# How to make by dilution a certain ion concentration

The question I'm trying to answer is this:

What volume of $0.25\ \mathrm{mol/l}$ calcium nitrate is required to make by dilution with water $500\ \mathrm{cm^3}$ of a solution with a nitrate ion concentration of $0.1\ \mathrm{mol/l}$?

My attempt: Using $n=cv$, we find the the number of moles of nitrate ions to be $0.05$ moles. Now from the chemical formula of calcium nitrate, we know that $1$ mole of calcium ions gives two moles of nitrate ions. Therefore $0.05$ moles of nitrate ions gives us $0.025$ moles of calcium ions. Now, using $v=\frac{n}{c}$, where $n$ is the no. of moles of calcium ions, we obtain the volume of $0.1$ litres.

My attempt actually managed to give me the correct but I don't understand a part of it; I can't understand why when I'm trying to find the volume of calcium nitrate I have to use the number of moles of calcium ions and not the number of moles calcium ions added to the number of moles of the nitrate ions.

(Note: I asked this question before, deleted it, changed it and re-published it as of now.)

• Hint - A molecule of $\ce{CaNO3}$ is 1:1 for anion and cation. So a 0.25 mol/l solution is 0.25 mol/l for both the cation and the anion.
– MaxW
Aug 24, 2016 at 19:04
• @MaxW But isn't the formula $Ca(NO_3)_2$? Aug 24, 2016 at 20:04
• Geez, never delete and repost! Questions can be edited and reopened. Aug 24, 2016 at 20:56
• @ReinhildVanRosenú - You're absolutely right. I had a major brain fart. A $\ce{CaNO3}$ solution is 1:2 for cation and anion. So a 0.25 mol/l solution of $\ce{CaNO3}$ is 0.25 mol/l for the cation $\ce{Ca^{+2}}$ and 0.50 mol/l for the anion $\ce{NO3^{-1}}$. Thanks for the correction.
– MaxW
Aug 24, 2016 at 22:21
• Noting that the solution is 0.50 mol/l for $\ce{NO3^{-1}}$, you can use the equation: $$m_1 * V_1 = m_2 * V_2$$ or $$0.50 mol/l * 0.500 l = 0.1 mol/l * x l$$ therefore $$x = \dfrac{0.25}{0.1} = 2.5 l$$
– MaxW
Aug 24, 2016 at 22:31

When you use the number of moles of calcium ions, you are actually using the number of moles of calcium nitrate. You have done the 1:2 conversion already in your question. You found the amount of nitrate ions needed to be 0.05. But then instead of converting that amount to moles of calcium ions, you were actually converting it to a number of moles of calcium nitrate, since the ratio of moles calcium ($\ce{Ca}$) to moles calcium nitrate ($\ce{Ca(NO3)2}$) is 1:1.