Oops, I accidently posted this as an answer to my previous question, when in fact it was another question.
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How do I convert mg/L NaCl to mg/L Na^+?
What I tried was:
ppm Na^+ = mg/L Na^+ = 1mg NaCl/L * (22.99gNa/35.45gCl) = 0.649 ppm Na^+
But this didn't seem right to me... I feel like I am missing something...
answered below.
To convert mg/L NaCl to mg/L Na^+, you need to consider the molar mass and charge of Na and Cl ions.
First, let's calculate the molar mass of NaCl. Na has a molar mass of approximately 22.99 g/mol, and Cl has a molar mass of approximately 35.45 g/mol. Adding these together, we get a molar mass of NaCl of approximately 58.44 g/mol.
Now, consider the stoichiometry of NaCl. One formula unit of NaCl contains one Na+ ion and one Cl- ion.
To convert mg/L NaCl to mg/L Na^+, you can use the following equation:
mg/L Na^+ = (mg/L NaCl) * (mm Na / mm NaCl)
Where:
- mg/L Na^+ is the concentration of Na^+ ions in mg per liter.
- mg/L NaCl is the concentration of NaCl in mg per liter.
- mm Na is the molar mass of Na, which is approximately 22.99 g/mol.
- mm NaCl is the molar mass of NaCl, which is approximately 58.44 g/mol.
For example, if you have a concentration of NaCl of 1 mg/L, you can calculate the concentration of Na^+ ions as follows:
mg/L Na^+ = (1 mg/L) * (22.99 g/mol / 58.44 g/mol)
mg/L Na^+ = 0.394 mg/L
So, the concentration of Na^+ ions would be approximately 0.394 mg/L.
In your calculation, you used the correct molar masses of Na and Cl, but your conversion factor should be (mm Na / mm NaCl), not (mm Na / mm Cl). By using the correct conversion factor, you would get the accurate concentration of Na^+ ions.