In fresh water systems, 1 mg/L is equivalent to 1 ppm, but when the specific gravity of the solution is not 1 this is not valid. The specific gravity of water is usually assumed to be ~ 1 but this can change because of temperature and amount of dissolved ions. For example, it is about 1.025 for sea water.
Note: specific gravity is related to density but they are not quite the same. Density is defined as mass per unit volume, like mg/L or kg/m3 (the international standard metric unit). Specific gravity is the ratio of a liquid's density with that of water at 4 °C, the temperature at which water is most dense, and is 999.974 kg/m3 or 0.999974 mg/L (1 mg/L = 1000 kg/m3). Since specific gravity is the density of a liquid divided by the density of water at 4 °C, it has no units because they cancel out; specific gravity is just a number. Density does have units like mg/L. A usually good approximation for fresh water is that it has a density of 1 mg/L or a specific gravity of 1.
If the specific gravity of the solution is not 1, the following is the relationship:
To convert from mg/L to mEq/L for an ion, you need to know the concentration in mg/L, its atomic mass (which is actually unitless because it is a ratio), and the valence (the absolute value of its charge).
For example:
Chloride - Its atomic mass is 35.4527 amu and its charge is negative (-1) which means its valence is 1. If your chloride concentration is 60 mg Cl/L, the mEq/L would be (60 mg/L) / (35.4527) = 1.6949152542372883 or 1.6949 mEq/L.
Sulfate (SO₄⁼) - Atomic mass is (32.066 + 4*(15.9994)) = 96.0636, Valance is 2, charge is negative (-2). If your result was 60 mg SO₄⁼/L, the mEq/L would be 1.249172423269584 or 1.2492 mEq/L.
Please note: 60 mg Cl/L is not the same electrochemically as 60 mg SO4/L. You may want to review the webpage related to conductivity and total dissolved solids.. This is critical to remember when you are trying to work with cation and anion mass balances and the electrochemistry of water.
Table 1 | Conversion Table from mg/L to mEq/L and vica versa
Water should be electrochemically neutral and you can use the cation and anion mass balance approach to get some idea of the potential quality of your data and attempt to understand if you are missing any specific large grouping of cations or anions.
This is an example of a cation-anion balance
Table 2 | Cation Anion Balance - Results and Calculations
FIrst calculate the Sum of the Cations - Sum anions (SA) or 3.77397 - 3.67496 = 0.102437.
Then calculate the Sum of the Cations + Sum anions (TSA) or 3.77397 + 3.67496 = 7.452357.
Finally, calculate the (SA/TA)100 = (0.102437/ 7.452357) 100 = 1.37 %
Ideally, there should be no difference between the sum of the cations (mEq/L) and the sum of the anions (mEq/L). If there is a difference, especially if it is a large difference, it means one of two things. Some significant ion was not included in the testing or the lab made a mistake and reported an incorrect result (something that happens surprisingly often even in certified labs). In the example above, the percent difference between the sums of the cations and anions is positive which might suggest that a minor anion is missing. Since the difference is < 2 %, this does strongly suggest the quality of the data is good. If the difference were 2 % or more, it would be very wise to conduct some additional testing.
The guidance on the Acceptable Percent Difference is a function of the Anion Sum. Standard Methods for the Examination of Water and Wastewater 18th edition recommends the following guidance:
Table 3 | Anion-Cation Balance: Acceptable Difference
Expressing results in other forms. In water treatment there is a tendency to express many things as the equivalent of CaCO3 or mg CaCO3/L. Also important to remember is that even though 1 mEq of CaCO3/L = 1 mEg of Ca⁺⁺, this does not mean 1 mg/L CaCO3 = 1 mg/L Ca⁺⁺.
1 mEq/L of Calcium as CaCO3
1 mEg/L of Calcium * (20.04) = 20.04 mg/L Calcium, but this does not equal 20.04 mg CaCO3.
1 mEq/L of Calcium * 50 mg CaCO3/mEq = 50 mg/L of Ca as CaCO3
(Read as 1 mEq of Calcium: Well 1 equivalent of CaCO3 has 50 mg/mEq, therefore 1*50 = 50 mg/L CaCO3)
Example,
2.2 mEq Ca/L * (50 meg CaCO3/ mEq) = 110.06 mg CaCO3
Table 4 | Convert mEq/l of one cation or anion into the mEq/L of something else
Note: Total PO₄ (mg/l as PO₄) x 0.3262 = Total P
At wt of P = 30.9737; of O = 15.9994
[TPO₄] * (30.9737/94.9713)= [TP]
Note: Total P (mg/L as P) x 3.066 = Total mg PO₄/L
At wt of P = 30.9737; of O = 15.9994
[TP] * (94.9713/30.9737) = [TP0₄]
Atomic weight Nitrogen (N) 14.0067 ; Oxygen is 15.9994
[TN] * (62.0049 / 14.0067) = mg NO3/L
Atomic weight Nitrogen (N) 14.0067 ; Oxygen is 15.9994
[mg NO3/L] * (14.0067 / 62.0049) = mg N/L
Note: mg/L x ft3/s x 2.446 = kg/day kg/day x 2.205 = lbs/day
2,000 lbs = one (English or long) ton - 1,000 kg = one tonne (metric or long ton) = 2,200 lbs
If a stream had a flow of 1 ft3/s, 1 mg N03-N/L, and 0.05 mg PO4-P/L,the stream would be transporting 5.39 lbs per day of Nitrate as Nitrogen and 0.269 lbs per day of Phosphate as Phosphorus.
