. Draw a balanced chemical equation for the reaction of sulfuric acid and sodium hydroxide. Based on that equation, if 9.5 mL of 0.406 M sulfuric acid is required to neutralized 26.0 mL of a sodium hydroxide solution, what is the concentration of the sodium hydroxide solution? (5 Points)
The Correct Answer and Explanation is:
The reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) can be represented by the following balanced chemical equation:
H₂SO₄ (aq) + 2 NaOH (aq) → Na₂SO₄ (aq) + 2 H₂O (l)
This equation indicates that one mole of sulfuric acid reacts with two moles of sodium hydroxide to form sodium sulfate and water.
Given:
- Volume of sulfuric acid (H₂SO₄) = 9.5 mL = 0.0095 L
- Molarity of sulfuric acid (H₂SO₄) = 0.406 M
- Volume of sodium hydroxide (NaOH) = 26.0 mL = 0.026 L
- Molarity of sodium hydroxide (NaOH) = ? (This is what we need to find)
Steps to solve:
- Calculate the moles of sulfuric acid:
The formula to calculate moles from molarity is: Moles of H₂SO₄=Molarity of H₂SO₄×Volume of H₂SO₄ (in L)\text{Moles of H₂SO₄} = \text{Molarity of H₂SO₄} \times \text{Volume of H₂SO₄ (in L)}Moles of H₂SO₄=Molarity of H₂SO₄×Volume of H₂SO₄ (in L) Moles of H₂SO₄=0.406 mol/L×0.0095 L=0.003857 mol\text{Moles of H₂SO₄} = 0.406 \, \text{mol/L} \times 0.0095 \, \text{L} = 0.003857 \, \text{mol}Moles of H₂SO₄=0.406mol/L×0.0095L=0.003857mol - Use the stoichiometry of the balanced equation:
According to the equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the moles of NaOH required to react with 0.003857 moles of H₂SO₄ are: Moles of NaOH=2×Moles of H₂SO₄\text{Moles of NaOH} = 2 \times \text{Moles of H₂SO₄}Moles of NaOH=2×Moles of H₂SO₄ Moles of NaOH=2×0.003857 mol=0.007714 mol\text{Moles of NaOH} = 2 \times 0.003857 \, \text{mol} = 0.007714 \, \text{mol}Moles of NaOH=2×0.003857mol=0.007714mol - Calculate the molarity of NaOH:
The formula to calculate molarity is: Molarity of NaOH=Moles of NaOHVolume of NaOH (in L)\text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH (in L)}}Molarity of NaOH=Volume of NaOH (in L)Moles of NaOH Molarity of NaOH=0.007714 mol0.0260 L=0.296 M\text{Molarity of NaOH} = \frac{0.007714 \, \text{mol}}{0.0260 \, \text{L}} = 0.296 \, \text{M}Molarity of NaOH=0.0260L0.007714mol=0.296M
Final Answer:
The concentration of the sodium hydroxide solution is 0.296 M.
Explanation:
This problem involves stoichiometry, where we use the balanced chemical equation to relate the moles of sulfuric acid to the moles of sodium hydroxide. By first calculating the moles of sulfuric acid, then using the mole ratio from the balanced equation to determine the moles of sodium hydroxide, we can finally find the concentration (molarity) of the sodium hydroxide solution.
