Biology Department

Furman University

Greenville, South Carolina


Active Learning from the Very Beginning
Keri Law, Rachel Lamb, and Min-Ken Liao


Time for a brief general chemistry review. The solutions you need for your research including antibiotics, buffers, etc., usually come in concentrations that must be altered according to the amount you need for the specific experiment you are performing. For example, a common solution used in our research is a phosphate buffer. It is used to wash bacterial cells and to provide a buffering condition in which the cells can carry out chemical reactions. The following is our "recipe" for the buffer solution:

434 mL of 0.2 M monobasic potassium phosphate (KH2PO4)
66 mL of 0.2 M dibasic potassium phosphate (K2HPO4)
500 mL distilled water

As with the other media and equipment involved in growing bacteria, the buffer solution must be autoclaved to ensure sterility. It can then be stored at room temperature.

The 0.2 M monobasic and 0.2 M dibasic potassium phosphate solutions here are referred to as the "stock solutions." When they are mixed together as instructed, the resulting phosphate buffer is known as the "working solution."

The "0.2 M" is the initial concentration or stock concentration. The resulting concentration is the working solution and is referred to as the final or working concentration. In most research articles, the final concentrations, not the stock concentrations, are given by the authors.


The monobasic potassium phosphate solution can be made by adding the appropriate amount of potassium phosphate to a certain volume of water necessary to achieve the desired concentration. The molecular weight of monobasic potassium phosphate is 136.09 g/mol. You can do some simple calculation and determine this value. However, there is an easier way to obtain this number: It is provided by the supplier and is printed on the side of the bottle. A 0.2 M solution has 0.2 moles of potassium phosphate in 1.0 L water. (0.2 mol/L x 1.0 L = ____mol potassium phosphate). Therefore in order to find how many grams of monobasic potassium phosphate to add to 1.0 L of water to make a 0.2 M solution, multiply the molecular weight of KH2PO4 by 0.2 moles. (136.09 g/mol x .2 mol = ______ grams).

The dibasic potassium phosphate solution can be prepared in the same manner only using the different molecular weight in the calculation. Begin by looking up this molecular weight. (_________g/mol). Again a 0.2 M solution is desired, so find how many moles of dibasic potassium phosphate this concentration requires (0.2 mol/L x 1.0 L = ______mol). Now find how many grams you need to make 1.0 L of a 0.2 M solution. Now we have our stock solutions.


If you are asked to find the final concentration of the working solution, you may be tempted to simply add the concentrations of the two stock solutions, 0.2 M + 0.2 M. However, you must also take into account the 500 ml of water added to dilute the working solution. A valuable formula to use when making such calculations is C1 x V1 = C2 x V2, where C1 and C2 are the initial and final concentrations of a particular stock solution, V1 is the initial volume of the stock solution, and V2 is the final volume of the working solution. For example, let's determine the final concentration of monobasic potassium phosphate in our working solution.

C1 = 0.2 M
V1 = 434 mL
V2 = 1000 mL (434 mL KH2PO4 + 66 mL K2HPO4 + 500mL dH2O)
C2 = (0.2 M x .434 L)/1.0 L = _______M

To ensure your understanding of this concept, perform the following exercises (only on paper):

1) Prepare a .3 M stock solution of monobasic potassium phosphate and a 0.4 M stock solution of dibasic potassium phosphate.

2) Find the final concentrations of the monobasic and dibasic potassium phosphate solutions in a working solution made by combining 300 mL .3 M KH2PO4, 125 mL .4 M K2HPO4, and 600 mL water.

3) A second stock solution often employed in the lab is an antibiotic such as Ampicillin. The concentration of our solution is 9 mg/mL. When preparing rich media, a concentration of 30 mg/mL is needed. When preparing a minimal media, the concentration of antibiotic must be 15 mg/mL. (m = micro, 10^-6)

a) What volume of antibiotic must be added to enough rich media to make 50 mL of a 30 mg/mL solution?

b) What volume of antibiotic must be added to enough minimal media to make 1.0 L of a 15 mg/mL solution?