A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid?
The Correct Answer and Explanation is:
To find the molarity of the sulfuric acid (H₂SO₄), we must first understand the neutralization reaction between sulfuric acid and sodium hydroxide (NaOH):
Balanced chemical equation: H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\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 tells us that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Step 1: Calculate moles of NaOH added
Given:
- Volume of NaOH = 35.08 mL = 0.03508 L
- Molarity of NaOH = 2.12 M
Moles of NaOH=M×V=2.12 mol/L×0.03508 L=0.07438 mol\text{Moles of NaOH} = M \times V = 2.12 \, \text{mol/L} \times 0.03508 \, \text{L} = 0.07438 \, \text{mol}
Step 2: Use the mole ratio to find moles of H₂SO₄
From the balanced equation, the mole ratio is: 1 mol H2SO42 mol NaOH=x0.07438⇒x=0.074382=0.03719 mol H2SO4\frac{1 \, \text{mol H}_2\text{SO}_4}{2 \, \text{mol NaOH}} = \frac{x}{0.07438} \Rightarrow x = \frac{0.07438}{2} = 0.03719 \, \text{mol H}_2\text{SO}_4
Step 3: Calculate molarity of H₂SO₄
Given:
- Volume of H₂SO₄ = 10.00 mL = 0.01000 L
Molarity of H2SO4=molesvolume=0.037190.01000=3.72 M\text{Molarity of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume}} = \frac{0.03719}{0.01000} = \boxed{3.72 \, \text{M}}
Conclusion
To determine the concentration of a sulfuric acid solution from an automobile battery, a titration with a standard sodium hydroxide solution was used. The balanced chemical reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide. In this titration, 10.00 mL of the acid required 35.08 mL of 2.12 M sodium hydroxide for complete neutralization.
First, we calculated the number of moles of NaOH used in the reaction: multiplying the volume in liters (0.03508 L) by the molarity (2.12 M) gave 0.07438 moles of NaOH. Since two moles of NaOH react with one mole of H₂SO₄, we divide the moles of NaOH by 2 to find the number of moles of H₂SO₄, which is 0.03719 mol.
Next, we determined the molarity of the sulfuric acid by dividing the number of moles of acid by its volume in liters (0.01000 L), resulting in a molarity of 3.72 M.
This titration illustrates a practical application of stoichiometry and molarity in real-world chemical analysis. Knowing the concentration of battery acid is crucial for evaluating battery health and preventing damage due to overly concentrated or diluted solutions. This method is widely used in analytical chemistry to determine unknown concentrations of acids or bases accurately.To find the molarity of the sulfuric acid (H₂SO₄), we must first understand the neutralization reaction between sulfuric acid and sodium hydroxide (NaOH):
Balanced chemical equation: H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\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 tells us that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Step 1: Calculate moles of NaOH added
Given:
- Volume of NaOH = 35.08 mL = 0.03508 L
- Molarity of NaOH = 2.12 M
Moles of NaOH=M×V=2.12 mol/L×0.03508 L=0.07438 mol\text{Moles of NaOH} = M \times V = 2.12 \, \text{mol/L} \times 0.03508 \, \text{L} = 0.07438 \, \text{mol}
Step 2: Use the mole ratio to find moles of H₂SO₄
From the balanced equation, the mole ratio is: 1 mol H2SO42 mol NaOH=x0.07438⇒x=0.074382=0.03719 mol H2SO4\frac{1 \, \text{mol H}_2\text{SO}_4}{2 \, \text{mol NaOH}} = \frac{x}{0.07438} \Rightarrow x = \frac{0.07438}{2} = 0.03719 \, \text{mol H}_2\text{SO}_4
Step 3: Calculate molarity of H₂SO₄
Given:
Step 1: Calculate moles of NaOH added
Given:
- Volume of NaOH = 35.08 mL = 0.03508 L
- Molarity of NaOH = 2.12 M
Moles of NaOH=M×V=2.12 mol/L×0.03508 L=0.07438 mol\text{Moles of NaOH} = M \times V = 2.12 \, \text{mol/L} \times 0.03508 \, \text{L} = 0.07438 \, \text{mol}
Step 2: Use the mole ratio to find moles of H₂SO₄
From the balanced equation, the mole ratio is: 1 mol H2SO42 mol NaOH=x0.07438⇒x=0.074382=0.03719 mol H2SO4\frac{1 \, \text{mol H}_2\text{SO}_4}{2 \, \text{mol NaOH}} = \frac{x}{0.07438} \Rightarrow x = \frac{0.07438}{2} = 0.03719 \, \text{mol H}_2\text{SO}_4
Step 3: Calculate molarity of H₂SO₄
Given:
- Volume of H₂SO₄ = 10.00 mL = 0.01000 L
Molarity of H2SO4=molesvolume=0.037190.01000=3.72 M\text{Molarity of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume}} = \frac{0.03719}{0.01000} = \boxed{3.72 \, \text{M}}
.