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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

Ask by Chang Black. in Nigeria
Jan 24,2025

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**Problem 1** - **a. Balanced Equation:** \[ \mathrm{H}_2\mathrm{SO}_4(aq) + 2\, \mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_2\mathrm{SO}_4(aq) + 2\, \mathrm{H}_2\mathrm{O}(l) \] - **b.i. Concentration of \( \mathrm{NaOH} \):** 0.100 mol dm⁻³

Solution

**Problem 1** In the titration experiment, we are given the following information: - **Volume of \( \mathrm{H}_2\mathrm{SO}_4 \)** required: \( 21.60 \, \mathrm{cm}^3 \) - **Volume of \( \mathrm{NaOH} \)** solution titrated: \( 25.0 \, \mathrm{cm}^3 \) - **Mass of alkali (assumed to be \( \mathrm{NaOH} \))** in \( 500 \, \mathrm{cm}^3 \) solution: \( 2.00 \, \mathrm{g} \) Let's address each part of the question step-by-step. --- ### **a. Balanced Equation for the Reaction** The reaction involves sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) reacting with sodium hydroxide (\( \mathrm{NaOH} \)). Sulfuric acid is a diprotic acid, meaning it can donate two protons (\( \mathrm{H}^+ \)), while sodium hydroxide is a monobasic base, donating one hydroxide ion (\( \mathrm{OH}^- \)) per molecule. **Balanced Chemical Equation:** \[ \mathrm{H}_2\mathrm{SO}_4 \, (aq) + 2 \, \mathrm{NaOH} \, (aq) \rightarrow \mathrm{Na}_2\mathrm{SO}_4 \, (aq) + 2 \, \mathrm{H}_2\mathrm{O} \, (l) \] --- ### **b. Calculation of the Concentration of \( \mathrm{NaOH} \) in \( \mathrm{mol} \cdot \mathrm{dm}^{-3} \)** To find the concentration of \( \mathrm{NaOH} \), we'll use the mass and volume information provided. 1. **Determine the Moles of \( \mathrm{NaOH} \):** - **Molar Mass of \( \mathrm{NaOH} \):** \[ \mathrm{Na} = 23.0 \, \mathrm{g/mol}, \quad \mathrm{O} = 16.0 \, \mathrm{g/mol}, \quad \mathrm{H} = 1.0 \, \mathrm{g/mol} \] \[ \text{Molar Mass of } \mathrm{NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \, \mathrm{g/mol} \] - **Moles of \( \mathrm{NaOH} \) in \( 500 \, \mathrm{cm}^3 \):** \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.00 \, \mathrm{g}}{40.0 \, \mathrm{g/mol}} = 0.0500 \, \mathrm{mol} \] 2. **Calculate the Concentration (\( C \)) of \( \mathrm{NaOH} \):** - **Volume in \( \mathrm{dm}^3 \):** \[ 500 \, \mathrm{cm}^3 = 0.500 \, \mathrm{dm}^3 \] - **Concentration:** \[ C = \frac{\text{Moles}}{\text{Volume}} = \frac{0.0500 \, \mathrm{mol}}{0.500 \, \mathrm{dm}^3} = 0.100 \, \mathrm{mol} \cdot \mathrm{dm}^{-3} \] **Answer:** #### a. The balanced chemical equation is \[ \mathrm{H}_2\mathrm{SO}_4(aq) + 2\, \mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_2\mathrm{SO}_4(aq) + 2\, \mathrm{H}_2\mathrm{O}(l) \] #### b.i. The concentration of NaOH is 0.100 mol dm⁻³

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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!

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