Question 1 In a titration experiment, \( 21.60 \mathrm{~cm}^{3} \) of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) (aq) was required to completely react with 25.0 \( \mathrm{~cm}^{3} \) of NaOH (aq) containing 2.00 g of the alkali in \( 500 \mathrm{~cm}^{3} \) of solution. a. Write a balanced equation for the reaction. b. Calculate the concentration of: i. \( \mathrm{NaOH}(\mathrm{aq}) \) in moldm

The balanced equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is: \[ \text{H}_2\text{SO}_4 (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \] This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce sodium sulfate and water. This stoichiometric relationship is crucial for determining the concentrations in your titration! Now, for the concentration calculation, start by finding the moles of NaOH. Given that 2.00 g of NaOH is present in 500 cm³ of solution, use the molar mass of NaOH (approximately 40 g/mol) to find: \[ \text{Moles of NaOH} = \frac{2.00 \text{ g}}{40 \text{ g/mol}} = 0.050 \text{ mol} \] Next, since this is in 500 cm³ (or 0.500 dm³) of solution, the concentration of NaOH is calculated as: \[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume in dm}^3} = \frac{0.050 \text{ mol}}{0.500 \text{ dm}^3} = 0.100 \text{ mol/dm}^3 \] So, the concentration of NaOH is \( 0.100 \, \text{mol/dm}^3 \). Happy titrating!