# Melting of ionic compound hydrates

My professor asked us the reason of the melting point of potash alum being so low, despite it being an ionic salt. He later went on to answer it by saying that melting point is a misnomer for the phenomenon that is happening here.

What's happening here is that the alum is losing its water of hydration and is then dissolving in that water itself. So it is actually a dissolution point rather than melting point.

I couldn't find any sources for what he said. Though what he said makes logical sense, I would like to read a bit more about it. Can you please provide some sources confirming what my professor said?

• A misnomer is a word that causes confusion by being used incorrectly. "Melting" in application to this process is something different. It is a word that causes confusion by being used correctly. What we have here is melting all right, but not of the sort we are used to seeing. – Ivan Neretin Aug 30 '18 at 7:20

Your professor's premise of dissolution is only true if the process is incongruent melting which by the way is still melting. Whereby the process proceeds as:

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][T1] KAl(SO4)2~_{(s)} + 12H2O_{(l)}\tag{\beta to \alpha + liquid}}$$ Then $$\ce{KAl(SO4)2~_{(s)} ->[\Delta][T_2 > T_1] KAl(SO4)2~_{(aq.)}\tag{\alpha to liquid}}$$

This concept is outlined in the generic binary phase diagram below where $\alpha$ is anhydrous potash alum, $\beta$ is the hydrated form, and liquid is potash alum solution. Above a certain temperature $\beta$ will decompose into $\alpha$ and liquid. As the temperature rises more potash alum will dissolve until eventually it will all be liquid again. Figure 1. Binary Phase Diagram with Peritectic melting

However if it is a direct conversion from liquid to solid then it is congruent melting and your professor is wrong. This can occur either as melting an eutectic (shown in Figure 2) or congruent melting of a compound (shown in Figure 2) and proceeds directly as follows.

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][92.5^\circ C] KAl(SO4)2~_{(aq.)} + 12H2O_{(l)}\tag{Interionic Melting}}$$ $$\ce{Sn_{(s)} + Pb_{(s)} ->[\Delta][183^\circ C] Sn_{(l)} + Pb_{(l)}\tag{Eutectic}}$$ $$\ce{Ti5Si3_{(s)} ->[\Delta][2130^\circ C] 5Ti_{(l)} + 3Si_{(l)}\tag{Intermetallic Melting}}$$ Figure 2. Binary Pahse Diagram with Eutectic. Figure 3. $\ce{Si\! -Ti}$ Binary Bhase Diagram with Compound at $\ce{Ti5Si3}$ That Undergoes Congruent Melting.

Usually differential scanning calorimetry is used to determine which melting regime you are in indicated by a range of temperature causing a spike in heating rate corresponding to incongruent melting or a single temperature where heating spikes which would indicate congruent melting.

In this case, however, we can confidently assume that $\ce{KAl(SO4)2.12H2O}$ is under going incongruent melting since it melts above the temperature of water and below the temperature of the anhydrate and there are no other hydrates that it forms, the process is the peritectic melting thus the process your professor describes is correct, but it is still considered a melting process.

• Excellent answer +1. But I have a query. Shouldn't the potash alum decompose into more discrete species on further heating or will it remain an aqueous solution? See this question for context. – Nilay Ghosh Aug 29 '18 at 19:02
• Yes, it will form the separate ions in solution, My point was the difference between a two step and one step melting – A.K. Aug 29 '18 at 19:06
• Also, you want to add the concept of "burnt alum" to your answer. Wikipedia states: When heated to nearly a red heat, it gives a porous, friable mass, which is known as "burnt alum". – Nilay Ghosh Aug 29 '18 at 19:09
• I could, but I felt like breaking out phase diagrams was far enough into the weeds already. – A.K. Aug 29 '18 at 19:10