Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution

Stoichiometry – The Mole and the Avogadro Constant

1️⃣ What is a Mole?

A mole is a way of counting atoms, ions or molecules. Think of it as a basket that can hold exactly \$6.022 \times 10^{23}\$ items. Just like a basket of 12 apples can be used to count apples, a mole lets chemists count tiny particles.

2️⃣ Avogadro Constant

The Avogadro constant, \$N_A\$, is the number of entities in one mole:

\$N_A = 6.022 \times 10^{23}\,\text{mol}^{-1}\$

It links the microscopic world (atoms) with the macroscopic world (grams).

3️⃣ Titration Basics

A titration is a laboratory technique where a solution of known concentration (the titrant) is added to a solution of unknown concentration (the analyte) until the reaction is complete. The point at which the reaction is finished is called the equivalence point.

• 🔬 Reactants: The substances that react.
• ⚖️ Stoichiometry: The ratio of reactants in a balanced equation.
• 📏 Volume measured: Usually with a burette.

4️⃣ Titration Data Table

5️⃣ Calculating Moles from Titration Data

The basic formula is:

\$n{\text{solute}} = C{\text{titrant}} \times V_{\text{titrant}}\$

Where:

• \$n_{\text{solute}}\$ = moles of the unknown solution.
• \$C_{\text{titrant}}\$ = concentration of the titrant (mol L⁻¹).
• \$V_{\text{titrant}}\$ = volume of titrant used (L).

If the reaction is not 1:1, multiply by the stoichiometric coefficient.

6️⃣ Example Problem

A student titrates 10.00 mL of a hydrochloric acid (HCl) solution with 0.100 M sodium hydroxide (NaOH). The titrant volume at the equivalence point is 25.00 mL.

1. Convert volumes to litres:

   • \$V_{\text{NaOH}} = 25.00\,\text{mL} = 0.02500\,\text{L}\$
   • \$V_{\text{HCl}} = 10.00\,\text{mL} = 0.01000\,\text{L}\$

2. Use the 1:1 stoichiometry of the reaction:

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

3. Calculate moles of NaOH (same as moles of HCl):

   \$n{\text{NaOH}} = C{\text{NaOH}} \times V_{\text{NaOH}} = 0.100\,\text{mol L}^{-1} \times 0.02500\,\text{L} = 0.00250\,\text{mol}\$

4. Determine concentration of HCl:

   \$C{\text{HCl}} = \frac{n{\text{HCl}}}{V_{\text{HCl}}} = \frac{0.00250\,\text{mol}}{0.01000\,\text{L}} = 0.250\,\text{mol L}^{-1}\$

5. Check with Avogadro’s number: 0.00250 mol × \$6.022\times10^{23}\$ = \$1.5055\times10^{21}\$ HCl molecules.

So the unknown HCl solution is 0.250 M.

7️⃣ Summary & Quick Tips

• 🧮 Mole concept: 1 mol = \$6.022\times10^{23}\$ entities.
• 📐 Stoichiometry: Use balanced equations to find ratios.
• 📊 Titration data: Volume × concentration = moles.
• 🔁 Check units: L for volume, mol L⁻¹ for concentration.
• 💡 Remember: If the reaction is not 1:1, adjust with the coefficient.

Practice with different titration scenarios and you’ll master stoichiometry in no time! 🚀