The superior convenience of using solutions of a known molar concentration is illustrated by the following simple example. Suppose we need to make batches of Cu(OH)2
in various quantities on a regular basis, several times a day. For the preparation, a solution of CuCl2
is treated with a solution of NaOH and the blue precipitate of insoluble Cu(OH)2
formed is separated by filtration, washed with water, and dried. CuCl2 + 2 NaOH = Cu(OH)2↓ + 2 NaCl
One way to deal with the assignment is each time we need to make a batch of Cu(OH)2
, we calculate the amounts of CuCl2
and NaOH needed for making the required quantity of the product, weigh out the reagents, and do the reaction.
There is, however, another, much more efficient and easy way of dealing with the task. We prepare two stock solutions
, one of NaOH and the other of CuCl2
. As follows from the equation, per each molecule (mole) of CuCl2
we need two molecules (moles) of NaOH for the reaction. It would be handy to have a 1 molar (1M) solution of CuCl2
and a 2 molar (2M) solution of NaOH. The prepared solutions can be stored on the shelf and used at any moment to make Cu(OH)2
by simply mixing equal volumes of each. Moreover, it is very easy to calculate what volume of each solution is needed to make a particular quantity of Cu(OH)2
. For example, if 10 g of Cu(OH)2
is needed to be made, we convert the mass amount to the molar amount by dividing 10 g by 97.6 g, the mass of 1 mole of Cu(OH)2
. The result, 10/97.6 = 0.102, is the number of moles in 10 g of Cu(OH)2
. Consequently, 0.102 mol of CuCl2
and 0.204 mol of NaOH are needed for the reaction.
Now that we know the number of moles of each reagent, we can calculate the volumes of the stock solutions to be mixed together to make the required amount of the product. Again, we think proportions. The molarity of our CuCl2
stock solution is 1M, meaning that 1 L of the solution contains 1 mole of CuCl2
. We need the fraction volume x
of this solution containing 0.102 mol of CuCl2
1 mol – contained in 1 L
0.102 mol – contained in x
= (1 x 0.102)/1 = 0.102 L = 102 mL. This is the volume of the CuCl2
stock solution to be used for the reaction.
Similarly, we come up with a proportion to find out the volume y
of the 2M NaOH stock solution.
2 mol – contained in 1 L
0.204 mol – contained in y
= (1 x 0.204)/2 = 0.102 L = 102 mL. It is not surprising that the same volume of the NaOH stock solution is needed because its molarity (2M) is twice that of the CuCl2
If a different gram quantity of Cu(OH)2
should be made, we just convert it to mol and calculate the volumes of the stock solutions to be used.
It is worth to reemphasize that molarity expresses the number of moles of a solute per volume
of solution. Why per volume, not mass, like for mass percent concentration? Because liquids are more convenient to measure by volume than by mass. The ease of accurately measuring volumes of solutions is the foundation for the broadly used important analytical method, called volumetric analysis