To solve the problem of how much sodium hydroxide (NaOH) is needed to neutralize a given volume of sulfuric acid (H₂SO₄), we need to follow a systematic approach. Let's break it down step-by-step. ### Step 1: Understand the Reaction The neutralization reaction between sodium hydroxide and sulfuric acid can be represented by the following balanced chemical equation: \[ \text{H}_2\text{SO}_4 (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \] From this equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. This stoichiometry is crucial for our calculations. ### Step 2: Calculate Moles of Sulfuric Acid We are given the concentration and volume of sulfuric acid. To find the number of moles of sulfuric acid, we can use the formula: \[ \text{Moles} = \text{Concentration (mol/L)} \times \text{Volume (L)} \] First, we need to convert the volume of sulfuric acid from cm³ to liters: \[ 10 \, \text{cm}^3 = 10 \times 10^{-3} \, \text{L} = 0.01 \, \text{L} \] Now, we can calculate the moles of sulfuric acid: \[ \text{Moles of H}_2\text{SO}_4 = 1.25 \, \text{mol/L} \times 0.01 \, \text{L} = 0.0125 \, \text{mol} \] ### Step 3: Determine Moles of Sodium Hydroxide Required From the balanced equation, we know that 1 mole of H₂SO₄ requires 2 moles of NaOH for complete neutralization. Therefore, the moles of NaOH required can be calculated as follows: \[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4 = 2 \times 0.0125 \, \text{mol} = 0.025 \, \text{mol} \] ### Step 4: Calculate the Volume of Sodium Hydroxide Needed Now that we know the moles of NaOH required, we can find the volume of the 0.5 M NaOH solution needed to provide these moles. We use the same formula for moles: \[ \text{Moles} = \text{Concentration (mol/L)} \times \text{Volume (L)} \] Rearranging this formula to find the volume gives us: \[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (mol/L)}} \] Substituting the values we have: \[ \text{Volume of NaOH} = \frac{0.025 \, \text{mol}}{0.5 \, \text{mol/L}} = 0.05 \, \text{L} \] Converting this volume back to cm³: \[ 0.05 \, \text{L} = 50 \, \text{cm}^3 \] ### Step 5: Review the Options Now, let's review the options provided: - A. 5 cm³ - B. 10 cm³ - C. 20 cm³ - D. 25 cm³ None of these options match our calculated volume of 50 cm³. However, if we consider the stoichiometry again, we realize that we need to check if the question was misinterpreted or if the options were incorrect. ### Conclusion The correct volume of 0.5 M sodium hydroxide required to neutralize 10 cm³ of 1.25 M sulfuric acid is **50 cm³**. Since this option is not listed, it suggests a potential error in the options provided. ### Revision Summary - The balanced equation shows that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. - Calculate moles of H₂SO₄ using concentration and volume. - Use stoichiometry to find moles of NaOH required. - Calculate the volume of NaOH needed using its concentration. - The final answer is 50 cm³, which is not among the provided options.