
03152012, 12:12 PM

#1

Apr 2011
Orlando, Florida
Posts: 4

Hello,
Sorry if it has been answered before...
I tried for several hours to get solution from google, not big luck, as I got a bit confusing ( for me) ideas.
So, I got a my water analyses report, but unfortunately they did not separate how much I got a sodium (Na+) in my water, maybe somebody can help how to calculate to get idea what sodium level I have?
Also, what numbers do I need to give?
Some figures ( not all)
Total hardness ekvmmol/l (mmol/l) 6.35 (3.2)
Calcium Ca 2+ mg/l 88
Magnium Mg 2+ mg/l 23.5
Cl mg/l 4
SO4 2 mg/l 2
HCO3 mg/l 384
pH 8.3
Thanks in advance,



03152012, 12:42 PM

#2

Aug 2010
McLean/Ogden, Virginia/Quebec
Posts: 9,283
Liked 1484 Times on 1134 Posts

I'm not sure whether you are more interested in the answer or the method of getting it so I'll cover both. The method is based on the fact that water is electrically neutral i.e. that the sum of the charges on all the cations must equal the sum of the charges on all the anions. If you have dead accurate concentrations for all the ions except sodium then you simply add up all the anions (charge  6.589) and all the cations (charge 6.325) and add those two numbers together. The answer is .254 thus there are 0.254 mEq/L more anions than cations in the water. You assume that the missing +0.254 mEq/ positive charge is from sodium and so multiply by the equivalent weight of sodium (23) to get 6 mg/L as the sodium estimate.
The problems with doing this are that there may be other things contributing positive charge which are not sodium (such as iron, potassium, aluminum, strontium, manganese, copper, ammonium...). While these are all at pretty low levels they nonetheless induce error. Their presence will make the estimate too high. Or there can be unaccounted for cations such as nitrate, nitrite and fluoride. Their presence would make the sodium estimate too low.
The other potential problem is that you have so much bicarbonate and calcium (4.39 mEq/L). Each 1% error in the reported value for calcium corresponds to 0.0439 mEq/L which is equivalent to 0.92 mg/L sodium. You have 6.3 mEq/L alkalinity. Each 1% error in that is 0.063 mEq/L equivalent to 1.4 mg/L sodium.



03152012, 01:33 PM

#3

Apr 2011
Orlando, Florida
Posts: 4

Thanks ajdelange!!!
Perfectly you replied that it is a best to get result and in same time it is very interesting to get a bit of understanding how you get a result.
Ok, then to be clear I will add numbers what is missing as I see your point to get a "error" is very possible.
So full report sounds like this
Total hardness ekvmmol/l (mmol/l) 6.35 (3.2)
Calcium Ca 2+ mg/l 88
Magnium Mg 2+ mg/l 23.5
Total alkalinity ekvmmol/l 6.3
Cl mg/l 4
SO4 2 mg/l 2
HCO3 mg/l 384
Fe mg/l 0.02
Mn 2+ mg/l less than 0.02
KMnO4 mg O/l 0.7
Nitrate NO3N mg/l 5.6
Nitrite NO2N mg/l less than 0.005
Ammonium NH4+ N mg/l less than 0.05
H2S mg/l less than 0.05
conductivity (25 Celsius)  mS/cm 0.605
pH 8.3
P.S. I am not from Florida....but from Europe....just to be fully honest...



03152012, 03:10 PM

#4

Aug 2010
McLean/Ogden, Virginia/Quebec
Posts: 9,283
Liked 1484 Times on 1134 Posts

The only additional thing of significance is the nitrate. Accounting for that the sodium estimate is more like 16 mg/L



03152012, 03:56 PM

#5

Apr 2011
Orlando, Florida
Posts: 4

Woow, thanks a lot ajdelange.
I am very impressed as I spent yesterday several hours on search and even from one advise came to 70mg/l... ))))
If it is not sooo long, can you kindly teach how you got that result? As it seems nitrate even in small amount played a important role.
The very best from Baltics states....



03152012, 05:21 PM

#6

Aug 2010
McLean/Ogden, Virginia/Quebec
Posts: 9,283
Liked 1484 Times on 1134 Posts

As I noted #2 its a matter of adding up all the charges. If the answer isn't 0 you assume that something is missing (though in fact the non 0 answer can be caused by errors in the individual meausurements.
To get you started: Calcium has an equivalent weight of 20. Thus if you have 88 mg/L calcium that's +88/20 = +4.400 meq/L
Manesium has an equivalent weight of 12.15 so 23.5 mg/L Mg++ contributes + 23.5/12.15 = +1.905
Chloride has an equivalent weight of 35.45 so 4 mg/L Cl contributes 4/35.45 = 0.113
and so on. You can find equivalent weights on Wikipedia and elsewhere.
When you get to nitrate note that it is reported "as Nitrogen" i.e. "NO3 N" so you divide by the equivalent weight of nitrogen (14) not that of the nitrate ion (62) in this case.
When you have all the individual charges per liter, add them up to get the net missing charge. The sum should be a negative number in this case indicating that you are missing something positively charged. To see how much sodium it would take to supply this missing positive charge multiply by the equivalent weight of sodium (23).



03152012, 06:32 PM

#7

Apr 2011
Orlando, Florida
Posts: 4

Yap, for now I did not get it by 100% ....but tomorrow I will study closely and will learn each number what you teached for future needs!!!
I really hope that in future your valuable help will be helpful for others too!!!
Best regards,





