Dilution Example Problems 1


A dilution is the process of adding solvent to a concentrated solution to create a new solution with less concentration.

In most laboratory settings, a stock solution is created when a compound is used over and over. This stock solution will have a high concentration. If lower concentrations are needed, a dilution is performed.

A dilution is a process where the concentration of a solution is lowered by adding solvent to the solution without adding more solute. These dilution example problems show how to perform the calculations needed to make a diluted solution.

The key idea behind a dilution is the number of moles of solute in the solutions does not change as the solvent is added.

moles of solute prior to dilution = moles solute after dilution

The concentration of a solution can be expressed in molarity (M).

M = moles per liter of solution
Dilution math 1
where M is the molarity and V is the concentration.

Solve for moles and get:

moles = MV

Since moles of solute prior to dilution = moles solute after dilution,

Mi = initial concentration
Vi = initial volume
MD = diluted concentration
VD = diluted volume

Example Problem 1:

Problem: What volume of 5 M NaOH is needed to create a 100 mL solution of 1 M NaOH?

Solution: Use the formula MiVi = MDVD.

Mi = 5 M
Vi = initial volume
MD = 1 M
VD = 100 mL

Solve for Vi
Dilution Math step
Dilution step 2
V= 20 mL

Answer: 20 mL of 5 M NaOH is needed to create a 100 mL solution of 1 M NaOH. Water is added to the 20 mL solution until there is 100 mL.

Example Problem 2:

Problem: If you have 300 mL of 1.5 M NaCl, how many mL of 0.25 M NaCl can you make?

Solution: Use the formula MiVi = MDVD.

Mi = 1.5 M
Vi = 300 mL
MD = 0.25 M
VD = final volume

Solve for VD
Dilution algebra step

Dilution Math 3
VD = 1800 mL = 1.8 L

Answer: You can make 1800 mL of 0.25 M NaCl solution from 300 mL of 1.5 M NaCl solution.


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