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,

M_{i}V_{i} = M_{D}V_{D}

where:

M_{i} = initial concentration

V_{i} = initial volume

M_{D} = diluted concentration

V_{D} = 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 M_{i}V_{i} = M_{D}V_{D}.

M_{i} = 5 M

V_{i} = initial volume

M_{D} = 1 M

V_{D} = 100 mL

Solve for V_{i}V_{i }= 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 M_{i}V_{i} = M_{D}V_{D}.

M_{i} = 1.5 M

V_{i} = 300 mL

M_{D} = 0.25 M

V_{D} = final volume

Solve for V_{D}

V_{D} = 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.