Lesson 5a: deliquescence of sodium nitrate

### Content

The deliquescence relative humidity of NaNO3 at 298.15 K is determined, and also the concentration of the saturated solution at that point.

### Entering the Data

We determine the deliquescence point by equilibrating 1.0 mole of NaNO3 to an atmosphere of increasing relative humidity and observe where water is first taken up by the salt:

 1st Calculation Select "Graph" as the form of output, and then enter the values and options under the following headings: Ambient Conditions Select "Relative humidity" as the variable. Start Value = 0.70 (i.e., 70%), End Value = 0.90 (i.e., 90%), Number of points = 60. Ionic Composition in Moles Sodium = 1.0, Nitrate = 1.0. Trace Gases There are no entries under this heading. 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

A page will appear in the other browser window which enables you to plot various quantities against each other by choosing the X and Y variables, their ranges, and scales (linear or log10) from the drop down lists. Instructions, and details of the variables, are given in the right frame.

First, plot a graph of moles of liquid phase water against relative humidity:

 1st Graph:  select the variables and enter the options as given below. X Variable: "relative humidity" Range: leave blank Scale: linear (the default) Y Variable: "moles of H2O(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.

Notice that the transition from the dry salt to a solution containing 5.25 moles of liquid water occurs at about 0.7375 (73.75%) relative humidity: this is the deliquescence point of sodium nitrate at 298.15 K. The salt thus remains a dry solid to a significantly higher relative humidity than does NH4NO3 (61 %) which was studied in Lesson 2.

Next we determine the composition of the aqueous solution of NaNO3 at the deliquescence point by plotting the molality of Na+(aq) against relative humidity:

 2nd Graph:  select the variables and enter the options as given below. X Variable: "relative humidity" Range: leave blank Scale: linear (the default) Y Variable: "mNa+(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.

At low relative humidities the molality of Na+ (aq) is zero, which corresponds to the dry particle. At the deliquescence relative humidity of 73.75% the particle takes up water, giving a molality of about 10.6 mol kg-1 in the liquid droplet. This value could also have been calculated from the moles of liquid water per mole of salt as follows: a saturated solution contains 5.25 moles or 5.25 × 0.0180152 = 0.09458 kg of water per mole of salt. This corresponds to a molality of 1 / 0.09458 = 10.6 mol kg-1.

The 10.6 kg-1 molality of the saturated solution compares with the value of about 26.5 mol kg-1 for NH4NO3 which deliquesces at 61% relative humidity. This is consistent with the general observation that solutions become more concentrated as relative humidity is decreased.

Proceed to Lesson 5b to learn how mixing different salts affects their deliquescence behaviour, or return to the main page for this lesson.