What is the resultant pH after mixing 30.0 mL of 0.50 M HCI with 20.0 mL of 0.90 M NaOH? Express your answer in decimal notation rounded to four significant figures.
What is the resultant pH after mixing 30.0 mL of 0.50 M HCI with 20.0 mL of 0.90 M NaOH? Express your answer in decimal notation rounded to four significant figures.
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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![**Problem Statement:**
What is the resultant pH after mixing 30.0 mL of 0.50 M HCl with 20.0 mL of 0.90 M NaOH? Express your answer in decimal notation rounded to four significant figures.
**Solution Explanation:**
To solve this problem, follow these steps:
1. **Calculate the moles of HCl and NaOH:**
- Moles of HCl = Volume (L) × Concentration (M) = \(0.030 \, \text{L} \times 0.50 \, \text{M} = 0.015 \, \text{moles}\)
- Moles of NaOH = Volume (L) × Concentration (M) = \(0.020 \, \text{L} \times 0.90 \, \text{M} = 0.018 \, \text{moles}\)
2. **Determine the limiting reactant:**
- NaOH is in excess since it has 0.018 moles compared to 0.015 moles of HCl.
3. **Calculate the moles of excess NaOH:**
- Moles of excess NaOH = 0.018 - 0.015 = 0.003 moles
4. **Calculate the concentration of the excess NaOH in the final solution:**
- Total volume of the solution = 30.0 mL + 20.0 mL = 50.0 mL = 0.050 L
- Concentration of NaOH = 0.003 moles / 0.050 L = 0.06 M
5. **Calculate the pH of the solution:**
- pH = 14 - pOH
- pOH = -log[OH⁻] = -log(0.06)
- pOH ≈ 1.2218
- pH = 14 - 1.2218 = 12.7782
Therefore, the resultant pH of the solution is **12.78** when rounded to four significant figures.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7398004-5a4c-4680-b931-919c50d280d3%2F2026d5ff-7397-4a00-93ad-24faa973d0bb%2Fw7eaw9a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
What is the resultant pH after mixing 30.0 mL of 0.50 M HCl with 20.0 mL of 0.90 M NaOH? Express your answer in decimal notation rounded to four significant figures.
**Solution Explanation:**
To solve this problem, follow these steps:
1. **Calculate the moles of HCl and NaOH:**
- Moles of HCl = Volume (L) × Concentration (M) = \(0.030 \, \text{L} \times 0.50 \, \text{M} = 0.015 \, \text{moles}\)
- Moles of NaOH = Volume (L) × Concentration (M) = \(0.020 \, \text{L} \times 0.90 \, \text{M} = 0.018 \, \text{moles}\)
2. **Determine the limiting reactant:**
- NaOH is in excess since it has 0.018 moles compared to 0.015 moles of HCl.
3. **Calculate the moles of excess NaOH:**
- Moles of excess NaOH = 0.018 - 0.015 = 0.003 moles
4. **Calculate the concentration of the excess NaOH in the final solution:**
- Total volume of the solution = 30.0 mL + 20.0 mL = 50.0 mL = 0.050 L
- Concentration of NaOH = 0.003 moles / 0.050 L = 0.06 M
5. **Calculate the pH of the solution:**
- pH = 14 - pOH
- pOH = -log[OH⁻] = -log(0.06)
- pOH ≈ 1.2218
- pH = 14 - 1.2218 = 12.7782
Therefore, the resultant pH of the solution is **12.78** when rounded to four significant figures.
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