### Diluting Molar Solutions

In the laboratory, you might use concentrated solutions of standard molarities, called stock solutions. These are very concentrated solutions that must be diluted before using.

An example is concentrated Hydrochloric acid (HCl) is 12M. Most of the time, we use a very diluted amount such as 1.0M or 0.1M.

Diluting molar solutions simply mean that we are “spreading” out the solutes with more solvent.

When we dilute a solution, the amount of solvent increases, while the amount of solute stays the same.

### Example 1:

What volume, in milliliters, of 2.00M calcium chloride (CaCl2) stock solution would you use to make 0.50L of 0.300M calcium chloride solution?

Given:

M1 = 2.00M     M2= 0.30M
V1 = ?              V2 = 0.50L

Work:

2.0 (V1) = (0.30)(0.50)

V1 = (0.3)(0.5)/(2.0)

V1 = 0.075L but the question asks for milliliters, therefore x 1000 = 75ml

What does this mean?

**To make the dilution, 75ml must be measured out from the stock solution and then dilute it enough water to make it 0.50L . ** This is important for them to understand why, instead of just the math. Draw a picture if you can.

### Example 2:

What volume of a 3.00M KI stock solution would you use to make 0.300L of a 1.25M KI solution?

Given:

M1 = 3.00M                 M2 = 1.25M
V1 = ?                           V2 = 0.300L

Work:

(3.00)(V1) = (1.25)(0.30)

V1 = (1.25)(0.30)/(3.00)

V1 = 0.125L

What does this mean? 125ml of the 3.00 stock solution is taken out, and diluted by adding enough water to make it to 0.300L. This solution will give a concentration of 1.25M

### Example 3:

If 0.50L of 5.00M stock solution of HCl is diluted to make 2.0L of solution, how much HCl, in grams is in the solution?

Given:

M1 = 5.00M                 M2 = ??
V1 = 0.50L                    V2 = 2.0L

Work:

(5.00)(0.5) = (M2)(2.0)
M2 = (5.00)(0.5)/(2.0)

M2 = 1.25M

Now to grams, we first need to find the moles.

M = n/L                       1.25 = n / 2.0

Number of moles = 2.5 moles of HCl
To grams = 2.5 moles of HCl x ~36g/mol = 90g of HCl will be in the diluted solution.