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Katie Owens
Product Management Coordinator

March 2017

1. How do I make ___ mL of a ___ molar solution?

Ex:  How do I make 100 mL of a 3M aqueous sodium hydroxide solution?

Determine molarity required: M = 3M solution = 3 moles NaOH/1 L of solution

Determine molar mass: MM of NaOH = 40 g/mol

Convert volume required to liters: V = 100 mL x 1 L/1000 mL = 0.1 L

Grams required = V x M x MMNaOH

0.1 L soln x 3 moles NaOH/1 L soln x 40 g NaOH/1 mole NaOH = 12 g NaOH

Prepare the solution:

### Key terms

Molarity: Concentration of a solution as the number of moles of solute per liters of solution.

Solution: A homogenous mixture composed of two or more substances.

Solute: A substance that has been dissolved into a solution.

Solvent: A substance that dissolves a solute to form a solution (typically a liquid).

Concentration: The relative amount of solute and solvent within a solution.

Molar Mass: The mass in grams per one mole of an element or compound.

Volume Percent Solution: The concentration of a solution as milliliters of solute per 100 mL of solution.

Hydrate: An inorganic salt that contains water molecules as part of its structure or crystallization.

Miscible: The ability of substances to form a homogenous mixture when added together.

1. Select a 100-mL flask, preferably volumetric.
2. Weigh out 12 grams of sodium hydroxide.
5. Mix until dissolved using a stirring bar and stir plate or by swirling gently.
6. Add water up to the 100-mL volume mark on the glassware and mix again.

2. How do I dilute a solution?

Ex: How do I make a 500-mL solution of 3M HCl acid from 12 M HCl acid?

M1= Initial Molarity = 12 M
V1= Initial Volume = ?
M2= Final Molarity = 3 M
V2= Final Volume = 500 mL

Calculate the volume of initial solution necessary for the dilution:

M1V1 = M2V2
(12 M)(V1) = (3 M)(500 mL)
V1 = 125 mL

125 mL of 12 M HCl acid is required to make 500 mL of 3 M HCl acid.

Prepare the solution:
Using a 500-mL volumetric flask and graduated cylinder, add 125 mL of concentrated acid to 250 mL of water. Make sure to stir slowly and then add remaining water to reach a final volume of 500 mL.

Note: If you need to prepare a very dilute solution from a stock solution, you may need to perform more than one dilution. By creating a more manageable and less concentrated solution, the final molarity will be more accurate. See our infographic on performing serial dilutions.

3. How do I convert to molarity from a percent concentration?

Ex: What is the molarity of a 10% potassium hydroxide solution?

100 mL of 10% KOH solution contains 10 g of potassium hydroxide.
MM of KOH = 56 g/mol

10 g KOH x 1 mol KOH/56 g KOH = 0.18 moles KOH

0.18 moles KOH/100 mL soln x 1000 mL soln/1 L soln = 1.8 moles KOH/L of soln = 1.8 M soln

### Teacher tips

1. Hydrates

Consider the formula weight of hydrates when calculating the concentration of solutions. The formula weight of hydrates will be higher due to the additional molecules of water. The attached water molecules contribute water to the final solution.

2. Always add acid to water—never the reverse.

Mixing strong acids and water causes an exothermic reaction. If you add water to acid, you form a very concentrated acid initially. The reaction gives off so much heat that the solution could boil out of the container. When you add acid to water, the solution slowly becomes more concentrated. If it does splatter, there will be more water and less chance of contact with acid.

3. Stock Solutions

Keeping stock solutions can conserve money and space! If needed, a stock solution can easily be diluted into a working solution of a desired concentration. Carolina offers a large variety of chemical grades, sizes, and concentrations.

4. Personal Protective Equipment

Always use proper personal protective equipment (PPE) when working with chemicals. Refer to SDS for safe usage, storage, and handling instructions. Common PPE for the lab includes:
• Safety glasses or goggles
• Lab coats and aprons
• Gloves (based on material handled)
• Face shield

5. Distilled or deionized?

Deionized
water has been treated to remove inorganic charged particles (ions). The process does not remove molecules or biological contaminants.

Distilled
water is free of almost all inorganic minerals, chemicals, and biological contaminants.

For most laboratory applications, deionized water is sufficient.