Moles, Solutions, and Masses

This section explains moles, solutions and masses, covering the concentration of solutions, acid-alkali titrations and the titrations formula.

Concentration of Solutions

Concentration in g/dm³

Concentration refers to how much solute (substance) is dissolved in a given volume of solution. It is measured in grams per cubic decimetre (g/dm³) and calculated using the formula:

$$\text{Concentration (g/dm}^3) = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (dm}^3)}$$ 

For example, if 10 g of sodium chloride (NaCl) is dissolved in 2 dm³ of water, the concentration is:

$$\frac{10}{2} = 5 \text{ g/dm}^3$$

Concentration in mol/dm³

Chemists often express concentration in terms of moles per cubic decimetre (mol/dm³), using the formula:

$$\text{Concentration (mol/dm}^3) = \frac{\text{Moles of solute (mol)}}{\text{Volume of solution (dm}^3)}$$ 

To convert between g/dm³ and mol/dm³, use the molar mass of the solute:

$$\text{Concentration (mol/dm}^3) = \frac{\text{Concentration (g/dm}^3)}{\text{Molar mass (g/mol)}}$$ 

For example, if the concentration of NaCl is 5 g/dm³, and the molar mass of NaCl is 58.5 g/mol, then:

$$\frac{5}{58.5} = 0.0855 \text{ mol/dm}^3$$ 

Acid-Alkali Titrations

What is a Titration?

Titrations are experiments used to determine the unknown concentration of an acid or alkali. A solution of known concentration (the titrant) is added to the unknown solution until neutralisation occurs. This point is identified using an indicator (e.g., phenolphthalein or methyl orange).

Titration Formula

The relationship between acid and alkali concentrations is given by:

$$\text{Moles of acid} = \text{Moles of alkali}$$ 

Since moles can be calculated using:

$$\text{Moles} = \text{Concentration (mol/dm}^3) \times \text{Volume (dm}^3)$$ 

We can use:

$$C_1 V_1 = C_2 V_2$$

Where:

• $C_1$​ = concentration of acid (mol/dm³)
• $V_1$​ = volume of acid (dm³)
• $C_2$​ = concentration of alkali (mol/dm³)
• $V_2$​ = volume of alkali (dm³)

Example Titration Calculation

A student titrates 25.0 cm³ of sodium hydroxide (NaOH) with an unknown concentration against 0.100 mol/dm³ of hydrochloric acid (HCl). The acid required 22.5 cm³ to neutralise the alkali. The balanced equation is:

$$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$ 

Since 1 mole of HCl reacts with 1 mole of NaOH, we apply:

$$C_1 V_1 = C_2 V_2$$

$$(0.100) \times (22.5/1000) = C_2 \times (25.0/1000)$$

$$0.00225 = C_2 \times 0.0250$$

$$C_2 = \frac{0.00225}{0.0250} = 0.0900 \text{ mol/dm}^3$$ 

Thus, the concentration of NaOH is 0.0900 mol/dm³.

Key Points to Remember for Titrations

• Use a pipette to measure the alkali and a burette to add the acid.
• Use a suitable indicator: 
• Phenolphthalein: Pink in alkali, colourless in acid.
• Methyl orange: Yellow in alkali, red in acid.
• The end point is when the indicator just changes colour.
• Perform multiple trials and take a mean for accuracy.

This guide provides a solid foundation for understanding quantitative chemistry related to solutions, concentration, and titrations.