# Predict melting point of water CaCl2 solution

I'm doing a science project for my school. I experimentally measured the melting point of water-salt solutions, and I want to compare my results to theoretical melting points, but I encountered a problem.

I used this equation to calculate the melting point depression: $\Delta T = b \cdot i \cdot K_\mathrm{f}$. I also compared my data to a phase diagram. I expected my measured melting point would be on the liquid line. But these two methods give me a different prediction. A phase diagram tells me that $\ce{CaCl2}$ would be much more effective at lowering the freezing point of water while the above equation tells me otherwise.

Here is a comparison: Purple indicates the melting point calculated with the equation. The black line is the liquid line on the phase diagram. Which method is correct and why is there such a big difference?

(I didn't encounter this problem with $\ce{NaCl}$ solutions, the two methods gave more or less the same predictions)

EDIT: my math was wrong on the graph i posted, here is the corrected graph: ## 1 Answer

I think there are three issues here. The first is that your purple line of calculating the freezing point doesn't appear to be correct. Let's arbitrarily select $b=1 ~ \mathrm{mol/kg}$ to compare. $i=3$ and $K_\mathrm{f}=1.86 ~ \mathrm{^{o}C/mol}$. The molecular weight of $\ce{CaCl2}$ is $111 ~ \mathrm{g/mol}$. Then, our mass concentration is $b·\mathrm{MW}/(b·\mathrm{MW}+1 ~ \mathrm{kg})=0.1 ~ \mathrm{kg}_\mathrm{salt}/\mathrm{kg}_\mathrm{total}$. The freezing point is $b·i·K_\mathrm{f}=1·3·1.86=5.58 ~ \mathrm{^{o}C}$. This is much closer to the red/black line than the purple one.

The second problem is that that relation is only valid at low concentrations. As the concentration of salt goes up, it has an increasing effect on the activity of the water. As we can see from the phase diagram, once you get to a 6:1 mol ratio of $\ce{Ca^{2+}}$ to $\ce{H2O}$, the interaction is extremely strong. I'm just speculating here, but I would guess six water molecules are forming a weak octahedral coordination complex with the $\ce{Ca^{2+}}$, with the electronegative oxygen ends oriented toward the ion. This effect becomes so strong, that the hydrate becomes the relevant element for consideration above that concentration.

Third, you're ignoring the existence of hydrates. If you find the melting points and the cryoscopic constants of the hydrates, you can actually calculate the right side of the plot the same way.

• Concerning the first issue, yes, my math was wrong. I have now fixed it and the graphs seem to only diverge at about 15% mass concentration. I don't fully understand the second and third issue you described. Should I consider the effect of hydrates at mass concentration lower than 30 percent (6:1 mol ratio as you said) or only above that? I'm only interested in the first segment of the graph (until 30%) but even after I corrected my math the graphs are still quite different (see my edit). Is that because of the hydrate issue? – Klemen Kersic May 1 '18 at 22:01
• Calculating the difference between the the ideal freezing point depression and the real one turns out to be a very hard problem. The way the UNIQUAC model (referenced in your graph) deals with it is by assigning each pair of ions and solvent molecules a some interaction coefficients. Some, like CaCl2, interact strongly and pull the real curve down. Others, like MgSO4 if I recall, interact weakly and push it up. Unfortunately, its not at all intuitive and your best bet is to look into more complete models. – ericksonla May 2 '18 at 2:47