A standard solution is prepared by dissolving an accurately known mass of a pure substance in a known volume of solvent. The concentration is usually expressed in moles per cubic decimeter (mol/dm³), also called molarity (M).

Steps to Prepare a Standard Solution

1. Calculate the mass required using the formula:

   $\text{Mass} = \text{Molarity} \times \text{Volume} \times \text{Molar mass}$

2. Weigh the calculated mass accurately using an analytical balance

3. Transfer the solid to a clean beaker and dissolve in some distilled water

4. Stir until completely dissolved

5. Transfer the solution to a volumetric flask using a funnel

6. Rinse the beaker and funnel with distilled water, adding rinsings to the flask

7. Add distilled water up to the calibration mark

8. Stopper the flask and invert several times to ensure uniform concentration

The Molarity Formula

The number of moles of a substance in solution is calculated using:

$\text{Number of moles } (n) = \text{Molarity } (M) \times \text{Volume } (V)$

Remember to convert volume to dm³:

• 1 dm³ = 1000 cm³
• To convert cm³ to dm³, divide by 1000

Molar Ratio in Titrations

The balanced chemical equation shows the molar ratio between reactants. This ratio is essential for calculations.

For example, in the reaction:

$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$

The molar ratio is 1:2 (1 mole of Na₂CO₃ reacts with 2 moles of HCl).

The Titration Formula

At the equivalence point:

$M_A V_A = M_B V_B$

Where:

• $M_A$ = molarity of analyte
• $V_A$ = volume of analyte
• $M_B$ = molarity of titrant
• $V_B$ = volume of titrant

This formula applies when the reaction ratio is 1:1. When the ratio is different, include the molar ratio:

$\frac{M_A V_A}{n_A} = \frac{M_B V_B}{n_B}$

Problem: 25.0 cm³ of sodium hydroxide (NaOH) solution is neutralized by 22.5 cm³ of 0.10 M hydrochloric acid (HCl). Calculate the molarity of the NaOH solution.

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

Solution:

Step 1: Identify the molar ratio The equation shows 1:1 ratio between NaOH and HCl, so $n_A = n_B = 1$

Step 2: Apply the titration formula

$M_A V_A = M_B V_B$

$M_{\text{NaOH}} \times 25.0 = 0.10 \times 22.5$

$M_{\text{NaOH}} = \frac{0.10 \times 22.5}{25.0}$

$M_{\text{NaOH}} = 0.090 \text{ M}$

The molarity of the NaOH solution is 0.090 mol/dm³.

When a solution is diluted by adding more solvent, the number of moles remains constant. The relationship is:

$M_1 V_1 = M_2 V_2$

Where:

• $M_1$ = initial concentration
• $V_1$ = initial volume
• $M_2$ = final concentration
• $V_2$ = final volume

Dilution Example

Problem: How much distilled water must be added to 50 cm³ of 2.0 M HCl to prepare a 0.50 M HCl solution?

Solution:

$M_1 V_1 = M_2 V_2$

$2.0 \times 50 = 0.50 \times V_2$

$V_2 = \frac{2.0 \times 50}{0.50} = 200 \text{ cm}^3$

Volume of water to add = $200 - 50 = 150 \text{ cm}^3$

In Tanzania, volumetric analysis is used in water quality testing at municipal water stations in cities like Dar es Salaam and Arusha. Technicians use titration to determine the acidity or alkalinity of drinking water and to check if harmful substances like chlorine are present at safe levels. This ensures the water supplied to households is safe for consumption, directly protecting public health across communities.