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

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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
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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