Name: ____________________________________ Band: _________ Date: _______________

Lab: Qualitative Analysis

Qualitative analysis is the detection, identification, and separation of cations and anions in solution. In this particular lab, we will be performing qualitative analysis on only a few drops of solutions. Remember that when using such small amounts, it is extremely important that the equipment you use is very clean. If the equipment is “dirty,” the solutions will become contaminated, and it is possible that you will identify the contaminant rather than the unknown. To avoid cross contamination, make sure to recap bottles with original caps and to clean your droppers very well if they are to be reused. Part 1 Solubility is your best friend when it comes to qualitative analysis. To illustrate this, consider an aqueous solution of sodium iodide, NaI(aq). Write the dissociation equation for this ionic salt. NaI(s) Na+(aq) + I-(aq)

Let’s say you want to separate the iodide ions from the solution. You can do this by taking advantage of those solubility rules you memorized: if you add an ion that will form a precipitate (an insoluble compound) with the iodide ions, you’ll effectively remove the iodide from the solution!

Here are the ions you have to choose from: • Pb2+(aq) • Zn2+(aq) • CrO4

2-(aq) • SO4

2-(aq) • OH-(aq) Which one of these ions will combine with iodide to form an insoluble compound? Make a

prediction. Lead will form an insoluble compound (precipitate) with iodide. Test your prediction as follows:

Place five separate drops of NaI(aq) on your plastic sheet. Add a different ion from the list above to the different drops of sodium iodide to create five different solutions. Record your observations in the table below.

Mixture Observations Pb2+(aq) + I-(aq) Zn2+(aq) + I-(aq) CrO4

2-(aq) + I-(aq) SO4

2-(aq) + I-(aq) OH-(aq) + I-(aq) Which solution formed a precipitate? How can you tell? Write the chemical formula of the precipitate formed. Solution became cloudy and colored when lead ions were added.

Part 2 Now you’ve been given an aqueous solution “A” containing Pb2+(aq), Fe2+(aq), and Cu2+(aq).

Which of the following would allow you to separate the lead from the solution? • Adding dilute Na2S(aq) solution • Adding dilute HCl(aq) solution • Adding dilute NaOH(aq) solution • Adding dilute HNO3(aq) solution Explain why your choice would separate the lead and why the other choices would not.

All sulfides and hydroxides will be insoluble, so all metal ions will precipitate—not good. Nitrate will form soluble compounds with all ions, so no metal ions will precipitate—also not good. Chloride will form an insoluble compound with lead but soluble compounds with iron and copper—perfect! Test your choice as follows:

Place five separate drops of “A” on a new section of your plastic sheet. Add a different solution from the list above to the different drops of “A” to create four different solutions. Record your observations in the table below.

Mixture Observations A(aq) + Na2S(aq) A(aq) + HCl(aq) A(aq) + NaOH(aq) A(aq) + HNO3(aq) Note: you will not be able to completely confirm your choice using the observations above, but you can certainly rule some choices out! Part 3

Answer the following qualitative analysis APC questions on a separate sheet of paper. 1. Answer the question below that relates to the five aqueous solutions at 298 K below.

Mixing solution one and solution two will produce the precipitate PbCl2. 2. You’ve been given five samples of salts: AgCl, BaCl2, CoCl2, NaCl, and NH4Cl.

a. Identify the salt that can be distinguished by its appearance alone. Describe the observation that supports your identification. Cobalt chloride is a colored solid. All of the other solids are white.

b. Identify the salt that can be distinguished by adding 10 mL of H2O to a small sample of the remaining unidentified salts. Describe the observation that supports your identification. Silver chloride is the only remaining salt that will not dissolve in water.

c. Identify the salt that can be distinguished by adding 1.0 M Na2SO4 to a small sample of each of the remaining unidentified salts. Describe the observation that supports your identification. All of the remaining salts will dissolve in the aqueous solution but the barium ions in barium chloride will form a precipitate with the sulfate ions, resulting in a cloudy solution.

3.

i) Ag2S ii) Solution three must be potassium chloride and solution two must be sodium sulfide in order

for there to be no reaction when those two solutions are mixed. Solution one is therefore silver nitrate, and combining silver nitrate with potassium chloride will result in a white precipitate (silver chloride).

iii) See above. 4.

a) i) K2CO3; ii) BaCO3 and Ag2CO3 b) i) R is Pb(NO3)2 and S is NaCl; ii) PbCO3 c) i) Lead nitrate will only produce a precipitate when mixed with a solution of barium chloride,

so add lead nitrate to both solutions and make observations; ii) Lead nitrate in barium chloride will appear cloudy; lead nitrate in silver nitrate will be colorless and transparent; iii) the cloudy mixture can be correctly ID’d as barium chloride and the transparent mixture can be correctly ID’d as silver nitrate.