 # Calculating of molar concentration for a concentrated solution

Three weeks ago my colleague and me conducted oral exams of student teachers (in chemistry). One of the tasks given to the students by me was calculating of the molar concentration for a concentrated solution of hydrochloric acid (36%) or sulfuric acid (about 100%), respectively.

Every chemists knows the big, heavy glass bottles comprising non–diluted solutions or pure fluids such as hydrochloric acid or sulfuric acid. Usually the concentration is given in percent. Indeed, it would be better in molar units. So, how do we calculate the concentration? Does this percentage refer to volume concentration (v/v), mass fraction (w/w, wt.%), or even mass concentration (w/v)? Unfortunately, it is not given on all bottles…

Let us start with the sulfuric acid and just assume it is pure (100%). We want to know the concentration, so our ansatz looks like this:

\begin{align} c_{H_2SO_4} & = \frac{n_{H_2SO_4}}{V_{solution}} = \frac{n_{H_2SO_4}}{V_{H_2SO_4}} \\ \end{align}

We can substitute the volume of the solution($$V_{solution}$$) by the volume of sulfuric acid ($$V_{H_2SO_4}$$) since our assumption from above (pure!). Next, we try to replace both the amount of substance and the volume by something we know. The amount of substance is given by the mass and molar mass, and the volume is given by the mass and the density of the fluid:

\begin{align} c_{H_2SO_4} & = \frac{n_{H_2SO_4}}{V_{H_2SO_4}} = \frac{\frac{m_{H_2SO_4}}{M_{H_2SO_4}}}{\frac{m_{H_2SO_4}}{\rho_{H_2SO_4}}} = \frac{m_{H_2SO_4} \cdot \rho_{H_2SO_4}}{m_{H_2SO_4} \cdot M_{H_2SO_4}}\\ \end{align}

Obviously, the mass can be cancelled out leaving only the molar mass and the density. We know both values for sulfuric acid (always use the density on the bottle!) and, therefore, can calculate the concentration:

\begin{align} c_{H_2SO_4} & = \frac{\rho_{H_2SO_4}}{M_{H_2SO_4}} = \frac{1.84~g \cdot mL^{-1}}{98.079~g~\cdot mol^{-1}} = \\ & = \frac{1840~g~\cdot~mol}{98.079~g~\cdot~L} = 18.76~\frac{mol}{L}\\ \end{align}

Thus, a high concentrated (pure) sulfuric acid solution/liquid has almost 19 M! That was easy, wasn’t it? What about the hydrochloric acid? We do not know what the percentage is referring to, yet! Let’s start like above:

\begin{align} c_{HCl} & = \frac{n_{HCl}}{V_{solution}} = … = \frac{m_{HCl} \cdot \rho_{solution}}{m_{solution} \cdot M_{HCl}} \\ \end{align}

In this case, the volume of the solution ($$V_{solv.}$$) is not the volume of the acid. Hence, we cannot substitute it. Similarly, we proceed as before (just be aware of the indices!). Since the mass of HCl is not the mass of the solution we cannot cancel it out, this time. Before we can proceed we need to know the meaning of the %’. We give the percentage the symbol f (for fraction). There are three different cases to consider:

• Mass fraction (w/w, wt.%) means $$f = \frac{m_{HCl}}{m_{solution}}$$; inserting as $$m_{solution} = \frac{m_{HCl}}{f}$$
• Volume concentration (v/v) means $$f = \frac{V_{HCl}}{V_{solution}}$$; inserting as $$V_{solution} = \frac{V_{HCl}}{f}$$
• Mass concentration (w/v) actually means $$\frac{m_{HCl}}{V_{solution}}$$, which is very missleading since it is not dimensionless. Indeed, it can be transformed to $$f = \frac{m_{HCl}}{V_{solution} \cdot \rho_{solution}}$$, which is just another expression for the first case.

Both the first and second case lead to similar but not identical formulas.

\begin{align} c^1_{HCl} & = f \cdot \frac{\rho_{solution}}{M_{HCl}}\\ c^2_{HCl} & = f \cdot \frac{\rho_{HCl}}{M_{HCl}} \end{align}

Independent of the case, we could now calculate the molar concentration. If you do a little research, you’ll discover that most percentages are given in mass concentrations (first case) and not in volume concentrations. The reason is simple: handling masses is much more easier than volumes, since the former are additive! Due to this and some information we can finally calculate the hydrochloride concentration:

\begin{align} c_{HCl} & = f \cdot \frac{\rho_{solution}}{M_{HCl}} = 37\% \cdot \frac{1.19~g \cdot mL^{-1}}{36.46~g~\cdot mol^{-1}} = \\ & = 37\% \frac{1190~g~\cdot~mol}{36.46~g~\cdot~L} = 12.08~\frac{mol}{L}\\ \end{align}

In the end, you just read the density given on the bottle, divide it by the molar mass of the compound, and multiply the result by the percentage/fraction also given on the bottle to get the concentration in molar units. Thats it!

A little suggestion: Remember some of the values of high concentrated reagents (e.g. about 18 M for sulfuric acid and about 12 M for hydrochlorid acid). It makes it easy to check if the formula in your mind is correct when calculating and also to guess molar concentrations for other percentages while reading papers or conducting discussions (10% hydrochloride acid? must be around 3 M!‘)

By the way: the student teachers did all well on this task (and they all passed)!