If 0.558 g KHP (molar mass 204.23 g/mol) were used to standardize 21.52 mL of a NaOH solution what is the concentration (M) of that NaOH solution given the following balanced reaction. KHP (s) + NaOH (aq) - NaKP (aq) + H2O (1) 0.254 0.127 O 2.54 0.558
If 0.558 g KHP (molar mass 204.23 g/mol) were used to standardize 21.52 mL of a NaOH solution what is the concentration (M) of that NaOH solution given the following balanced reaction. KHP (s) + NaOH (aq) - NaKP (aq) + H2O (1) 0.254 0.127 O 2.54 0.558
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:
If 0.558 g KHP (molar mass 204.23 g/mol) were used to standardize 21.52 mL of a NaOH solution, what is the concentration (M) of that NaOH solution given the following balanced reaction:
\[ \text{KHP (s)} + \text{NaOH (aq)} \rightarrow \text{NaKP (aq)} + \text{H}_2\text{O (l)} \]
### Multiple Choice Options:
- ○ 0.254
- ○ 0.127
- ○ 2.54
- ○ 0.558
### Explanation of the Balanced Reaction:
In this reaction, potassium hydrogen phthalate (KHP) reacts with sodium hydroxide (NaOH) to form sodium potassium phthalate (NaKP) and water (H₂O). The stoichiometry of the reaction is 1:1, meaning that one mole of KHP reacts with one mole of NaOH.
### Calculations:
1. **Determine the number of moles of KHP used:**
- Mass of KHP = 0.558 g
- Molar mass of KHP = 204.23 g/mol
- Moles of KHP = \(\frac{0.558 \text{ g}}{204.23 \text{ g/mol}} \approx 0.00273 \text{ mol}\)
2. **Calculate the concentration of the NaOH solution:**
- Volume of NaOH solution = 21.52 mL = 0.02152 L
- Since the stoichiometry is 1:1, moles of NaOH = moles of KHP = 0.00273 mol
- Concentration \( M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.00273 \text{ mol}}{0.02152 \text{ L}} \approx 0.127 \text{ M} \)
### Correct Answer:
○ 0.127](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcffe6fb8-a8b1-4738-951c-b41ccaeecf4e%2Fc9ca02a5-6f62-4f85-ad75-ca90c69223f8%2Fg0zzym_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
If 0.558 g KHP (molar mass 204.23 g/mol) were used to standardize 21.52 mL of a NaOH solution, what is the concentration (M) of that NaOH solution given the following balanced reaction:
\[ \text{KHP (s)} + \text{NaOH (aq)} \rightarrow \text{NaKP (aq)} + \text{H}_2\text{O (l)} \]
### Multiple Choice Options:
- ○ 0.254
- ○ 0.127
- ○ 2.54
- ○ 0.558
### Explanation of the Balanced Reaction:
In this reaction, potassium hydrogen phthalate (KHP) reacts with sodium hydroxide (NaOH) to form sodium potassium phthalate (NaKP) and water (H₂O). The stoichiometry of the reaction is 1:1, meaning that one mole of KHP reacts with one mole of NaOH.
### Calculations:
1. **Determine the number of moles of KHP used:**
- Mass of KHP = 0.558 g
- Molar mass of KHP = 204.23 g/mol
- Moles of KHP = \(\frac{0.558 \text{ g}}{204.23 \text{ g/mol}} \approx 0.00273 \text{ mol}\)
2. **Calculate the concentration of the NaOH solution:**
- Volume of NaOH solution = 21.52 mL = 0.02152 L
- Since the stoichiometry is 1:1, moles of NaOH = moles of KHP = 0.00273 mol
- Concentration \( M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.00273 \text{ mol}}{0.02152 \text{ L}} \approx 0.127 \text{ M} \)
### Correct Answer:
○ 0.127
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