Experiment 11: Properties of Solutions |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Page 80: Objectives 1. Classify compounds according to their behavior in dilute water solution. |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
A. Electrolytes The word, "Electrolyte" should be explained more. "-lyte" is from a Greek word, Lytos, meaning loosened or soluble. Electrolytes are soluble in water meaning they are pulled apart (loosened) by the water. Remember, water has partial negative and partial positive charges. "-lyte" also indicates that this is the substance that can undergo "Electrolysis". "Lysis" means about the same thing. It is the pulling apart or decomposing of something. Electrolysis means the breaking apart of something using electricity. Remember the lab where water underwent electrolysis and the water was decomposed into hydrogen and oxygen gases. Electrolytes are vulnerable to electrolysis. Put them in water and attach wires from a 9 volt battery and they will decompose. Copper (II) chloride (CuCl2) put in water will dissolve into copper ions and chloride ions. Let the water evaporate and you back the copper chloride. However, expose it to a battery and the negative chlorine ions will be attracted to the positive side of the battery and the ions turn into chlorine gas and bubble away. The positive copper ions will be attracted to the negative side of the battery and will plate out on that wire as copper metal. So the copper chloride was decomposed. In this lab we are not really decomposing the electrolytes using electrolysis. We are just dissolving them in water. (Not as much fun but easier to do). If a substance dissolves in water, it gives you insight as to the make up of the that substance. In other words it tells you that there are charges or partial charges on the substance that water is pulling on. If the solution conducts electricity, it tells you more. It says that the substance has split apart into ions. Remember chemistry has the disadvantage of trying to study things that are too small to see. So we depend on these indirect methods of "looking at" the substances. For example, table salt (NaCl) in water conducts electricity, so we can "see" the sodium ions (Na+) moving independent of the chloride ions (Cl-). |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Page 82: Exp 11 A. Electrolytes
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Page 83
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
3. Solubility Curve
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
B. Solubility tables and graphs are usually prepared using the number of grams of solute dissolved in 100g of solvent. [Note: Above it was 100 grams of solution (solute=KBr+ solvent=H2O) . In these tables, they want "per 100 grams" of just water (solvent). ] This can be calculated from %KBr. Example: For a 21% KBr solution, 100 g of solution contains 21 g of KBr and 79 g of water. This gives us 21 g KBr per 79 g of water. To find g KBr per 100g of water: 21 g KBr x 100 = 26.58 g KBr |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Fill in your %KBr and temperature in the table below and calculate the solubility of KBr in g KBr/100g H2O. Using the information in the table, prepare a graph of g KBr/100g H2O (y-axis) versus temperature C (x-axis)
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
The graph on the right was generated with Excel. If you use the graph paper, it will look different. I made another chart below that is based on the graph paper in the lab manual. |
![]() |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
For the vertical axis, I made each square an increment of 3. That way every five squares would be 15. For the horizontal axis, I made each square an increment of 2.5 so that every 4 squares equaled 10.
|
![]() |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Exp 11: page 85 1. Classify each of the following as an electrolyte or nonelectrolyte a) solid Ca(NO3)2 nonelectrolyte b) rubbing alcohol (C3H8O) ??? c) H2O nonelectrolyte d) KCl solution ??? |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
2. Determine if a mixture of equal amounts of 1M H2SO4 and 1M Ba(OH)2 will give an electrolyte or a nonelectrolyte (hint: check the solubility on page 67). H2SO4(aq) + Ba(OH)2(aq) --> HOH(l) + BaSO4(s) Since barium sulfate is not soluble, it forms a precipitate (solid). So that doesn't provide any ions for conduction of electricity. The other product is water and pure water doesn't conduct electricity. So the final solution is a nonelectrolyte. If there were unequal amounts of sulfuric acid and barium hydroxide, then there would be some ions left over and the solution would be an electrolyte. But since they said equal amounts and equal concentrations, then no ions are left over. |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
3. Based on the following dissociation equations, identify which one is a strong acid, HA or HB. HA < --> H+(aq) + A-(aq) HB > H+(aq) + B-(aq) A strong acid will dissociate into H+ ions and the accompanying negative ions. The longer bottom arrow of the HA equation shows that this compound mostly wants to exist as HA, meaning the H+ wants to stay attached to A-. The small arrow on top means that only a small amount of HA will dissociate into H+ and A-. The HB equation has one arrow that points to the right. So HB completely breaks into H+ and B-. |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
4. Refer to experiment section B and list the compounds that can be used for cold packs. These would be the ones that are endothermic (absorbs energy) when they dissolve in water. So the ones where it felt cooler or the thermometer showed a drop in temperature are the ones on this list |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
5. From the solubility information in the table below, prepare a solubility curve (graph) for BaCl2.
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
The graph is 35 squares tall and 24 squares wide. We needed to go to almost 60 in the grams of BaCl2. So with each square incrementing by 2, that allows a height of 70. So that's works. The width needs to go to 100 C. With 24 squares, 4 degrees per square would only be 88 degrees, so I went with 5 degrees per square giving us (5x24=120 degree range). The line is basically straight but solubility does not need to rise in a linear (straight-line) fashion. So a curve is OK. |
![]() |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
5. One compound mentioned earlier is sodium sulfate. It's solubility curve is rather strange. It starts out normal with the solubility going up as the temperature goes up, but around 32 degrees C, it actually gets less soluble as the temperature goes up. |
![]() |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
c.) What would you expect to happen if you placed a saturated solution of BaCl2 at 80°C into a refrigerator and waited until the next lab during the following week to determine the solubility of the solution? According to our BaCl2 solubility graph and data, 52.6 grams of BaCl2 should be in the solution at 80°C. Refrigerator temperatures are around 40°F or 4°C. Looking at the graph, we can estimate that about 32.5 grams of BaCl2 will dissolve in 4°C water. So that means about 20 grams of the BaCl2 will have precipitated from the solution forming a solid at the bottom. Even without looking at the data or graph, when a saturated solution is cooled, some of the dissolved solute will come out of solution. If it doesn't come out, the solution is called supersaturated, which means it is holding more of the solute than it normally can. In those cases a few crystals of that compound will cause a quick crystallization of the excess solute in the solution. |
Number of different visitors (not just
hits) since 11-12-08