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
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
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter4: Chemical Reactions
Section: Chapter Questions
Problem 4.24QP: Three acid samples are prepared for titration by 0.01 M NaOH: 1 Sample 1 is prepared by dissolving...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda7693ac-42ba-43da-875a-8ee9e0eebed9%2F7b9f4e83-b6f2-47b9-ae9f-496d87ab2512%2Fl8ta8pf_processed.jpeg&w=3840&q=75)
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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