Solubility, Temperature, and Crystallization Lab Report Answers

Answers

1. Plot the supplied solubility data

Units are mol/L. Plot temperature on the horizontal axis and molar solubility on the vertical axis.

2. Initial molarities

Using common classroom molar masses, 8.5 g NaNO3 is approximately 0.100 mol and 7.5 g KCl is approximately 0.101 mol. With the exercise's assumed 0.025 L volume:

$$[NaNO_3]=\frac{8.5/85.00}{0.025}\approx4.0\ \mathrm{M}$$ $$[KCl]=\frac{7.5/74.55}{0.025}\approx4.0\ \mathrm{M}$$

Because the salts dissociate, the starting solution contains about 4.0 M each of Na+, K+, Cl−, and NO3−. It is therefore also about 4.0 M with respect to each possible ion pairing NaNO3, KCl, NaCl, and KNO3 for the graph comparison. This notation does not imply that a net ionic reaction occurred on mixing.

3. First crop at about 10 °C

Linear interpolation between 0 and 20 °C gives approximate 10 °C solubilities of 5.4 M NaCl, 7.15 M NaNO3, 3.7 M KCl, and 2.25 M KNO3. The approximately 4.0 M solution most strongly exceeds the KNO3 curve, so the predicted first crop is KNO3.

Record the observed crystal shape from the actual crop. A shape description such as needles, cubes, plates, or rhombs is an observation, not something the numerical table can determine by itself.

4. Second crop during hot evaporation

Evaporation raises all ion concentrations. At high temperature, the KNO3 and NaNO3 curves rise steeply, while NaCl remains near 5.5 M. The hot concentrated solution therefore reaches the NaCl limit first. The predicted second crop is NaCl.

Enter the actual filtrate temperature and observed crystal shape from the experiment. They are not provided by the problem.

5. Third crop after cooling again to 10 °C

After hot NaCl removal, cooling the smaller volume sharply lowers KNO3 solubility. The predicted dominant third crop is KNO3 again. A small amount of another salt can contaminate a real crop if evaporation, filtration temperature, or washing differs from the idealized sequence, so connect the identity to both the curve and observed crystals.

6. Why mixing does not produce a net ionic equation

All four salts are soluble under the initial warm conditions. The molecular exchange can be written formally as:

$$NaNO_3(aq)+KCl(aq)\rightarrow NaCl(aq)+KNO_3(aq)$$

But every species remains as aqueous ions, so the complete ionic equation cancels entirely. There is no net ionic reaction on mixing. Crystallization happens later because cooling or evaporation pushes a particular salt beyond its solubility limit.

Conclusion

The graph predicts KNO3 as the first crop, NaCl as the hot-evaporation crop, and KNO3 as the dominant final cooling crop. The measured temperatures and crystal-shape descriptions must still come from the student's experiment.