Lab #11  Reactions of Solutions of Known Concentration 
General Discussion
Solutes in solutions react in the same mole and mass ratios as they do in any other state.  The ratios in which they react may be determined from a balanced equation. The volume of solution which contains a given quantity of solute may be calculated from the concentrations of the solution.
 
In this experiment you will prepare solutions of known concentration for two reactants.  By carefully weighing the amount of solid reactant and accurately measuring the volume of the solution, you will obtain data which will enable you to calculate the molar concentrations of the solutions.  You will then allow measured volumes of the two solutions to react.  One of the products formed in the reaction will be an insoluble substance, a precipitate.  You will vary the quantity of one reagent by using different volumes in successive trials. The quantity of the other reagent will be held constant.  The quantity of precipitate formed will indicate when the reactants are present in the mole ratio indicated by the coefficients in the balanced equation. As soon as the fixed reagent is consumed, no more significaat amount of precipitate will form, regardless of how much of the other reactant is added.  At this point the ratio of the moles of each reactant in the reaction is equal to the mole ratio expressed by the coefficients in the balanced equation.
 
Finally, you will filter, dry, and weigh the precipitate obtained from one trial. This experimenally determined quantity may then be compared with the calculated value from the balanced equation.
 
Objectives:
1.  To prepare solutions of known molar concentration.
2.  To become acquainted with the simpler quantitative aspects of reactions involving solutions.
3.  To verify that chemical substances react in definite mole ratios.
 
Materials: Lead(II)  nitrate, sodium iodide.
 
Equipment: Buret, test tubes, balance, filter, filter funnel, wash bottle, alcohol.
 
Procedure:
1. Before you arrive for the lab, calculate the grams of lead(II)  nitrate needed to prepare 50 mL of a 0.50 M solution.  Also calculate the number of grams of sodium iodide needed to prepare 50 mL of 0.50 M solution.   Have your teacher check your calculations before any chemicals are opened.
2.   Weigh out the amounts of the above chemicals to the nearest 0.01 g.
3.   In separate beakers add enough distilled water to make a total of 50 mL of solution per beaker.
4.   Rinse a clean buret with 4 or 5 mL of the lead(II)  nitrate soltuion.  Be sure to run a few drops of this solution out through the tip of the buret.  Pour the remaining lead nitrate solution into this buret.
5.   Obtain six clean, dry test tubes and number them 1 through 6.  Add 6 mL of NaI  solution to each of these tubes.
6.  To test tube 1, add 1.0 mL of the lead(II) nitrate solution. Then add 2.0 mL, 3.0 mL, 4.0 mL, 5.0 mL and 6.0 mL respectively to test tubes 2, 3, 4, 5 and 6.    Mix the solutions well and allow the precipitate of PbI2 to settle 8 to 10 minutes.  Measure the height of the precipitate in each test tube in cm and record these data in Table I.
7.  Weigh several pieces of filter paper and determine the average mass of a single piece.
8.   Filter the precipitate from test tube 3.  Rinse any remaining PbI2 from the test tube into the filter paper using a wash bottle.  Avoid using excess amounts of water.
9.  Wash the PbI2 precipitate with 4 to 5 mL of alcohol.
10.  Place the filter paper containing the precipitate on the counter, labeled with your name on it, to dry overnight.
11.  Weigh the dry precipitate of PbI2.  Subtract the mass of the filter paper and record the mass of precipitate in Table II.
 
Data Table I   Mole Relationships in a Chemical Reaction
Test
Tube
No.
Volume of
0.50 M
NaI (mL)
Volume of
0.50 M
Pb(NO3)2
(mL)
  Moles 
of
NaI
Moles 
of 
Pb(NO3)2
   Moles 
of PbI2
(calc.)
Mass in
  gram of 
PbI2
(calc.)
Mass of 
PbI2
(exp.)
Height of 
PbI2
ppt.
(cm)
1
6.0
1.0
            
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2
6.0
2.0
            
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3
6.0
3.0
            
  
4
6.0
4.0
            
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5
6.0
5.0
            
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6
6.0
6.0
            
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Data Table II   Mass of PbI2 Precipitate

                                                                   Trial Number
 
 
     1     
     2     
     3    
     4     
     5     
     6     
Mass of filter paper + PbI2                  
Mass of filter paper                  
Mass of PbI2 precipitate                  

 
Followup Discussion
The molecular equation for the reaction in this experiment is:
 
Pb(NO3)2(aq)   +   2 NaI(aq)   ----->  2 NaNO3(aq)  +   PbI2(ppt)
Followup Questions
1.  Identify the reactant in excess in tubes 1, 2, 4, 5, and 6.
2.  Calculate the number of moles of each reactant in excess.
3.  How many grams of lead nitrate are present in 6 mL of 0.5 M lead nitrate solution?
4.  How do you account for the relative heights of the precipitate in the 6 test tubes?
5.  Why didn't the quantity of precipitate in tubes 4, 5 and 6 increase appreciably when additional lead nitrate is added?
6.   Make a general statement regarding which reactant in a reaction will determine the quantity of product formed.
7.   How many grams of PbI2 will be formed from 4.0 mL of 0.5 M NaI and 2.0 mL of 0.50 M lead(II)  nitrate?
8.   How does the number of grams of  lead(II) iodide calculated for tube 3 compare with your experimentally determined value?  What is the percentage error.
9.  List some of the sources of error in the experiment.