# Assumptions: Part 1 — Preparing solutions

I am certain that all instructors have had discussions with their students about assumptions. When a new topic is introduced, it is impossible to completely cover all the various aspects, without resorting to a “simplification” to make it easier for the students to understand the topic. In this first article of the series, I would like to discuss the assumptions behind the preparation of solutions; i.e., why are volumes not always additive? In this discussion I will not be considering the “corrected concentrations/activities” of solutes.

It is very easy to convince students that volumes are additive by having them take 25 mL of water, and then add it to another 25 mL of water. They expect, and are gratified to observe, that the final result is indeed 50 mL of water. When mixing a solution of NaCl with a solution of MgCl2, and then asking them to determine the [Cl-] in the new solution, they will happily perform the calculation on the basis that the volume of the new solution will be the additive sum of the two original solutions. A similar situation would involve the mixing of dilute solutions of HCl and NaOH, and asking them to determine the [Cl-] in the new solution. These are simple exercises, but gives the students practice with mole ratios and the dilution of solutions.

One can now ask the students if they should be worried that perhaps the volumes were not additive, as they contained different compounds? The intermolecular forces present in the original solutions will not be the same as the new solution, so that one cannot be certain that the total volume occupied by the molecules and ions in the new solution will be equal to, less than or greater than the sum of volumes of the original solutions. An excellent demonstration is now in order,1 in which 250 mL of water (250 g) and 250 mL of ethanol (198 g) are added together. Most students would initially predict that the new solution should have a volume of 500 mL; i.e., the two volumes are additive. Obviously, the masses will be additive, but the students will be surprised to discover that the final volume of the new solution is only 480 mL, instead of the expected 500 mL The reference cited also contains a detailed explanation of the intermolecular forces between individual molecules in pure water and pure ethanol, as well as between the molecules in the solution formed after the pure liquids are mixed together.

At this point it is a good time to allow the students to discuss and propose possible explanations for this discrepant event. Since the volume of the final solution is less than expected, that means that the density of the new solution must be larger than one would have expected if the volumes were indeed additive. This implies that the molecules in the two original volumes must now be closer together in the new solution.

What about the situation where a chemical reaction occurs when two solutions are mixed? This is an opportunity to perform another excellent demonstration2 involving the reaction of HCl with NaOH. When dilute solutions of acids and bases are mixed, the volume of water formed during the chemical reaction is normally ignored, as it is negligible compared to the initial volumes of the acid and the base. Again the volume is not simply additive.

In the instructions for the above demonstration,2 there is a cautionary note for the preparation of the two solutions; i.e., each solution must be allowed to cool down to room temperature before they are mixed together to initiate the acid/base neutralization. Why does each solution become warm in the first place? In the case of the NaOH, the energy it takes to dissociate NaOH ions is less than the energy of hydration (attraction between the ions and water molecules). Therefore the result is a negative DH for the formation of the solution. This is also the basis for the instructions when preparing a dilute solution to add a small volume of a concentrated acid to a much larger volume of water, not the other way around. With a small volume of water, the heat released by the dilution process would be sufficient to boil the water, a dangerous situation.

At this point it is a good time to allow the students to discuss and propose possible explanations for this discrepant event. Since the volume of the final solution is less than expected, that means that the density of the new solution must be larger than one would have expected if the volumes were indeed additive. This implies that the molecules in the two original volumes must now be closer together in the new solution.

What about the situation where a chemical reaction occurs when two solutions are mixed? This is an opportunity to perform another excellent demonstration2 involving the reaction of HCl with NaOH. When dilute solutions of acids and bases are mixed, the volume of water formed during the chemical reaction is normally ignored, as it is negligible compared to the initial volumes of the acid and the base. Again the volume is not simply additive.

In the instructions for the above demonstration,2 there is a cautionary note for the preparation of the two solutions; i.e., each solution must be allowed to cool down to room temperature before they are mixed together to initiate the acid/base neutralization. Why does each solution become warm in the first place? In the case of the NaOH, the energy it takes to dissociate NaOH ions is less than the energy of hydration (attraction between the ions and water molecules). Therefore the result is a negative DH for the formation of the solution. This is also the basis for the instructions when preparing a dilute solution to add a small volume of a concentrated acid to a much larger volume of water, not the other way around. With a small volume of water, the heat released by the dilution process would be sufficient to boil the water, a dangerous situation.

Once the students have a good understanding of the issues involved in preparing solutions and mixing solutions together, the following example may be used to allow them to see the volume difference when a solution is prepared. In this example, the volume of the KOH and water used to make a 1 L solution of 44% KOH is determined. The values for KOH, and many other solutes, may be found in the CRC Handbook of Chemistry and Physics.

Data Required for KOH(aq)

 Concentration (mass %) of the solution: 44.0% Density of the solution: 1.436 g/mL Density of solid KOH: 2.04 g/mL

Solution

Assume that one has 1.00 L of the solution (1436 g); therefore, the solution will contain:

1436 g solution x 0.440 =  632 g KOH

Using the density of solid KOH, one can determine the volume of KOH used: Therefore, if DV = 0, then the volume of water — using conservation of mass — needed to prepare the solution is:

1436 g solution – 632 g KOH  =  804 g water

804 g water = 804 mL water

Therefore, if DV = 0, the final volume of the prepared solution would be:

804 mL water  +  310 mL KOH  =   1114 mL = 1.11 L

But, as the actual volume of the solution was originally assumed to be 1.00 L:

• The volume of the prepared solution is less than the expected value.
• On average the molecules and ions are closer together than expected in the prepared solution.

In this case, one would observe a temperature increase during the formation of the solution and therefore there will be a negative DH for the formation of the solution.

I hope that instructors, and their students, find the discussion in this paper to be useful.

## Student questions

1. Determine the molarity of the 44% by mass KOH solution at 20 °C. (Answer: 11.3 mol/L KOH)
2. The handymath.com — a website for “solutions for technicians” — gives the density of 30% mass KOH solution as 1.281 g/L at 20 °C.3  Repeat the process above to determine the volume of the KOH and water used to make a 1.00 L solution. (Answer: 188 mL of KOH and 897 mL water)