# Solving Dilution Equation Problems

Solving Equations
Dr. MJ Patterson

The dilution equation is one of the the first examples we have encountered of a mathematical equation used to describe a chemical process.  You are expected to be able to solve for any one of the variables in this equation if given values for all of the rest.  The following procedure will help you to solve equations like this.

1. Make a variable list.  Write out all of the variables in the equation and figure out which ones you have been given values for.  This step will help you identify what to solve for, as it will be the only variable without a numerical value.  Be sure to include units in this list.
2. Check units.  Make sure that the units on all of the variables are compatible.
3. Plug into the equation.  Take all of the values and plug into the equation in place of each variable.
4. Solve.  We want to isolate the unknown.  To get it by itself, you have to perform the opposite of any mathematical operations to both sides of the equation.  If the unknown is multiplied by 6, divide both sides by 6.  If the unknown is divided by pi, multiply both sides by pi.

Example:
A 125 mL sample of a14.0 M solution is diluted to 500 mL.  What is the concentration of the final solution?

Solution:
Since this is a dilution problem, the equation we need is:

M1V1 = M2V2

There are 4 variables, so we need to make a list of each variable and the value we have for it.  For anything that is unknown, just put a question mark for its value.  It is conventional to use the subscript 1 for the undiluted, or starting, solution, and the subscript 2 for the final or diluted solution.  Just keep in mind that every variable with a subscript 1 needs to refer to the same solution, and every variable with a subscript 2 also needs to refer to the same solution.

Variable list:
M1 = 14.0 M
V1 = 125 mL
M2 = ?
V2 = 500 mL

From this variable list we can see that there is one variable that is unknown, M2.  We can plug these values into the dilution equation, and solve for the unknown M2.

M1V1 = M2V2
(14.0 M)(125 mL) = (M2)(500 mL)

Since the units mL are on both sides, they will cancel out.

(14.0 M)(125 mL) = (M2)(500 mL)
(14.0 M)(125) = (M2)(500)

Since the unknown M2 is multiplied by 500, we need to do the opposite operation to both sides of the equation to solve for M2.  In other words, divide both sides of the equation by 500.

(14.0 M)(125)/500 = (M2)(500)/500

The 500's cancel on the right.  Using a calculator on the left hand side, we are left with:

3.5 M = M2

This means that after the solution is diluted, the concentration will be 3.5 M.