# Preparing Chemical Solutions

Lab experiments and types of research often require preparation of chemical solutions in their procedure. We look at preparation of these chemical solutions by weight (w/v) and by volume (v/v). The glossary below cites definitions to know when your work calls for making these and the most accurate molar solutions.

To this we add information designed for understanding how to use the pH scale when measuring acidity or alkalinity of a solution.

### Glossary, basic terms to understand...

**Solute** - The substance which dissolves in a solution

**Solvent** - The substance which dissolves another to form a solution. For example, in a sugar and water solution, water is the solvent; sugar is the solute.

**Solution** - A mixture of two or more pure substances. In a solution one pure substance is dissolved in another pure substance homogenously. For example, in a sugar and water solution, the solution has the same concentration throughout, ie. it is homogenous.

**Mole** - A fundamental unit of mass (like a "dozen" to a baker) used by chemists. This term refers to a large number of elementary particles (atoms, molecules, ions, electrons, etc) of any substance. 1 mole is 6.02 x 10^{23} molecules of that substance. (Avogadro's number).M

### Introduction to preparation of solutions.

Many experiments involving chemicals call for their use in solution form. That is, two or more substances are mixed together in known quantities. This may involve weighing a precise amount of dry material or measuring a precise amount of liquid. Preparing solutions accurately will improve an experiment's safety and chances for success.

### Solution 1: Using percentage by weight (w/v)

### Formula

The formula for weight percent (w/v) is: [Mass of solute (g) / Volume of solution (ml)] x 100

### Example

A 10% NaCl solution has ten grams of sodium chloride dissolved in 100 ml of solution.

### Procedure

Weigh **10g** of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about **80ml** of water. Once the sodium chloride has dissolved completely (swirl the flask gently if necessary), add water to bring the volume up to the final 100 ml. Caution: Do not simply measure **100ml** of water and add 10g of sodium chloride. This will introduce error because adding the solid will change the final volume of the solution and throw off the final percentage.

### Solution 2: Using percentage by volume (v/v)

When the solute is a liquid, it is sometimes convenient to express the solution concentration as a volume percent.

### Formula

The formula for volume percent (v/v) is: [Volume of solute (ml) / Volume of solution (ml)] x 100

### Example

Make 1000ml of a 5% by volume solution of ethylene glycol in water.

### Procedure

First, express the percent of solute as a decimal: 5% = 0.05

Multiply this decimal by the total volume: 0.05 x 1000ml = 50ml (ethylene glycol needed).

Subtract the volume of solute (ethylene glycol) from the total solution volume:

1000ml (total solution volume) - 50ml (ethylene glycol volume) = 950ml (water needed)

Dissolve **50ml** ethylene glycol in a little less than **950ml** of water. Now bring final volume of solution up to **1000ml** with the addition of more water. (This eliminates any error because the final volume of the solution may not equal the calculated sum of the individual components).

So, 50ml ethylene glycol / 1000ml solution x100 = 5% (v/v) ethylene glycol solution.

### Solution 3: Molar Solutions

Molar solutions are the most useful in chemical reaction calculations because they directly relate the moles of solute to the volume of solution.

### Formula

The formula for molarity (M) is: moles of solute / 1 liter of solution or gram-molecular masses of solute / 1 liter of solution.

### Examples

The molecular weight of a sodium chloride molecule (NaCl) is 58.44, so one gram-molecular mass (=1 mole) is 58.44 g. We know this by looking at the periodic table. The atomic mass (or weight) of Na is 22.99, the atomic mass of Cl is 35.45, so 22.99 + 35.45 = 58.44.

If you dissolve **58.44g** of NaCl in a final volume of **1 liter**, you have made a **1M NaCl** solution, a 1 molar solution.

### Procedure

To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:

**0.1M NaCl** solution requires **0.1 x 58.44 g of NaCl** = **5.844g**

**0.5M NaCl** solution requires **0.5 x 58.44 g of NaCl** = **29.22g**

**2M NaCl **solution requires **2.0 x 58.44 g of NaCl** = **116.88g**