# Calculating the Molarity of Water

Calculating the Molarity of Water

Chemistry teachers often like to challenge their students with this one. It combines the use of concentration, density, and molar mass all in one problem, along with an added twist. (What could a truly warped chemistry teacher enjoy better than that?) Several assumptions are necessary as well, but they are commonly made assumptions that students should be in the habit of making, whether they realize they are doing it or not.

Understanding the “twist” is probably the toughest part of the problem, so let us begin there. Molarity is a measure of concentration, calculated as moles of solute divided by liters of solution. The tricky part in this problem is that there is no distinct solute (the chemical being dissolved) or solvent (what the solute is dissolved in), there is only water. In short, we are not really dealing with a solution, so calculating the Molarity is just a formal exercise, taking water as both solute and solvent.

Since water is the only chemical we are dealing with (yes, I know, assumptions are noted below), we need to find the moles and liters of water to calculate the Molarity. Simple enough, right? In fact, since concentration is constant for a given solution, no matter how much of it there is, we can simply pick any volume we like. To make life as simple as possible, let us choose a volume of one liter. We could, of course, choose 57.896 liters just as easily, and our final answer would be exactly the same. (Try it out after we’ve done it the easy way, if you don’t believe me.)

Now we just need to find the moles of water. Many students get stuck here – after all, how can you find moles if you don’t know how much water there is? But we do know – we picked the amount ourselves. We know that there is exactly one liter of water (or whatever sinister volume you might have chosen). But from liters, how does one get to moles? There is no molar volume for liquids like there is for gasses. About the only useful thing a student will know that involves volume is density. Density is as good a starting place as any. It relates mass and volume, and we know volume. The density of water is known. (See assumptions.) It is 1.00 g / mL, or 1000 g / L if you prefer. What this means is that our liter of water has a mass of 1000 grams.

Grams aren’t moles, of course, but grams and moles are related to one another by the molar mass of a molecule – water in this case. Digging out the periodic table, we find that two hydrogens and an oxygen add up to 18.02 g / mole. Our grams divided by our molar mass gives us moles – the moles of water in one liter of the same. (No, I’m not giving you the value – you can do some of the work on your own.)

Finally, we have moles of water and we have liters of water. Dividing the moles by the liters gives us the Molarity of water – or at least a good approximation of it, since we did not take into account the factors listed below. (High school students generally won’t be expected to. College on the other hand…)

Assumptions:
1. We are operating at standard temperature and pressure, so that the density of water actually is 1.00 g / mL. At higher temperatures or lower pressures, water expands, giving a slightly lower value for density. At colder temperatures or higher pressures, the density increases. These values can be found, as they have been studied and tabulated for many years.

2. We are ignoring the fact that water undergoes auto-ionization. (That’s the process by which water molecules behave as acid and base, forming hydronium and hydroxide ions. While the overall ratio of hydrogen to oxygen in water is still 2:1, not everything is actually water, so the concentration is slightly lowered. (Very slightly, as Kw is 10^-14, thankfully.)

3. We also ignore the fact that gasses can dissolve in water, resulting in small volume changes. We could certainly assume that we are working with degassed water, but in most real-life situations, this will not be the case.

4. We also ignore the fact that gasses can react with water as acid and base, as with carbon dioxide which forms carbonic acid in water. This is a justified assumption, however, as the question states “water”, and not “a solution of carbonic acid”.