Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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## Determining the Molarity of a Sulfuric Acid Sample

**Question 6:** 
A 10.00 mL sample of sulfuric acid is titrated with 27.25 mL of a 0.1420 M lithium hydroxide solution. Determine the molarity of the sulfuric acid sample.

**Your Answer:** _______________

### Explanation:

To determine the molarity of the sulfuric acid (H₂SO₄) sample, we can use the titration formula and the balanced chemical equation. The balanced chemical reaction between sulfuric acid and lithium hydroxide (LiOH) is:

\[ \text{H}_2\text{SO}_4 + 2\text{LiOH} \rightarrow \text{Li}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

From the equation, we see that one mole of sulfuric acid reacts with two moles of lithium hydroxide.

1. **Calculate the moles of LiOH used in the titration:**
   \[
   \text{Moles of LiOH} = M_{\text{LiOH}} \times V_{\text{LiOH}}
   \]
   Where:
   - \(M_{\text{LiOH}} = 0.1420 \text{ M}\)
   - \(V_{\text{LiOH}} = 27.25 \text{ mL} = 0.02725 \text{ L}\)
   
   \[
   \text{Moles of LiOH} = 0.1420 \text{ M} \times 0.02725 \text{ L} = 0.0038695 \text{ moles}
   \]

2. **Determine the moles of H₂SO₄ reacted:**
   From the balanced chemical equation, 1 mole of H₂SO₄ reacts with 2 moles of LiOH. Therefore:
   \[
   \text{Moles of H₂SO₄} = \frac{\text{Moles of LiOH}}{2} = \frac{0.0038695 \text{ moles}}{2} = 0.00193475 \text{ moles}
   \]

3. **Calculate the molarity of the H₂SO₄ sample:**
   \[
   M_{\text{H₂SO
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Transcribed Image Text:## Determining the Molarity of a Sulfuric Acid Sample **Question 6:** A 10.00 mL sample of sulfuric acid is titrated with 27.25 mL of a 0.1420 M lithium hydroxide solution. Determine the molarity of the sulfuric acid sample. **Your Answer:** _______________ ### Explanation: To determine the molarity of the sulfuric acid (H₂SO₄) sample, we can use the titration formula and the balanced chemical equation. The balanced chemical reaction between sulfuric acid and lithium hydroxide (LiOH) is: \[ \text{H}_2\text{SO}_4 + 2\text{LiOH} \rightarrow \text{Li}_2\text{SO}_4 + 2\text{H}_2\text{O} \] From the equation, we see that one mole of sulfuric acid reacts with two moles of lithium hydroxide. 1. **Calculate the moles of LiOH used in the titration:** \[ \text{Moles of LiOH} = M_{\text{LiOH}} \times V_{\text{LiOH}} \] Where: - \(M_{\text{LiOH}} = 0.1420 \text{ M}\) - \(V_{\text{LiOH}} = 27.25 \text{ mL} = 0.02725 \text{ L}\) \[ \text{Moles of LiOH} = 0.1420 \text{ M} \times 0.02725 \text{ L} = 0.0038695 \text{ moles} \] 2. **Determine the moles of H₂SO₄ reacted:** From the balanced chemical equation, 1 mole of H₂SO₄ reacts with 2 moles of LiOH. Therefore: \[ \text{Moles of H₂SO₄} = \frac{\text{Moles of LiOH}}{2} = \frac{0.0038695 \text{ moles}}{2} = 0.00193475 \text{ moles} \] 3. **Calculate the molarity of the H₂SO₄ sample:** \[ M_{\text{H₂SO
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