Post-lab Question #5-1: Ka for hypochlorous acid, HCIO, is 3.0 x 10-8. Calculate the pH after 10.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 M HCIO. 8.00 7.52 7.05 10.11
Post-lab Question #5-1: Ka for hypochlorous acid, HCIO, is 3.0 x 10-8. Calculate the pH after 10.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 M HCIO. 8.00 7.52 7.05 10.11
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Post-lab Question #5-1: Calculation of pH for a Buffer Solution**
**Hypochlorous Acid and Sodium Hydroxide Reaction**
The question requires calculating the pH after mixing two solutions. Here's the detailed problem:
Ka for hypochlorous acid, HClO, is \(3.0 \times 10^{-8}\). Calculate the pH after 10.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 M HClO.
**Multiple Choice Options:**
1. \( \circ \) 8.00
2. \( \circ \) 7.52
3. \( \circ \) 7.05
4. \( \circ \) 10.11
### Explanation:
We need to find out the pH of the resulting solution after adding a base (NaOH) to an acid (HClO). This involves calculating the concentrations of the acid and its conjugate base, then using the Henderson-Hasselbalch equation to find the pH.
The equation for the reaction between hypochlorous acid and sodium hydroxide is:
\[ \text{HClO} + \text{NaOH} \rightarrow \text{NaClO} + \text{H}_2\text{O} \]
1. **Determine moles of HClO and NaOH:**
- Moles of HClO before reaction:
\[
\text{Volume} \times \text{Concentration} = 40.0 \text{ mL} \times 0.100 \text{ M} = 0.00400 \text{ moles}
\]
- Moles of NaOH added:
\[
\text{Volume} \times \text{Concentration} = 10.0 \text{ mL} \times 0.100 \text{ M} = 0.00100 \text{ moles}
\]
2. **After NaOH is added, a buffer solution is formed with HClO and its conjugate base (ClO\(^-\)):**
- Remaining moles of HClO after reaction:
\[
0.00400 \text{ moles} - 0.00100 \text{ moles} = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcffe6fb8-a8b1-4738-951c-b41ccaeecf4e%2Fae098f42-2b9d-494d-80cd-0e49054d4c4f%2Fug5c2w9_processed.png&w=3840&q=75)
Transcribed Image Text:**Post-lab Question #5-1: Calculation of pH for a Buffer Solution**
**Hypochlorous Acid and Sodium Hydroxide Reaction**
The question requires calculating the pH after mixing two solutions. Here's the detailed problem:
Ka for hypochlorous acid, HClO, is \(3.0 \times 10^{-8}\). Calculate the pH after 10.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 M HClO.
**Multiple Choice Options:**
1. \( \circ \) 8.00
2. \( \circ \) 7.52
3. \( \circ \) 7.05
4. \( \circ \) 10.11
### Explanation:
We need to find out the pH of the resulting solution after adding a base (NaOH) to an acid (HClO). This involves calculating the concentrations of the acid and its conjugate base, then using the Henderson-Hasselbalch equation to find the pH.
The equation for the reaction between hypochlorous acid and sodium hydroxide is:
\[ \text{HClO} + \text{NaOH} \rightarrow \text{NaClO} + \text{H}_2\text{O} \]
1. **Determine moles of HClO and NaOH:**
- Moles of HClO before reaction:
\[
\text{Volume} \times \text{Concentration} = 40.0 \text{ mL} \times 0.100 \text{ M} = 0.00400 \text{ moles}
\]
- Moles of NaOH added:
\[
\text{Volume} \times \text{Concentration} = 10.0 \text{ mL} \times 0.100 \text{ M} = 0.00100 \text{ moles}
\]
2. **After NaOH is added, a buffer solution is formed with HClO and its conjugate base (ClO\(^-\)):**
- Remaining moles of HClO after reaction:
\[
0.00400 \text{ moles} - 0.00100 \text{ moles} = 0.
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