# Can I get some pointers on what to do in calculating the hardness of water?

Information provided: $\pu{1.64 g}$ of $\ce{Ca(NO3)2}$ and $\pu{0.94 g}$ of $\ce{MgCl2}$ are dissolved in $\pu{10 L}$ of water, calculate the hardness of water.

I have used these two equations: $m = n \cdot M$ and $c = \frac nV$.

What I have done is simply insert the molar mass of both $\ce{MgCl2}$ and $\ce{Ca(NO3)2}$ in the first equation, isolated $n$, then used that answer to isolate $c$ in the next equation. But I need to find the mass for $\ce{Ca^2+}$ and $\ce{Mg^2+}$ only, how do I do that? Have I found the concentration of the wrong things?

Please explain in simple language. Thanks!

The hardness of water is defined as the sum of the dissolved $\ce{Ca^2+}$ and $\ce{Mg^2+}$ species and can be reported in different ways (moles/L, g/L, and so on). The one I have seen most commonly is mg/L, so that is the calculation presented below.

You need to determine how many milligrams of $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are present and then divide that number by the solution volume (here, 10 L).

For $\ce{Ca^2+}$:

$$1.64\;\mathrm{g}\;\ce{Ca(NO3)2}\cdot{1\;\mathrm{mol}\;\ce{Ca^2+}\over 164.09\;\mathrm{g}}\cdot{40078\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Ca^2+}} = 400\;\mathrm{mg}$$

For $\ce{Mg^2+}$:

$$0.94\;\mathrm{g}\;\ce{Mg(Cl)2}\cdot{1\;\mathrm{mol}\;\ce{Mg^2+}\over 95.2\;\mathrm{g}}\cdot{24305\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Mg^2+}} = 240\;\mathrm{mg}$$

Thus, you have 640 mg $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in 10 L of solution so the hardness is:

$${640\;\mathrm{mg}\over 10\;\mathrm{L}} = 64\;\mathrm{mg}\cdot\mathrm{L}^{-1}$$