Experiment 10

Topic: Solubility Product Determination

Introduction:

In this experiment you will determine the solubility

product of Ca(OH)2. The hydroxide ions formed when

Ca(OH)2(s) dissolves can be measured by titration with an

acid. From the balanced equilibrium equation, we can see

that we do not need to measure the Ca2+ ions, since they are

equal to one half of the concentration of the OH- ions. In

this procedure, a dilute solution of HCl is titrated with a

saturated solution of Ca(OH)2.

Apparatus and Materials:

0.05 mol dm-3 HCl, saturated Ca(OH)2 solution, burette, 20

cm3 pipette, filter paper, filter funnel, conical flasks,

beakers, methyl red indicator.

Procedure:

1. Filter about 100 cm3 of saturated Ca(OH)2 solution into a beaker. Because Ca(OH)2 reacts with the CO2 in the air to form insoluble CaCO3(s), you may noticed the formation of “skim” across the top of the solution. To avoid this as much as possible, filter your solution as close as possible to the time it will be used and do not let it stand around in the air for long periods of time.

2. Fill the burette to an appropriate calibration mark with HCl solution.

3. Using a pipette, place 20 cm3 of the filtered Ca(OH)2 solution into a conical flask and add two or three drops of methyl red.

4. Titrate to a permanent red end point with the HCl solution. Record the volume of HCl used.

5. Repeat steps 3 and 4 until burette readings are accurate.

Treatment:

1. Calculate the [OH-] from your titration data.2. From the balanced equation, calculate the [Ca2+].3. Substitute into the Ksp expression and calculate the value

of Ksp.4. Discuss sources of error in this experiment.