Solubility Product Constant Lab

[pic 1]

Introduction:

When a salt of low solubility is added to water, only a few ions are formed, and most of the salt remains undissolved.  In a saturated solution, the solid dissociates into ions at the same rate that the ions reunite to form the solid.  In other words, a dynamic equilibrium exists between the undissolved solid and the ions in solution.  An undetermined amount of dissolved salt exists in equilibrium with the solid.  The concentration of the undissolved solid remains a constant.

The solubility product constant, Ksp, of a salt can be used to determine the concentration of ions in a saturated solution.

For example, suppose that a certain salt, A3B2, is dissolved in water.  The solid is in equilibrium with the ions

                        A3B2(s)  ↔  3A2+(aq)  +  2B3-(aq)

The expression for the equilibrium constant for this reaction is:

Kc = [A2+]3[B3-]2

                                   [A3B2]

Since A3B2 is a solid, its concentration is always the same.  Therefore a new constant can be used that considers only the concentration of the ions.  The new constant is the solubility product constant, Ksp.

Ksp =[A2+]3[B3-]2

To calculate the solubility product constant, the ionic concentrations must always be in moles per litre (mol/L or M).

All low solubility salts will exhibit similar mathematical relationships.  The position of the equilibrium is dependent upon the type of solid and the temperature.  One such salt, PbCl2, will be investigated in this lab.  Its solubility equilibrium and Ksp expression are given below:

Solubility Equilibrium:          PbI2(s)   ↔   Pb2+(aq)  +  2Cl-(aq)

Ksp for PbCl2 is:                        Ksp = [Pb2+][Cl-]2

Note that substances that have a large Ksp value have a higher solubility (more dissolved ions); and substances with a small Ksp value have a lower solubility (few dissolved ions).  

A molecular view of a saturated solution of lead(II)chloride would look similar to the following, with clumps of undissolved lead(II)chloride at the bottom of the beaker in equilibrium with dissolved Pb2+ and Cl- ions (although their would be twice as many Cl- ions than Pb2+ ions):

[pic 2]

Every substance that forms a saturated solution will have a Ksp.   However, for very soluble substances like NaCl, the value is so large that the concept is rarely used.  In low solubility substances, the value of Ksp is a useful quantity that lets us predict and calculate solubilities of substances in solution. 

In this experiment you will start with a saturated solution of a slightly soluble salt, PbCl2.  You will then find the concentration of the Pb2+ ions by reacting them with iodide ions, I- to form an insoluble precipitate.  Finally, you will use the [Pb2+] to calculate the solubility product constant for the original salt PbCl2.

Prior to performing this experiment, it may be helpful to view the following animations to observe the dissolving of a salt and the formation of a precipitate.  This will aid in drawing molecular diagrams as you proceed through the experiment.

[pic 3]

Purpose:

In this experiment you will,

• Determine the moles of Pb2+ ions in 100 mL of saturated PbCl2 solution by precipitation.
• Convert the number of Pb2+ ions per 100 mL to moles of Pb2+ ions per L, and
• Calculate the solubility product constant for PbCl2.

Materials:

• Electronic balance
• 2-250 mL beakers
• Funnel and filter paper
• Universal stand with ring
• Hot plate
• Stirring rod
• 100 mL graduated cylinder
• 0.5 mol/L KI solution
• Saturated PbCl2 solution
• Distilled water

Procedure:

Caution!  Lead compounds are poisonous and cause skin irritations.  Handle with care!

1. Measure the mass of the filter paper and enter this result in the data table provided.  
2. Set up filtration apparatus (place funnel and fluted filter paper on universal stand with ring).

1. In a 100 mL graduated cylinder, carefully decant 100 mL of clear supernatant from the saturated PbCl2 solution.  Avoid disturbing the crystals of PbCl2.  Add this solution to a 250 mL beaker.  Draw a diagram of the components (i.e. ions and molecules) in this beaker.
2. Add 20 mL of 0.5 mol/L KI solution to the beaker containing the PbCl2 solution.
3. Draw a diagram of the components (i.e. ions and molecules) of the 250 mL beaker prepared in steps 3 and 4 above.  Circle the ions that will form a precipitate when mixed together.  Also, symbolically write the equation for the chemical reaction that is taking place.

1. Write the net ionic equation for the reaction in #5 above and indicate which two species are undergoing a chemical reaction.  Which species are spectator ions?

1. Heat this solution on a hot plate to the boiling point, stirring occasionally.  (Why are we heating the solution?)
2. Allow the mixture to cool for 5 minutes.  Also allow any solid material to settle to the bottom of the beaker.  
3. Once the precipitate has settled, carefully decant the liquid through the filter.  Try to retain most of the solid in the beaker.  
4. Wash the precipitate three times with distilled water and decant the liquid through the filter.  On the final wash, swirl the flask and empty the entire contents of the flask into the filter.  Rinse the flask one more time with water and swirl and empty the contents into the filter.
5. Discard the filtrate and retain the filter paper containing the precipitate PbI2.  State which ions and molecules are in the filtrate.  Why don’t we need this solution?
6. After the precipitate, PbI2, has been thoroughly filtered, place the filter paper in a drying oven or let it dry overnight.  (Why must the precipitate be dried?)
7. Measure the mass of the filter paper and precipitate PbI2 and enter this result in the data table provided.
8. Subtract the mass of the filter paper from the value in #13 above and enter this result in the data table provided.

Data:

Analysis:

1.  Write and balance the equation for the reaction between saturated PbCl2 and KI solution.

1. From your data, calculate the number of moles of lead chloride precipitate formed.

1. How many moles of Pb2+ ions are in each mole of PbI2 precipitate?

1. Calculate the concentration of the Pb2+ ions in moles per L for the original solution (the original volume was 100 mL).

1. What was the concentration of the Cl- ions in moles per L for the original solution?  (Don’t forget that two chloride ions are produced for every mole of PbCl2).

1. Write the solubility equilibrium equation for the dissolution of PbCl2 in water.  Include physical states.

1. Write the Ksp expression and calculate the Ksp for PbCl2 at room temperature (25oC) using the concentrations obtained from your data.  Why is the chloride ion concentration squared in the Ksp expression?

1. If the actual Ksp for PbI2 at 25oC is 8.5x10-9, what is the percentage error for this lab?

%error = experimental Ksp – Actual Ksp   X   100  =  [pic 4]

                                                                    Actual Ksp 

1. Was this experiment actually done at 25oC?  If it was done at less than 25oC, how would your result (i.e Ksp value) be affected?  What if the temperature was warmer than 25oC?  What happens to the ion concentrations of Pb2+ and Cl- when the temperature is raised?  What would happen to the value of the Ksp as a result of the temperature increase?

1. The Ksp values for PbCl2 and PbI2 at room temperature (25oC) are given below.

Ksp (PbCl2) = 1.6 x 10-5

Ksp (PbI2) =   7.1 x 10-9

Why was the I- used to remove the Pb2+?  Can you name an ion that would have removed the Pb2+ ion more completely? (Hint: look up the Ksp values of some other lead salts).