__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 12*M*. Most of the time, we use a very diluted amount such as 1.0*M* or 0.1*M*.

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 2.00*M* calcium chloride (CaCl_{2}) stock solution would you use to make 0.50L of 0.300*M* calcium chloride solution?

Given:

M_{1} = 2.00M M_{2}= 0.30M

V_{1} = ? V_{2} = 0.50L

Work:

2.0 (V_{1}) = (0.30)(0.50)

V_{1} = (0.3)(0.5)/(2.0)

V_{1} = 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:

M_{1} = 3.00M M_{2} = 1.25M

V_{1 }= ? V_{2 }= 0.300L

Work:

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

V_{1} = (1.25)(0.30)/(3.00)

V_{1} = 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:

M_{1} = 5.00M M_{2} = ??

V_{1 }= 0.50L V_{2} = 2.0L

Work:

(5.00)(0.5) = (M_{2})(2.0)

M_{2} = (5.00)(0.5)/(2.0)

M_{2} = 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.

### Topics Covered

- Describe and give examples of various types of solutions. Include: all nine possible types
- Describe the structure of water in terms of electronegativity and the polarity of its chemical bonds.
- Explain the solution process of simple ionic and covalent compounds, using visual, particulate representations and chemical equations. Include: crystal structure, dissociation, hydration
- Interpreting a a solubility curve of a pure substance in water and differentiate among saturated, unsaturated, and supersaturated solutions.
- Explain how a change in pressure affects the solubility of gases.
- Explain freezing-point depression and boiling-point elevation at the molecular level.
*Examples: antifreeze, road salt…*

- Include: grams per litre (g/L), % weight-weight (% w/w), % weight-volume (% w/v), % volume/volume (% v/v), parts per million (ppm), parts per billion (ppb), moles per litre (mol/L) (molarity)
- Prepare a solution, given the amount of solute (in grams) and the volume of solution (in millilitres), and determine the concentration in moles/litre. (Molarity)
- Solve problems involving the dilution of solutions.

Include: dilution of stock solutions, mixing common solutions with different volumes and concentrations