(5.7: Similar to For practice 5.11)The titration of a 20.0 mL sample of an H₂SO4 solution of unknown concentration requires 18.88 mL of a 0.203 M KOH to reach the equivalence point. What is the concentration (in M) of the unknown H₂SO4 solution? (Hint: Write the balanced reaction equation first.) 0.0958 M O 0.383 M O 0.767 M O 0.192 M
(5.7: Similar to For practice 5.11)The titration of a 20.0 mL sample of an H₂SO4 solution of unknown concentration requires 18.88 mL of a 0.203 M KOH to reach the equivalence point. What is the concentration (in M) of the unknown H₂SO4 solution? (Hint: Write the balanced reaction equation first.) 0.0958 M O 0.383 M O 0.767 M O 0.192 M
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Titration Problem Example (Similar to Practice 5.11)**
In this example, we are given a problem involving the titration of a 20.0 mL sample of an H₂SO₄ solution with an unknown concentration. To reach the equivalence point, 18.88 mL of a 0.203 M KOH solution is required. Our task is to determine the concentration (in M) of the unknown H₂SO₄ solution.
**Hint: Write the balanced reaction equation first.**
**Options:**
- 0.0958 M
- 0.383 M
- 0.767 M
- 0.192 M
**Step-by-step approach:**
1. **Write the balanced reaction equation:**
\[ \text{H}_2\text{SO}_4 (aq) + 2 \text{KOH} (aq) \rightarrow \text{K}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \]
2. **Determine the moles of KOH used:**
\[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity (M)} \]
\[ \text{Moles of KOH} = 0.01888 \, \text{L} \times 0.203 \, \text{M} = 0.00383264 \, \text{mol KOH} \]
3. **Use the stoichiometry of the balanced equation to find moles of H₂SO₄:**
According to the equation, 1 mole of H₂SO₄ reacts with 2 moles of KOH.
\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of KOH}}{2} = \frac{0.00383264}{2} = 0.00191632 \, \text{mol H}_2\text{SO}_4 \]
4. **Calculate the concentration of H₂SO₄:**
\[ \text{Concentration (M)} = \frac{\text{Moles}}{\text{Volume (L)}} \]
\[ \text{Concentration of H}_2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6aa13af-b0b4-4b84-a222-8b6a152716db%2F8a2d6247-c914-4d8d-968e-913ad4d49dbc%2Fphz953e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Problem Example (Similar to Practice 5.11)**
In this example, we are given a problem involving the titration of a 20.0 mL sample of an H₂SO₄ solution with an unknown concentration. To reach the equivalence point, 18.88 mL of a 0.203 M KOH solution is required. Our task is to determine the concentration (in M) of the unknown H₂SO₄ solution.
**Hint: Write the balanced reaction equation first.**
**Options:**
- 0.0958 M
- 0.383 M
- 0.767 M
- 0.192 M
**Step-by-step approach:**
1. **Write the balanced reaction equation:**
\[ \text{H}_2\text{SO}_4 (aq) + 2 \text{KOH} (aq) \rightarrow \text{K}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \]
2. **Determine the moles of KOH used:**
\[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity (M)} \]
\[ \text{Moles of KOH} = 0.01888 \, \text{L} \times 0.203 \, \text{M} = 0.00383264 \, \text{mol KOH} \]
3. **Use the stoichiometry of the balanced equation to find moles of H₂SO₄:**
According to the equation, 1 mole of H₂SO₄ reacts with 2 moles of KOH.
\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of KOH}}{2} = \frac{0.00383264}{2} = 0.00191632 \, \text{mol H}_2\text{SO}_4 \]
4. **Calculate the concentration of H₂SO₄:**
\[ \text{Concentration (M)} = \frac{\text{Moles}}{\text{Volume (L)}} \]
\[ \text{Concentration of H}_2
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