Lesson 2c: salt solubility and deliquescence

### Content

The solubility of the salt NH4NO3 in water is calculated, and this property is related to the deliquescence point.

### Entering the Data

In the previous part of this lesson we saw that the deliquescence point of ammonium nitrate is at about 61% relative humidity at 298.15 K. This point is a transition between solid salt (at lower relative humidities) and aqueous solution (at higher relative humidities).

Now we relate deliquescence behaviour to the solubility of the salt in water, by adding liquid water to a fixed amount of solid NH4NO3(s) and determining the state of the system. Note that all the water in the system is forced to remain in the aqueous phase (as are the trace gases), thus the calculation is equivalent to simply mixing water and salt in a container and determining the state and thermodynamic properties of the system.

Use the back button on the right hand browser window to return to the Model III data input page for variable "relative humidity, or total water" calculations, or select this link. Fill in the form as given below.

 1st Calculation Select "Graph" as the form of output, and then enter the values and options under the following headings: Ambient Conditions Select "Total water" as the variable. Start Value = 0.5 (i.e., 1.5 moles), End Value = 5.0 (i.e., 5 moles), Number of points = 50. Check the box below to ensure all water remains in the aqueous phase. Ionic Composition in Moles Ammonium = 1.0, Nitrate = 1.0. Trace Gases Click on "HNO3" and "NH3" to prevent these gases being partitioned into the vapour phase. Solid Phases There are no entries under this heading. Click on the "Run" button at the end of the page to do the calculation. Note:  the above should be entered on the "variable relative humidity or total water" parametric calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/mod3rhw.php).

### Viewing and Interpreting the Results

First, plot the moles of solid NH4NO3(s) against moles of liquid water:

 1st Graph:  select the variables and enter the options as given below. X Variable: "moles H2O(aq)" Range: leave blank Scale: linear (the default) Y Variable: "moles NH4NO3" Range: leave blank Scale: linear (the default) Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame.

Notice that the moles of solid salt in the system is zero for amounts of water greater than about 2.1 moles - where we have an aqueous solution in which all the salt is dissolved - but increases linearly as the amount of water decreases below 2.1 moles. This region of the graph corresponds to a mixture of aqueous solution and solid salt.

Next, plot the molality of the NO3(aq) ion:

 2nd Graph:  select the variables and enter the options as given below. X Variable: "moles H2O(aq)" Range: leave blank Scale: linear (the default) Y Variable: "mNO3-(aq)" Range: leave blank Scale: linear (the default) Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame.

In this graph we see that the system exists as an aqueous solution with a molality of about 11 mol kg-1 for 5.0 moles of liquid water (actually 1/5 × (1000/18.0182) = 11.102 mol kg-1). As the total amount of water decreases the molality rises until we reach about 2.1 moles of added water and a molality of just over 26 mol kg-1. Below this point the molality is constant. From the previous graph we know that below 2.1 moles of added water the solid salt is present, so clearly in this region of the graph we have a saturated solution in equilibrium with solid NH4NO3.

We saw in Lesson 2b that the deliquescence point of NH4NO3 - the transition between solid salt and aqueous solution - occurs at 2.1 moles of added water. Here the aqueous solution must be saturated with respect to the salt, thus the results we have plotted confirm that the deliquescence point corresponds to a saturated aqueous solution.

It was also found in Lesson2b that the deliquescence relative humidity was just over 61%. Next, plot the equilibrium relative humidity against moles of added water for the present calculation:

 3rd Graph:  select the variables and enter the options as given below. X Variable: "moles H2O(aq)" Range: leave blank Scale: linear (the default) Y Variable: "relative humidity" Range: leave blank Scale: linear (the default) Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame.

As the number of moles of water is decreased the solution becomes more concentrated and the equilibrium relative humidity also decreases. Below 2.1 moles of added water we know, from the previous graph, that the solution is saturated (and of constant molality). This present plot shows that the equilibrium relative humidity of this solution is just over 61%, which is the same as the deliquescence relative humidity.

These results demonstrate, first, that the molality of the solution at the deliquescence point is the same as the solubility of the salt at that temperature. Second, the deliquescence relative humidity is consequently the same as that over the saturated solution.

You should now review the conclusions on the main page for this lesson.