If one sample of water has a certain concentration (g/L) of calcium ions, and another sample as that same concentration of magnesium ions, will one solution have a greater "hardness than the other"? In other words, does one ion have more influence over water hardness than the other?
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1$\begingroup$ There are other ions that contribute to hardness, particularly iron and aluminum, just FYI. $\endgroup$– Jason PattersonFeb 5, 2016 at 23:51
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2 Answers
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The one with the magnesium would be harder (assuming no other hardness-inducing ions) because magnesium has a lower molecular mass, so the molarity of magnesium would be higher. Assuming a 1:1 molar ratio of magnesium:calcium would give the same hardness (i.e. the same $ppm\ \ce{CaCO3}$ rating).
answered Feb 5, 2016 at 16:14
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The permanent hardness of water is typically given in one of three types of measurements: grains per gallon, milligrams per liter (mg/L), or parts per million (ppm) of "calcium carbonate" in the water. Since calcium carbonate has a mw of 100 g/mole the equivalents of calcium carbonate would be: $$ \text{g }\ce{CaCO3} = 100 \text{ g/mole }\times \text{(}\ce{[Ca^{2+}] +[Mg^{2+}]}\text{)}$$ where $\ce{[Ca2+]}$ is the molarity of calcium and $\ce{[Mg2+]}$ is the molarity of magnesium. So in the sense of molarity, calcium and magnesium are equal.
However you could also measure calcium and magnesium as mass of the cation per volume in which case $$ \text{mass of }\ce{CaCO3} = 2.5 \times \text{(mass of }\ce{Ca^{2+}} \text{)} + 4.1 \times \text{(mass of }\ce{Mg^{2+}} \text{)}$$ so in the sense of the mass of cations, then 1 g/l of magnesium is harder than 1 g/l of calcium.
