Dilution Calculations

by Todd A. Carlson, Grand Valley State University, Allendale, MI 49401

Dilution calculations are common in all areas of chemistry, particularly biochemistry. They also prove to be particularly difficult for many students. You may find this worksheet helpful as you work through some of the problems you will be doing this semester. Consider the following samples problems:

A protein solution is prepared by dissolving 40.0 mg of protein in 10.0 mL of water. A 0.500 mL sample of this solution is diluted to a volume of 20.0 mL. How many mg of protein will be in a 2.00 mL sample of the diluted solution? This a relatively complex dilution calculation. The difficulty of dilution calculations comes from the fact that all problems are a little bit different. Therefore, you can not learn how do solve these types of problems by memorizing a single equation or by practicing one type of problem. However, all dilution problems can be solved by applying three simple rules:
• Rule 1: The concentration of a solution is the amount of solute in the solution divided by the total volume of the solution.
• Rule 2: When a solution is diluted, the amount of solute in solution remains constant.
• Rule 3: A sample of a solution has the same concentration as the original solution.
• Rule 1 is the definition of volume based concentration units. Rule 2 is a statement of the law of conservation of matter. Rule 3 is derived from the definition of a solution (a liquid is a homogeneous mixture). This approach to solution calculations is more versatile than memorizing equations because the rules are based on fundamental basics of chemistry. By applying these rules, you will understand the process and chemistry of dilutions and solutions.

The definition given in Rule 1 provides the basic equation for calculating concentration and the two variations used to calculate volume and amount.

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This approach to dilution calculations is based on the fact that all problems are based on very simple fundamentals outlined in Rules 1 through 3 The difficulty lies in the organization of the problem. Rules 1 through 3 are applied using a dilution chart. The chart provides visual clues to help you organize the problem. The chart has three rows (one each for concentration, amount and volume) and a column for each different solution in the problem. In this example there are four solutions: the original solution, a sample of the original solution, the diluted solution, and a sample of the dilution. Identifying the different solutions that need to be analyzed is probably the most difficult but also the most important aspect of the problem. All of the values given in the problem are written in the appropriate box of the chart. The box for the desired answer should be highlighted. The initial dilution chart for the sample problem above is shown in Table 1. The three rules above are then applied as shown in the example shown below. Table 2 shows the completed dilution chart with arrows indicating the procession of steps.

This procedure may seem tedious at first. There are short-cut methods that will give the answer with less work. If you can get the right answer using a different approach, then this chart method may not be necessary for you. However, all solution to this problem will ultimately use the same calculations organized differently. With practice, many students find that they can do these types of problems without the chart as the concepts of dilution calculations become intuitive. You do not have to use this method. However, I have found that for many students, the primary obstacle in problem solving is organization. If you find yourself saying "I don't know where to begin" when solving problems, then this method may help.

A Step By Step Solution To The Sample Problem:

A protein solution is prepared by dissolving 40.0 mg of protein in 10.0 mL of water. A 0.500 mL sample of this solution is diluted to a volume of 20.0 mL. How many mg of protein will be in a 2.00 mL sample of the diluted solution?

1. Begin by preparing a chart which includes all the information given in the problem (Table 1).
2. Use Rule 1 to calculate the concentration in the first column. The chart provides a visual clue for the application of the Rule: Any time a column has two numbers filled in, you can calculate the third.
3. Since column 2 is a sample of column 1, then (according to Rule 3) the concentrations of columns 2 and 1 must be the same.
4. Use Rule 1 to calculate mg of protein in column 2.
5. Since column 3 is a dilution of column 2, then (according to Rule 2) the amounts of protein in columns 2 and 3 must be the same.
6. Use Rule 1 to calculate the concentration of column 3.
7. The concentrations of column 3 and 4 are the same (Rule 3).
8. The final answer is calculated with Rule 1.
Table 1: The starting chart for the sample problem.  The highlighted box indicates the desired answer.

 original solution sample of original diluted solution sample of dilution concentration (mg/mL) amount (mg) 40.0 volume (mL) 10.0 0.500 20.0 2.00

Table 2: The completed chart for the sample problem.  The arrows indicate the progression of the calculation.

 original solution sample of original diluted solution sample of dilution concentration (mg/mL) 4.00 ® 4.00 ¯ 0.100 ® 0.100 ¯ amount (mg) 40.0 ­ 2.00 ® 2.00 ­ 0.200 volume (mL) 10.0 0.500 20.0 2.00