Example: Calculation of the Activity Product of Diethylaminium Nitrate in a Saturated Aqueous Solution

This document shows how to calculate the equilibrium activity products of amine (i.e., aminium) salts in saturated aqueous solution. They are needed to predict the formation of the solid, and should be entered into E-AIM on the data entry page of the amine. We use as an example the nitrate salt of diethylamine (DEA), which is one of the amines that can be selected from the Available Compounds page.

The formation of the solid nitrate salt is described by:

(DEA+)NO3(s) = DEA+(aq) + NO3(aq)    (1)

The activity product KS in molality based units is given by:

KS = mDEA+ · γDEA+) · mNO3 · γNO3    (2)

where prefix m indicates molality, and γ the molality-based activity coefficient. We require the values of these quantities for a solution that is saturated with respect to the salt. The solubility of diethylaminium nitrate – i.e., the molalities in equation (2) above – are listed in Table 5 of Ge et al. (2011) for a series of temperatures from 0 °C to 50 °C. These molalities, m, are given in the second column of the table below.

T m f xH2O KS
263.15 8.37 0.2294 0.7683 2.176
273.15 15.27 0.2170 0.6451 4.569
293.15 26.85 0.2462 0.5083 11.29
298.15 33.5 0.2522 0.4531 14.65
313.15 41.67 0.2801 0.3998 21.77
323.15 66.18 0.2899 0.2955 32.14

We need to determine, using E-AIM, the molal activity coefficients γ corresponding to each of these solubilities so that KS can be calculated. This is done in the following steps:

  1. Select diethylamine on the Available Compounds page so that it is included in E-AIM calculations.
  2. Go to the Aqueous Solution page for Model II. You will see that, below the boxes for entering the molalities of the inorganic ions, there is now a section in which: (i) the molality of diethylamine can be entered; (ii) the compound can be constrained to exist in one of two possible liquid phases; (iii) the dissociation reaction of the amine cation can be switched off; and (iv) the UNIFAC parameter set (for interactions between water and the neutral amine molecule) can be chosen.

    For this calculation select "Aqueous only" under (2), and leave amine dissociation "on" and the standard UNIFAC parameter set selected.

  3. Carry out a calculation for the saturated aqueous solution at 298.15 K, in which the molality of (DEA+)NO3 is 33.5 mol kg−1. There is no entry box on the E-AIM Model II page for the DEA+ ion, only for the uncharged molecule diethylamine. We therefore enter 33.5 mol kg−1 of diethylamine, and 33.5 mol kg−1 each of H+ and NO3 which is the amount of nitric acid required to neutralise the amine to (DEA+)NO3.

    Do the calculation by pressing the "Submit" button, to obtain the following results:

    Species      Moles         Grams        Molality     Mole Frac.   Act. Coeff.   Act. Eqn. 
    H(aq)        0.20976E-05   0.2114E-05   0.2098E-05   0.1712E-07   0.2776E+02    PSC       
    DEA+(aq)     0.33500E+02   0.2484E+04   0.3350E+02   0.2734E+00   0.2522E+00    PSC       
    NO3(aq)      0.33500E+02   0.2077E+04   0.3350E+02   0.2735E+00   0.2522E+00    PSC       
    OH(aq)       0.15267E-12   0.2597E-11   0.1527E-12   0.1246E-14   0.3061E+04    PSC       
    H2O(aq)      0.55509E+02   0.1000E+04   0.5551E+02   0.4531E+00   0.1230E+01              
    DEA(aq)      0.20976E-05   0.1534E-03   0.2098E-05   0.1712E-07   0.2207E+01    UNIFAC    

    The molalities of the DEA+ and NO3 ions are equal to the 33.5 mol kg−1 that was input (because the molality of the uncharged DEA is very small, at 0.2098 × 10−5 mol kg−1). The activity coefficients of the ions are both 0.2552. Remember that E-AIM lists mole fraction values of activity coefficients (f), and these must be converted to a molality basis (γ) for use in equation (2). This is done by multiplying each activity coefficient by the mole fraction of water in the solution (0.4531 from the results above), so that:

    γDEA+ = fDEA+ × xH2O

    = 0.2522 × 0.4531 = 0.11427

    The analogous relationship applies for NO3, also yielding γNO3 = 0.11427.

    The value of KS at 25 °C can now be calculated from equation (2):

    KS(25 °C) = (33.5 × 0.11427) × (33.3 × 0.11427) = 14.654 mol2 kg−2

  4. Repeat this procedure to determine KS from the solubilities at the other temperatures in the table at the top of this page. The mole fraction activity coefficients of both ions (f), mole fractions of water (xH2O), and KS from these calculations are listed there.

The next thing to do is use the values of KS we have determined to obtain best-fit values of KS(25 °C), ΔHo and (possibly) ΔCp for the reaction so that E-AIM will be able to calculate the formation of the salt from aqueous solutions at any temperature. The equation for KS as a function of temperature is:

R · ln(KS(T)) = R · ln(KS(Tr)) + ΔHo(Tr)(1/Tr - 1/T) + ΔCpo(Tr) (Tr/T - (1 + ln(Tr/T)))     (3)

where R (8.3144 J mol−1 K−1) is the gas constant and Tr is equal to 298.15 K. This equation should be fitted to the natural logarithms of the tabulated KS with ln(KS(Tr)), ΔHo(Tr) and ΔCpo(Tr) as unknowns.

Such a fit yields the following result: ln(KS(Tr)) = 2.655 ± 0.045, ΔHo(Tr) = 28.41 ± 1.4 kJ mol−1, and ΔCpo(Tr) = -320 ± 119 J mol−1 K−1. Clearly the uncertainty in ΔCpo(Tr), which is related to the second differential of ln(KS(T)) with respect to T, is large. The absolute value is also large compared to the -101.2 J mol−1 K−1 obtained by Clegg et al. (1998) for NH4NO3(s) (see their Table 2).

Further test fits of the data show that if the measurement for 263.15 K is omitted then the level of significance of the fitted ΔCpo(Tr) is so low that it can be set to zero. Thus, it appears that the data do not extend over a wide enough temperature range, and/or are too imprecise, to reliably determine ΔCpo(Tr). Such a fit yields ln(KS(Tr)) = 2.598 ± 0.030, and ΔHo(Tr) = 28.13 ± 1.3 kJ mol−1. One further fit, fixing ΔCpo(Tr) to -101.2 J mol−1 K−1 and using all the data, yields ln(KS(Tr)) = 2.595 ± 0.039 and ΔHo(Tr) = 30.07 ± 1.3 kJ mol−1. Values of ln(KS(Tr)) and ΔHo(Tr) from both of these fits agree to within the uncertainties in the fitted parameters and either set of results could be used in the model.

How are the equilibrium constant and thermal properties included in E-AIM? On the data entry page for diethylamine, the last part of section 6 has the heading "Formation of inorganic aminium salts". If "nitrate" is checked then input boxes for KS(Tr), ΔHo(Tr) and ΔCpo(Tr) will appear. Enter the values that have been fitted. There is also a box for the molar volume of the salt. This can be entered, if it is known, otherwise it can be left blank.

Next, save the data using the button in section 10, selecting "Overwrite existing data" from the browser page that appears. If you return to one of the E-AIM problem pages, such as the one used to calculate KS, it will be possible to calculate the formation of the salt. The ability to switch off the formation of the salt will also have been added (there will be a checkbox for (DEA+)NO3(s) in the Solids section of the page).

Chloride and Sulphate Salts

The procedure for determining KS and the associated thermal properties for chloride salts of singly dissociating amines is exactly the same as described above for the nitrate salt, except that HCl is used as the neutralising acid instead of HNO3. For the sulphate, H2SO4 is used, remembering that the stoichiometry of the reaction is different:

(DEA+)2SO4(s) = 2DEA+(aq) + SO42−(aq)    (4)


KS = (mDEA+ · γDEA+)2 · mSO42− · γSO42−    (5)

Thus, if the solubility of the diethylaminium sulphate salt were 1.0 mol kg−1 then the molalities entered for the calculation of the activity coefficients would be: 2.0 mol kg−1 H+, 1.0 mol kg−1 SO42−, and 1.0 mol kg−1 diethylamine. The conversion of the activity coefficients from mole fraction to a molality basis, and equation (3), are exactly the same as for the nitrate and chloride salts.

Diamines and Their Salts

Diamines form doubly charged cations, so that the stoichiometries of the solubility reactions differ from those given above. For example, for a diaminium nitrate salt we would have:

H3NRNH3(NO3)2(s) = +H3NRNH3+(aq) + 2NO3(aq)     (6)

KS = a(+H3NRNH3+) · (aNO3)2 = m(+H3NRNH3+) · (mNO3)2 · γ(+H3NRNH3+) · (γNO3)2     (7)

If the solubility of the salt in water were 10 mol kg−1 then, to calculate KS, we would enter 20.0 mol kg of H+, 20.0 mol kg−1 of NO3 and 10.0 mol kg−1 of the diaminium salt on the E-AIM problem page.

Sources of Further Information


K. S. Carslaw, S. L. Clegg, and P. Brimblecombe (1995) J. Phys. Chem. 99, 11557-11574.
X. Ge, A. S. Wexler and S. L. Clegg (2011) Atmos. Environ. 45, 561-577.