A solution is a homogeneous mixture of at least two components. Homogeneous means only one phase/state is visible. For example sugar water; you only see one liquid phase when looking at the glass even though both sugar and water molecules are present. Sugar water is a solution of two components, sugar which is termed the solute and water which is the solvent. The solute is the component present in a lesser amount while the solvent is the component present in a greater amount.
Solubility is the maximum amount of solute (sugar) that can dissolve in a given amount of solvent (water). The units for solubility are typically reported as g of solute/ 100mL solution. If we add too much sugar to the water, and exceed the solubility, the extra sugar will not dissolve but simply fall to the bottom of the glass. How can we get the extra sugar to dissolve? We can stir the solution and/or heat the solution both of which add heat energy to the system which increases the solubility limit.
In today’s experiment you will start with a mixture of two ionic compounds. One of the compounds will be much more soluble than the other. We will use this difference in order to isolate and purify the least soluble compound. At higher temperatures both compounds will dissolve, however as the temperature drops, the solubilities drop. The less soluble compound will crystallize, become a solid, at this lower temperature while the more soluble compound will remain dissolved. Then we can filter the mixture and isolate the less soluble compound as crystals.
To the left you can see a solubility chart for the salts that will be used for today’s lab. Compare the two sets of salts that you will be separating: set 1 – KNO3/Cu(NO3)2-3H2O, set 2 – KClO3/KCl
1. Label two clean and dry 50 mL beakers one as “NO3” the other as “Cl”. Weigh, record (1).
2. Add 11 to 12g of KNO3/Cu(NO3)2∙3H2O mixture to the beaker labeled “NO3” and add 4 to 4.5g of KClO3/KCl mixture to the beaker labeled “Cl” . Record the mass of each beaker with mixture (2).
3. Add 5 mL of distilled water to the nitrates, “NO3”, and 10 mL of distilled water to the potassium salts, “Cl”. Record amount of water added (4).
4. Add two drops of HNO3 to the “NO3” solution and two drops of HCl to the “Cl” solution.
5. Prepare an ice bath using a 400 mL beaker, filling half of it with ice then water.
6. Begin heating one of the solutions on the hot plate or if there is enough room heat both.
• DO NOT HEAT HIGHER THAT 60 ̊C to 70 ̊C.
7. Stir the heating solution(s) until all or nearly all of the crystals have dissolved.
8. Cool each beaker in room temperature water first to gradually reduce the temperature of the solution.
9. Then immerse the beaker into the ice water bath until the solution reaches 3 ̊C to 5 ̊C.
• DO NOT stir or swirl the solution.
10. As the solution cools, set up a vacuum filtration apparatus (view picture):
a. Use a 50 mL filter flask and a Buchner funnel
b. Clamp the 50 mL filter flask to a support stand to ensure that it does not fall over and spill your product.
c. Use a vacuum hose and connect one end to the filter flask and the other to the vacuum (VAC knob) by the sink.
d. Place Buchner funnel on top of filter flask and place the filter paper flat in the bottom of the funnel. The filter paper must fit perfectly in the bottom without cutting or folding the paper. Wet the filter paper with distilled water.
11. Once the solution cools to the appropriate temperature, record the exact temperature (5).
12. Add 10 mL of distilled water to a graduated cylinder and place the cylinder in the ice bath.
13. Turn on the vacuum
14. Use a spatula to help pour one of the mixtures onto the filter paper as