Molar Concentration

Divide the dissolved amount of substance by the volume of the solution: $c =\frac{ n}{V}$, with n in moles and V in litres. If the weighed mass is given instead of the amount, first convert with $n =\frac{ m}{M}$. Example: 5.85 g of table salt $(M =\frac{ 58.44 g}{mol})$ are dissolved in water and filled up to 500 mL. Then $n =\frac{ 5.85}{58.44} = 0.100 mol$ and $c =\frac{ 0.100}{0.500} =\frac{ 0.200 mol}{L}$. Two pitfalls: the volume must be in litres, so $500 mL = 0.500 L$, and what counts is the volume of the finished solution, not the amount of water added. That is why you fill up to the calibration mark in a volumetric flask.

A 1-molar solution contains exactly one mole of the dissolved substance per litre of solution, i.e. $c =\frac{ 1 mol}{L}$. The phrase "molar" (abbreviated M) is common in the lab: a 0.1-molar hydrochloric acid has c(HCl) = 0.1 mol/L. The reference to the solution volume matters: for a 1-molar NaCl solution you weigh out 58.44 g of NaCl, dissolve it in some water and then fill up to exactly one litre. You do not add one litre of water, because the dissolved substance changes the volume. Also beware of the term molality: it refers to the mass of the solvent in $\frac{mol}{kg}$ and is a different quantity.

When diluting, the dissolved amount of substance stays the same, only the volume grows. From $n = c\cdot{}V$ it follows that $c_{1}\cdot{}V_{1} = c_{2}\cdot{}V_{2}$. Example: you need 250 mL of hydrochloric acid with $c =\frac{ 0.1 mol}{L}$ and have a stock solution with $c =\frac{ 1.0 mol}{L}$. Then $V_{1} =\frac{ c_{2}\cdot{}V_{2}}{c_{1}} =\frac{ 0.1\cdot{}0.250}{1.0} = 0.025 L$. So you pipette 25 mL of stock solution into a 250 mL volumetric flask and fill up to the mark with water. The formula applies to pure dilution without reaction. Practical rule for acids: always add the acid to the water, never the other way round, because of the heat of dilution.

The molar concentration $c =\frac{ n}{V}$ counts moles per litre, the mass concentration $\beta{} =\frac{ m}{V}$ counts grams per litre. Both refer to the solution volume but differ by the molar mass: $\beta{} = c\cdot{}M$. A saline solution with $c =\frac{ 0.15 mol}{L}$ thus has $\beta{} = 0.15\cdot{}58.44 \approx{}\frac{ 8.8 g}{L}$; that roughly corresponds to physiological saline $(\frac{9 g}{L})$. For chemical calculations c is the more useful quantity, because reactions proceed in ratios of amounts; on packaging and in medicine, however, you often encounter β. In tasks, therefore, check carefully which concentration is meant and convert via M if necessary.

In a titration you determine an unknown concentration from the consumption of a standard solution of known concentration. At the equivalence point, for a 1:1 reaction such as HCl + NaOH: n(acid) = n(base), i.e. $c_{1}\cdot{}V_{1} = c_{2}\cdot{}V_{2}$. Example: 20.0 mL of hydrochloric acid of unknown concentration consume 12.5 mL of sodium hydroxide with $c =\frac{ 0.1 mol}{L}$. Then n(NaOH) = 0.1·0.0125 = 1.25×10⁻³ mol = n(HCl), and c(HCl) = 1.25×10⁻³/0.0200 = 0.0625 mol/L. For other stoichiometries, such as sulfuric acid with two protons, you must build in the ratio from the equation: n(NaOH) = 2·n(H₂SO₄). This is exactly where most exam errors happen.