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2.5: Preparing Solutions

Preparing a solution of known concentration is perhaps the most common activity in any analytical lab. The method for measuring out the solute and solvent depend on the desired concentration unit and how exact the solution’s concentration needs to be known. Pipets and volumetric flasks are used when a solution’s concentration must be exact; graduated cylinders, beakers and reagent bottles suffice when concentrations need only be approximate. Two methods for preparing solutions are described in this section.

2.5.1 Preparing Stock Solutions

A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid and diluting to a known volume. Exactly how this is done depends on the required concentration unit. For example, to prepare a solution with a desired molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring to the desired volume. To prepare a solution where the solute’s concentration is a volume percent, you measure out an appropriate volume of solute and add sufficient solvent to obtain the desired total volume.

Example 2.8

Describe how to prepare the following three solutions:

  1. 500 mL of approximately 0.20 M NaOH using solid NaOH;
  2. 1 L of 150.0 ppm Cu2+ using Cu metal; and
  3. 2 L of 4% v/v acetic acid using concentrated glacial acetic acid (99.8% w/w acetic acid).


(a) Since the concentration is known to two significant figures the mass of NaOH and the volume of solution do not need to be measured exactly. The desired mass of NaOH is

\[\mathrm{\dfrac{0.20\: mol\: NaOH}{L} \times \dfrac{40.0\: g\: NaOH}{mol\: NaOH} \times 0.50\: L = 4.0\: g}\]

To prepare the solution, place 4.0 grams of NaOH, weighed to the nearest tenth of a gram, in a bottle or beaker and add approximately 500 mL of water.

(b) Since the concentration of Cu2+ has four significant figures, the mass of Cu metal and the final solution volume must be measured exactly. The desired mass of Cu metal is

\[\mathrm{\dfrac{150.0\: mg\: Cu}{L} \times 1.000\: L = 150.0\: mg\: Cu = 0.1500\: g\: Cu}\]

To prepare the solution we measure out exactly 0.1500 g of Cu into a small beaker and dissolve using small portion of concentrated HNO3. The resulting solution is transferred into a 1-L volumetric flask. Rinse the beaker several times with small portions of water, adding each rinse to the volumetric flask. This process, which is called a quantitative transfer, ensures that the complete transfer of Cu2+ to the volumetric flask. Finally, additional water is added to the volumetric flask’s calibration mark.

(c) The concentration of this solution is only approximate so it is not necessary to measure the volumes exactly, nor is it necessary to account for the fact that glacial acetic acid is slightly less than 100% w/w acetic acid (it is approximately 99.8% w/w). The necessary volume of glacial acetic acid is

\[\mathrm{\dfrac{4\: mL\: CH_3COOH}{100\: mL} \times 2000\: mL = 80\: mL\: CH_3COOH}\]

To prepare the solution, use a graduated cylinder to transfer 80 mL of glacial acetic acid to a container that holds approximately 2 L and add sufficient water to bring the solution to the desired volume.

Exercise 2.5

Provide instructions for preparing 500 mL of 0.1250 M KBrO3.
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2.5.2 Preparing Solutions by Dilution

Solutions are often prepared by diluting a more concentrated stock solution. A known volume of the stock solution is transferred to a new container and brought to a new volume. Since the total amount of solute is the same before and after dilution, we know that

\[C_\ce{o} \times V_\ce{o} = C_\ce{d} \times V_\ce{d}   \tag{2.2}\]

where Co is the stock solution’s concentration, Vo is the volume of stock solution being diluted, Cd is the dilute solution’s concentration, and Vd is the volume of the dilute solution. Again, the type of glassware used to measure Vo and Vd depends on how exact the solution’s concentration must be known.


Equation 2.2 applies only when the concentration are written in terms of volume, as is the case with molarity. Using this equation with a mass-based concentration unit, such as % w/w, leads to an error. See Rodríquez-López, M.; Carrasquillo, A. J. Chem. Educ. 2005, 82, 1327-1328 for further discussion.

Exercise 2.6

To prepare a standard solution of Zn2+ you dissolve a 1.004 g sample of Zn wire in a minimal amount of HCl and dilute to volume in a 500-mL volumetric flask. If you dilute 2.000 mL of this stock solution to 250.0 mL, what is the concentration of Zn2+, in μg/mL, in your standard solution?
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Example 2.9

A laboratory procedure calls for 250 mL of an approximately 0.10 M solution of NH3. Describe how you would prepare this solution using a stock solution of concentrated NH3 (14.8 M).



Substituting known volumes in equation 2.2

\[\mathrm{14.8\: M \times \mathit{V}_o= 0.10\: M \times 0.25\: L}\]

and solving for Vo gives 1.69 × 10-3 liters, or 1.7 mL. Since we are making a solution that is approximately 0.10 M NH3 we can use a graduated cylinder to measure the 1.7 mL of concentrated NH3, transfer the NH3 to a beaker, and add sufficient water to give a total volume of approximately 250 mL.

As shown in the following example, we can use equation 2.2 to calculate a solution’s original concentration using its known concentration after dilution. 

Example 2.10

A sample of an ore was analyzed for Cu2+ as follows. A 1.25 gram sample of the ore was dissolved in acid and diluted to volume in a 250-mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipet to a 50-mL volumetric flask and diluted to volume. An analysis of this solution gave the concentration of Cu2+ as 4.62 μg/L. What is the weight percent of Cu in the original ore?



Substituting known volumes (with significant figures appropriate for pipets and volumetric flasks) into equation 2.2

\[\mathrm{( g/L\: Cu^{2+})_o \times 20.00\: mL = 4.62\: g/L\: Cu^{2+} \times 50.00\: mL}\]

and solving for (μg/L Cu2+)o gives the original solution concentration as 11.55 μg/L Cu2+. To calculate the grams of Cu2+ we multiply this concentration by the total volume

\[\mathrm{\dfrac{11.5{\color{Red} 5}\: g\: Cu^{2+}}{mL} \times 250.0\: mL \times \dfrac{1\: g}{10^6\: g} = 2.88{\color{Red} 8}\times10^{-3}\: g\: Cu^{2+}}\]

The weight percent Cu is

\[\mathrm{\dfrac{2.88{\color{Red} 8}\times 10^{-3}\: g\: Cu^{2+}}{1.25\: g\: sample} \times 100 = 0.231\%\: w/w\: Cu^{2+}}\]