How Many Moles of Na+ Are Needed to Soften 1.0×10^3 L of Hard Water?

[David112234]

Water “softeners” remove metal ions such as Ca^2+ and Fe^3+ by replacing them with enough Na+ ions to maintain the same number of positive charges in the solution. If 1.0×10^3 L of “hard” water is 0.015 MCa2+ and 0.0010 M Fe3+, how many moles of Na+ are needed to replace these ions?

Ok, So this is how I would do it, Since Ca is 2+ ion you will need twice as much Na to replace its charge since Na is 1+. And for Fe 3+ 3 time as much. So if you have 0.015moles of Ca^2+ you would need 0.03 moles of Na. For Fe^3+ , 0.003 moles of Na
0.015 * 2 = 0.03
0.0010 * 3 = 0.003

But when I googled this question, I found this answer, where someone multiplied the moles of Ca, and Fe times Liters of water. (https://answers.yahoo.com/question/index?qid=20081023184058AAPL21u)
I don't understand why? Is my version correct? Or theirs?

 

[Borek]

You don't have 0.015 moles of Ca²⁺. You have 0.015 moles of Ca²⁺ per liter.

That's the definition of the concentration. Concentration is an intensive property, amount is an extensive property (check what these mean if you have not heard about them these are very important concepts).

 

[David112234]

You don't have 0.015 moles of Ca²⁺. You have 0.015 moles of Ca²⁺ per liter.

That's the definition of the concentration. Concentration is an intensive property, amount is an extensive property (check what these mean if you have not heard about them these are very important concepts).

Oh, I did not realize it was giving me the concentration, I thought it was giving me just the amount in total. Thank you