Solution Found!
Water “softeners” remove metal ions such as Ca2+ and Fe3+
Chapter 4, Problem 19P(choose chapter or problem)
Questions & Answers
QUESTION: Water “softeners” remove metal ions such as \(\mathrm{Ca}^{2+}\) and \(\mathrm{Fe}^{3+}\) by replacing them with enough \(\mathrm{Na}^{+}\) ions to maintain the same number of positive charges in the solution. If \(1.0 \times 10^{3} \mathrm{~L}\) of “hard” water is \(0.015\mathrm{\ M\ Ca}^{2+}\) and \(0.0010\ M\mathrm{\ Fe}^{3+}\), how many moles of \(\mathrm{Na}^{+}\) are needed to replace these ions?
ANSWER:
Step 1 of 2
Here we have to find out how many moles of \(\mathrm{Na}^{+}\) are needed to replace ions such as \(\mathrm{Ca}^{2+}\) and \(\mathrm{Fe}^{3+}\).
Given:
Volume of hard water = \(1.0 \times 10^{3} \mathrm{~L}\)
Molarity of \(\mathrm{Ca}^{2+}\) = 0.015 M
Molarity of \(\mathrm{Fe}^{3+}\) = 0.0010 M