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
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,
MiVi = MDVD
where:
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
Vi = 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
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.