EXAMPLE Making Solutions from More Concentrated Solutions: 

As stockroom technician for a college chemistry department, you are asked to make the solutions for an experiment that calls for each student to use 15.0 mL of 1.50 M HCl. The lab class has 22 students and you want to have some extra solution available, so you decide to make 500 mL of 1.50 M HCl. How would you make this solution from a hydrochloric acid solution that is 12.0 M HCl?

See the solution below.


We are given a lot of information in this problem, so it might be a good idea to write a list of the data that we think will be necessary for producing the answer. This is a good way of separating essential from inessential information and focusing on the tools we will use to work the problem. We want to make 500 mL of 1.50 M HCl solution from 12.0 M HCl. We will use the following expressions in our calculation:


Each molarity is written as a ratio, with the units clearly identified to the point of distinguishing pure solute from solution and dilute solution from concentrated solution. Because milliliters are mentioned in the problem, we write our molarity ratios in the form that has milliliters instead of liters.

The first step in our procedure is to measure a specific volume of concentrated acid, but we are not told how much to measure. Thus, we begin by calculating the volume of 12.0 M HCl needed to make 500 mL of 1.50 M HCl.

There are two general approaches to working this problem. We will first work it using a dimensional analysis format, but you will see that there is also a simple shortcut.

                           = 62.5 mL conc HCl solution

Now that we know what volume of HCl well need, we follow these steps to make the solution.

  • Carefully add 62.5 mL of 12.0 M HCl to a 500-mL volumetric flask.
  • Add water to the flask until the total volume reaches the 500-mL mark on the flask.

Click here to see a sample study sheet for this task.

Click here to see an exercise that will allow your to try this task yourself.

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