Over on the TritonUS forum there has been some discussion about normalizing test results to a standard salinity of, say, 35 PPT (SG of 1.0264 @ 25C). The Triton results do not currently return any value for salinity, although the Na value is used as a rough salinity indicator. Of the various macro elements that contribute to salinity, the ones absent from the Triton ICPOES results are Cl (chlorine), C (carbon as carbonate, HCO_{3}), and F (fluoride). So, the question is, can a reasonable estimate of a sample's salinity be calculated from the macro elements that are included in the test, namely Na, Mg, Ca, K, Sr, S, Br, and B?
I submit that the answer is "yes", if we make a couple of reasonable assumptions. The first implicit assumption is that the Triton results are accurate to the precision reported, and in the units reported.
Setting aside the question of exactly how salinity is defined, I choose to use as a starting off place the concentrations of the Major Ions in seawater as shown in Table 41 found in Chapter 4 of An Introduction to the Chemistry of the Sea (Pilson, 1998), which you can find here: http://www.ocean.washington.edu/courses/oc400/Lecture_Notes/CHPT4.pdf
The principle of electroneutrality tells us that any aqueous solution (salts dissolved in water) must contain an even number of cations (+) and anions (). Let's take the NSW levels published by Pilson, and demonstrate this. I have taken the values from Pilson, and rearranged them into the following table:
Note that I have replaced the HCO_{3} and B(OH)_{3} with a single value for "Alkalinity", and that the value for "Alkalinity" is taken from the lower portion of the chart. As Pilson explains in the footnote, "In order to make a charge balance, however, the total alkalinity concentration was substituted for HCO_{3}^{} and the sodium ion adjusted accordingly." So, you can see that when you multiply the molarity of the cations by their respective charges, and add them up, and then do the same for all the anions, they balance each other out. The tiny error of 0.0008 is due to the extra precision in the fluoride value; if all the other ions had been reported with the same precision as fluoride, then they would exactly balance out.
Now, let's take these same numbers, and report them as though they had been presented in a Triton report. For this exercise, we will use the Pilson units of mg/kg instead of the Triton units of mg/L (we will address this difference in units later). Presenting Pilson's numbers like Triton would requires us to convert the sulphate (SO_{4}^{}) as sulfur (S), and boric acid (H(BO)_{3}) as boron (B), so we multiply each of these values by the atomic weight of the element divided by the molecular mass of the ion, i.e., to convert SO_{4}^{} to S we multiply by 32.06/96.07, and to convert H(BO)_{3} we multiply by 10.81/61.83. The resulting report would show the following values:
Now is the time for us to start making some reasonable assumptions. Assumption #1 will be that the water will contain the NSW concentration of fluoride, so we will use a value of 1.30 mg/kg for F. Assumption #2 will be that the sample contains the same Alkalinity as NSW, so we will use a value of 2.32 for Alkalinity. Assumption #3 will be that all of the S from the Triton report is in the form of SO_{4} in our solution. Now, all we have to do is calculate the total cations in our sample, the total amount of "known" anions, based on these three assumptions, and then we can calculate the missing Cl value. Before we can do that, we have to convert the mg/kg into mmol/kg by dividing the mg/kg value by the atomic weight of each element:
Not surprisingly, we end up with the exact same numbers as in Pilson's table, except for very minor differences due to rounding errors for Na and Br. We can now plug these into the same table we used above to calculate the missing Cl molarity:
I submit that the answer is "yes", if we make a couple of reasonable assumptions. The first implicit assumption is that the Triton results are accurate to the precision reported, and in the units reported.
Setting aside the question of exactly how salinity is defined, I choose to use as a starting off place the concentrations of the Major Ions in seawater as shown in Table 41 found in Chapter 4 of An Introduction to the Chemistry of the Sea (Pilson, 1998), which you can find here: http://www.ocean.washington.edu/courses/oc400/Lecture_Notes/CHPT4.pdf
The principle of electroneutrality tells us that any aqueous solution (salts dissolved in water) must contain an even number of cations (+) and anions (). Let's take the NSW levels published by Pilson, and demonstrate this. I have taken the values from Pilson, and rearranged them into the following table:
Ion  mmol/kg  Charge  Ionic strength  



 



 



 



 



 

 



 



 



 



 



 

 


Note that I have replaced the HCO_{3} and B(OH)_{3} with a single value for "Alkalinity", and that the value for "Alkalinity" is taken from the lower portion of the chart. As Pilson explains in the footnote, "In order to make a charge balance, however, the total alkalinity concentration was substituted for HCO_{3}^{} and the sodium ion adjusted accordingly." So, you can see that when you multiply the molarity of the cations by their respective charges, and add them up, and then do the same for all the anions, they balance each other out. The tiny error of 0.0008 is due to the extra precision in the fluoride value; if all the other ions had been reported with the same precision as fluoride, then they would exactly balance out.
Now, let's take these same numbers, and report them as though they had been presented in a Triton report. For this exercise, we will use the Pilson units of mg/kg instead of the Triton units of mg/L (we will address this difference in units later). Presenting Pilson's numbers like Triton would requires us to convert the sulphate (SO_{4}^{}) as sulfur (S), and boric acid (H(BO)_{3}) as boron (B), so we multiply each of these values by the atomic weight of the element divided by the molecular mass of the ion, i.e., to convert SO_{4}^{} to S we multiply by 32.06/96.07, and to convert H(BO)_{3} we multiply by 10.81/61.83. The resulting report would show the following values:
Na  10781 
Ca  411.9 
Mg  1284 
K  399 
Br  67.3 
B  4.49 
Sr  7.94 
S  905.0 
Now is the time for us to start making some reasonable assumptions. Assumption #1 will be that the water will contain the NSW concentration of fluoride, so we will use a value of 1.30 mg/kg for F. Assumption #2 will be that the sample contains the same Alkalinity as NSW, so we will use a value of 2.32 for Alkalinity. Assumption #3 will be that all of the S from the Triton report is in the form of SO_{4} in our solution. Now, all we have to do is calculate the total cations in our sample, the total amount of "known" anions, based on these three assumptions, and then we can calculate the missing Cl value. Before we can do that, we have to convert the mg/kg into mmol/kg by dividing the mg/kg value by the atomic weight of each element:
Element  mg/kg  atomic wt.  mmol/kg 
Na  10781  22.9898  468.95 
Ca  411.9  40.078  10.28 
Mg  1284  24.305  52.83 
K  399  39.0983  10.21 
Br  67.3  79.904  0.842 
Sr  7.94  87.62  0.0906 
S  905  32.06  28.23 
Not surprisingly, we end up with the exact same numbers as in Pilson's table, except for very minor differences due to rounding errors for Na and Br. We can now plug these into the same table we used above to calculate the missing Cl molarity: