Preparing Aqueous Solutions of a Given Molarity 

Molarity is a useful way of describing the concentrations of solutes in solution, as well as a source of useful conversion factors for converting between volume (an easily measurable property of solutions) and moles of solute.

Now we will see how solutions of a specific molarity can be made.

Imagine that you are preparing to do an experiment that calls for two solutions, 0.500 M Na2CO3 and 1.50 M HCl. The typical chemistry stockroom would probably not store solutions of these kinds, so you would need to make them up from the stock thats available. The stockroom probably has pure sodium carbonate, so the 0.500 M Na2CO3 solution can be made from pure Na2CO3 and water. The stockroom would almost certainly not have pure hydrogen chloride, however, because it is a gas that is difficult and dangerous to store. Instead, you would be likely to find a concentrated hydrochloric acid solution that is 12 M HCl. This solution can be diluted to 1.50 M HCl.

The discussion above suggests that there are two general ways to make solutions of a given molarity:

  • The solutions can be made starting with pure solute and water. 
  • The solutions can be made from more concentrated solutions. 

Making Solutions from Pure Solids

The following steps describe the procedure for making a solution of a specific molarity from a pure, solid substance.

  • First, weigh out the correct mass of solute. 
  • Dissolve the solute in water, keeping the volume less than the desired total volume of solution.
  • Dilute the solution to the desired total volume of solution.

Click here to see Example 1.

Click here to see a sample study sheet for problems that ask you to describe how to make an aqueous solution starting with a pure solid.

Click here to see Exercise 1.

Making Solutions from More Concentrated Solutions

Because it is not practical for the typical lab to make hydrochloric acid solutions from pure HCl, a chemist is more likely to order hydrochloric acid solution rather than pure hydrogen chloride from a chemical supply company. The hydrochloric acid solutions a chemist is likely to use in reactions have concentrations of 1 M HCl or 6 M HCl, but solutions of this concentration would take up an inconvenient amount of storage space in a chemistry stockroom. Therefore, a concentrated hydrochloric acid solution, usually about 12 M HCl, is purchased and stored, and diluted with water whenever a more dilute solution is needed. The concentrations of typical concentrated acid solutions and for the base ammonia are listed in the table below.

Typical Concentrations of Concentrated Acids and Ammonia

Substance

Molarity

Mass Percentage

Hydrochloric acid, HCl(aq)

12.1 M

38.7%

Sulfuric Acid, H2SO4

17.8 M to 18.4 M

95%-98%

Nitric Acid, HNO3

15.8 M

69.71%

Acetic acid, HC2H3O2

17.4 M

99.7%

Ammonia, NH3(aq)

14.8 M

28%

The following is the general procedure for diluting a relatively concentrated solution to form one that is more dilute. (There is a slightly different procedure for diluting acids that are almost pure, like concentrated sulfuric acid, which is 95-98% H2SO4. We will restrict our examples to the dilution of solutions that already contain a significant amount of water.)

  • Add the correct volume of the more concentrated solution to a volume‑measuring instrument, like a volumetric flask. (See Example 15.7 for an example of how this volume is calculated.)
  • Add water until the volume reaches the desired total.

Click here to see Example 2.

As was mentioned in Example #2, a shortcut can be used for solving problems in which a relatively concentrated solution must be diluted to produce a solution that is less concentrated. When a more concentrated solution is diluted by mixing with water, the amount of solute does not change. Thus, the number of moles of solute in the concentrated solution is equal to the number of moles of solute in the diluted solution.

#moles solute in concentrated solution  =  #moles solute in dilute solution

Moles of solute can be calculated by multiplying the volume of solution times the molarity of solution.

or 

 

or  Vconc soln Mconc soln = Vdil soln Mdil soln  

This knowledge leads to an easily memorized equation that we can use for solving problems of this type. We will call these problems dilution problems.

MCVC = MDVD   

MC = molarity of the more concentrated solution

VC = volume of the more concentrated solution

MD = molarity of the more dilute solution

VD = volume of the more dilute solution

If you know this equation and can recognize the situations to which it applies, it can save you quite a lot of time. For example, it can be used instead of dimensional analysis to work Example #3.

Click here to see an alternative solution to Example 2.

a sample study sheet for dilution problems.

Click here to see Example 3.

Click here to see Exercise 2.

Click here to see Example 4.

Click here to see a general sample study sheet for problems that ask you to describe how to make solutions of a specific molarity.

Click here to see Exercise 3.

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