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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![**Chemistry Problem: Calculating the pH of a Sodium Hydroxide Solution**
**Problem Statement:**
A chemist dissolves 672 mg of pure sodium hydroxide in enough water to make up 150 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.)
**Instructions:**
Round your answer to 3 significant decimal places.
**Answer Box:**
[Input Box]
**Additional Options:**
- [Checkbox] x10
- [Button] x
- [Button] ↺ (reset)
- [Button] ? (help)
**Explanation:**
To solve this problem, follow the steps below:
1. **Calculate the molarity (M) of the NaOH solution:**
- Convert milligrams to grams:
\( 672 \text{ mg} = 0.672 \text{ g} \)
- Determine the molar mass of NaOH (Sodium Hydroxide):
\( \text{NaOH} = 22.99 \text{(Na)} + 15.999 \text{(O)} + 1.008 \text{(H)} = 39.997 \text{ g/mol} \)
- Calculate the number of moles of NaOH:
\( \text{moles of NaOH} = \frac{0.672 \text{ g}}{39.997 \text{ g/mol}} \approx 0.0168 \text{ moles} \)
- Convert mL to L:
\( 150 \text{ mL} = 0.150 \text{ L} \)
- Calculate the molarity (M) of the solution:
\( \text{Molarity (M)} = \frac{0.0168 \text{ moles of NaOH}}{0.150 \text{ L}} \approx 0.112 \text{ M} \)
2. **Calculate the concentration of hydroxide ions (OH\(^-\)):**
Since NaOH is a strong base and dissociates completely in water:
\( [\text{OH}^-] = 0.112 \text{ M} \)
3. **Calculate the pOH:**
\( \text{pOH} = -\log[\text{OH}^-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf399169-ef14-4107-9621-c5ec1888471d%2Fc9041cb2-9351-406b-ba4d-c9f46ad946bf%2Fohx4bbi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemistry Problem: Calculating the pH of a Sodium Hydroxide Solution**
**Problem Statement:**
A chemist dissolves 672 mg of pure sodium hydroxide in enough water to make up 150 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.)
**Instructions:**
Round your answer to 3 significant decimal places.
**Answer Box:**
[Input Box]
**Additional Options:**
- [Checkbox] x10
- [Button] x
- [Button] ↺ (reset)
- [Button] ? (help)
**Explanation:**
To solve this problem, follow the steps below:
1. **Calculate the molarity (M) of the NaOH solution:**
- Convert milligrams to grams:
\( 672 \text{ mg} = 0.672 \text{ g} \)
- Determine the molar mass of NaOH (Sodium Hydroxide):
\( \text{NaOH} = 22.99 \text{(Na)} + 15.999 \text{(O)} + 1.008 \text{(H)} = 39.997 \text{ g/mol} \)
- Calculate the number of moles of NaOH:
\( \text{moles of NaOH} = \frac{0.672 \text{ g}}{39.997 \text{ g/mol}} \approx 0.0168 \text{ moles} \)
- Convert mL to L:
\( 150 \text{ mL} = 0.150 \text{ L} \)
- Calculate the molarity (M) of the solution:
\( \text{Molarity (M)} = \frac{0.0168 \text{ moles of NaOH}}{0.150 \text{ L}} \approx 0.112 \text{ M} \)
2. **Calculate the concentration of hydroxide ions (OH\(^-\)):**
Since NaOH is a strong base and dissociates completely in water:
\( [\text{OH}^-] = 0.112 \text{ M} \)
3. **Calculate the pOH:**
\( \text{pOH} = -\log[\text{OH}^-
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