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 dont understand why? Is my version correct? Or theirs?