Give students the appropriate lab safety gear: latex gloves, goggles and lab coats. Inform them of the important protocols involved with handling chemicals in this experiment. They should avoid sniffing or otherwise ingesting any of the chemicals involved, and should inform you if they spill any substances on themselves.
Pour about 25 mL of the cloudy ammonia solution into a 250 mL beaker.
Add 50.00mL of 0.100M HCl to the beaker. The HCl will react with the ammonia in the solution, following the equation NH3 + HCl ' NH4Cl. Thus, the HCl removes ammonia from the solution; leftover HCl will make the solution acidic. Add 4 drops of bromothymol blue, which turns blue in the presence of an acidic solution.
React the excess HCl with 0.050M Na2CO3. Na2CO3, following the equation 2HCl + Na2CO3 ' 2NaCl+ CO2+ H2O, reacts with HCl and makes the solution less acidic. Continue to add Na2CO3, drop by drop, until the bromothymol blue turns green, indicating that the solution has become neutral. Record the amount of Na2CO3 that you had to add to obtain a neutral solution, indicating complete removal of HCl.
Calculate the number of moles of Na2CO3 that you had to use to completely remove HCl from the solution. This is equal to the volume of Na2CO3 that you added, multiplied by its molarity, M = 0.050 molL-1. For instance, if you needed 20 mL of Na2CO3, then you would use 0.001 moles in total.
Determine the number of moles of HCl that reacted with Na2CO3, using stoichiometry. Since two moles of HCl react with every mole of Na2CO3, in this example, 0.002 moles of HCl reacted with Na2CO3. This also indicates that 0.002 moles of HCl remained from the original reaction with ammonia.
Calculate the original number of moles of HCl that you added to the ammonia solution. This is equal to the molarity of HCl* the volume of HCl, or .05 L * .100 moles/L. Thus, originally, you added .005 moles of HCl to the solution.
Calculate the total number of moles of HCl that actually reacted with the ammonia. Subtract the number of moles that didn't react (from Step 6) from the total number of moles added (from Step 7). This gives you .005-.002=.003 moles.
Refer to the equation in Step 2 for the reaction of HCl with NH3. It shows that 1 mole of HCl reacts with 1 mole of NH3. Since .003 moles of HCl reacted here, there must be .003 moles of NH3 in the solution.