions in one liter of water.
A second thing which is required in the use of a primary standard is a means of knowing when all of the (in this case) hydrogen ions from the standard have been neutralized with the base.

One way of doing this is to use an indicator dye.

These are organic chemicals which are also weak acids and bases. On their ionization or protonation, they undergo a chemical change which also results in a change in color.

One such dye is phyenylphtalene, a close relative of the primary standard you are using. This dye is clear in acidic solutions, slightly pink in neutral pH solutions, and dark purple in basic solutions.

Addition of a small amount of the dye to the solution you are testing allows you to determine when an equivalent number of acids and bases were present.

AS level chemistry practical (GCE) Titration

Titration

Titration that involve H2SO4 and NaOH is an examples of acid-base titration of a strong base acid and strong base, both the titrant and the analyte are completely ionized. The balanced reaction and its ionic reaction are:

H2SO4 + 2NaOH -> Na2SO4 + 2H2OH+ + SO42- + 2 Na+ + 2OH- -> 2Na+ + SO42- + 2H2OH+ + + 2OH- -> H2O

The H+ combined with OH- to form H2O, and other ions (Na+ and SO42-) remain unchanged.

This is an example of neutralization reaction, the net result of this neutralization is conversion of H2SO4 into a neutral solution Na2SO4 in the equivalent point.

A titration curve is constructed by plotting the pH of the solution as a function of the volume of titrant added.

The volume changes during titration must be employed for determining the concentrations of the species in the solutions (H+, OH- ). This is the example of the titration curve of 0.1 M H2SO4 100 mL versus 0.1 M of NaOH.
We began with 0.1 mL H2SO4 100 mL in Erlenmeyer flask and then 0.1 mL of NaOH in the buret.
At the start of the titration there is only H2SO4 in the solution so the pH of this solution is:
[H2SO4] = 0.1 M[H+] = 2 x 0.1 = 0.2 MpH = - log [H+] = -log(0.2) = 0.699

as the titration begins, NaOH solution is added from the burette into the H2SO4 solution in the Erlenmeyer.

The pH of the solution when the 20 mL NaOH 0.1 M is added to the solution is:
moles of H2SO4 = 0.1 M x 100 mL = 10 mmolmoles of NaOH = 0.1 M x 20 mL = 2 mmol

AS level chemistry practical (GCE)

A level chemistry practicals( GCE)
1. Titration -Acid base titration – Hcl AND NaOH
Purpose:

To determine the molarity of an unknown acid solution by titration using phenolphthalein as an indicator.

Materials:

Burets
Clamp
Stand
Beakers
1M HCl solution
1M NaOH solution
Unknown HCl solution
Phenolphthalein indicator solution

Procedure:

Clean and dry the burets and beaker, and clamp the two burets to the ring stand. Fill one of the two burets with 1M HCl solution, and the other with the NaOH solution.
Use the buret to measure out 20 mL of HCl into an empty beaker. Add 2-3 drops of the indicator solution.
Titrate slowly with the NaOH solution, with constant swirling, until one single drop of NaOH causes a permanent pink color that does not fade onswirling. Record the volume of NaOH used.
Use the formula M1V1=M2V2 to determine the concentration of the NaOH solution. This solution may now be used to titrate the unknown acid sample.
Replace the buret containing the 1M HCl with the buret containing the HCl solution of unknown concentration. Refill the NaOH buret, and wash out the beaker.
Repeat the titration from steps 2-4 using 20 mL of the unknown acid solution to determine the concentration of the HCl solution.