Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H'] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1×10¹ Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: [H]=Mc-Man V V/ LV (c) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x 10-¹4 MXV NOW VO+VY [H"]= ABERDEELC titrant: NaOH, 0.IM Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b. 12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79c. 12.81 d. 12. 78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H'] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1×10¹ Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: [H]=Mc-Man V V/ LV (c) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x 10-¹4 MXV NOW VO+VY [H"]= ABERDEELC titrant: NaOH, 0.IM Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b. 12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79c. 12.81 d. 12. 78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
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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![## Titration Curve Construction and Analysis
### Exercise:
**Objective:**
To construct a titration curve by plotting pH versus the volume of NaOH added.
**Chemical Reaction:**
\[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} + \text{NaCl}_{(aq)} \]
**Graph Data:**
| Volume of NaOH (mL) | \([H^+]\) | pH |
|---------------------|----------|-----|
| 0.00 | \( 1.0 \times 10^{-1} \) | 1.00 |
| 5.00 | \( 7.9 \times 10^{-2} \) | 1.10 |
| 10.00 | \( 6.3 \times 10^{-2} \) | 1.20 |
| 20.00 | \( 4.0 \times 10^{-2} \) | 1.40 |
| 30.00 | \( 2.5 \times 10^{-2} \) | 1.60 |
| 40.00 | \( 1.6 \times 10^{-2} \) | 1.80 |
| 45.00 | \( 1.3 \times 10^{-2} \) | 1.89 |
| 50.00 | \( 1.0 \times 10^{-7} \) | 7.00 |
| 55.00 | \( 1.3 \times 10^{-2} \) | 12.89 |
| 60.00 | \( 2.5 \times 10^{-2} \) | 12.60 |
| 70.00 | \( 4.0 \times 10^{-2} \) | 12.40 |
| 80.00 | \( 6.3 \times 10^{-2} \) | 12.20 |
| 90.00 | \( 7.9 \times 10^{-2} \) | 12.10 |
| 95.00 | \( 8.9 \times 10^{-2} \) | 12.04 |](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5106e58f-ed35-4dd8-8e36-19670cace218%2F09f0d42a-5893-4260-bc32-1b2c498486f9%2F5zyzjq4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Titration Curve Construction and Analysis
### Exercise:
**Objective:**
To construct a titration curve by plotting pH versus the volume of NaOH added.
**Chemical Reaction:**
\[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} + \text{NaCl}_{(aq)} \]
**Graph Data:**
| Volume of NaOH (mL) | \([H^+]\) | pH |
|---------------------|----------|-----|
| 0.00 | \( 1.0 \times 10^{-1} \) | 1.00 |
| 5.00 | \( 7.9 \times 10^{-2} \) | 1.10 |
| 10.00 | \( 6.3 \times 10^{-2} \) | 1.20 |
| 20.00 | \( 4.0 \times 10^{-2} \) | 1.40 |
| 30.00 | \( 2.5 \times 10^{-2} \) | 1.60 |
| 40.00 | \( 1.6 \times 10^{-2} \) | 1.80 |
| 45.00 | \( 1.3 \times 10^{-2} \) | 1.89 |
| 50.00 | \( 1.0 \times 10^{-7} \) | 7.00 |
| 55.00 | \( 1.3 \times 10^{-2} \) | 12.89 |
| 60.00 | \( 2.5 \times 10^{-2} \) | 12.60 |
| 70.00 | \( 4.0 \times 10^{-2} \) | 12.40 |
| 80.00 | \( 6.3 \times 10^{-2} \) | 12.20 |
| 90.00 | \( 7.9 \times 10^{-2} \) | 12.10 |
| 95.00 | \( 8.9 \times 10^{-2} \) | 12.04 |
![**Exercise: Construct a titration curve by plotting pH vs. \( V_{\text{NaOH}} \) added.**
**Reaction:**
\[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} + \text{NaCl}_{(aq)} \]
**Volume NaOH (mL) | [H⁺] | pH**
- 0.00 | \(1.0 \times 10^{-1}\)| 1.00
- 5.00 | \(7.9 \times 10^{-2}\) | 1.10
- 15.00 | \(3.2 \times 10^{-2}\) | 1.50
- 20.00 | \(1.8 \times 10^{-2}\) | 1.74
- 49.00 | \(2.0 \times 10^{-4}\) | 3.70
- 49.90 | \(1.0 \times 10^{-5}\) | 5.00
- 49.99 | \(1.0 \times 10^{-7}\) | 7.00
- 50.01 | \(5.0 \times 10^{-11}\) | 10.30
- 50.50 | \(1.1 \times 10^{-10}\) | 10.95
- 52.00 | \(2.5 \times 10^{-10}\) | 10.60
- 55.00 | \(1.8 \times 10^{-10}\) | 10.74
- 90.00 | \(5.1 \times 10^{-11}\) | 11.29
- 95.00 | \(3.2 \times 10^{-11}\) | 11.50
**Hints:**
- **The pH of the solution is calculated using the following equations:**
- **(a) Initial conditions (\( V_{\text{NaOH}} = 0 \)):** \([H^+] = [\text{HCl}]\)
- **(b) After initial addition of NaOH and before the end point:**
\[ [H^+] = \frac{M \times V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5106e58f-ed35-4dd8-8e36-19670cace218%2F09f0d42a-5893-4260-bc32-1b2c498486f9%2Fvw8k3ai_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exercise: Construct a titration curve by plotting pH vs. \( V_{\text{NaOH}} \) added.**
**Reaction:**
\[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} + \text{NaCl}_{(aq)} \]
**Volume NaOH (mL) | [H⁺] | pH**
- 0.00 | \(1.0 \times 10^{-1}\)| 1.00
- 5.00 | \(7.9 \times 10^{-2}\) | 1.10
- 15.00 | \(3.2 \times 10^{-2}\) | 1.50
- 20.00 | \(1.8 \times 10^{-2}\) | 1.74
- 49.00 | \(2.0 \times 10^{-4}\) | 3.70
- 49.90 | \(1.0 \times 10^{-5}\) | 5.00
- 49.99 | \(1.0 \times 10^{-7}\) | 7.00
- 50.01 | \(5.0 \times 10^{-11}\) | 10.30
- 50.50 | \(1.1 \times 10^{-10}\) | 10.95
- 52.00 | \(2.5 \times 10^{-10}\) | 10.60
- 55.00 | \(1.8 \times 10^{-10}\) | 10.74
- 90.00 | \(5.1 \times 10^{-11}\) | 11.29
- 95.00 | \(3.2 \times 10^{-11}\) | 11.50
**Hints:**
- **The pH of the solution is calculated using the following equations:**
- **(a) Initial conditions (\( V_{\text{NaOH}} = 0 \)):** \([H^+] = [\text{HCl}]\)
- **(b) After initial addition of NaOH and before the end point:**
\[ [H^+] = \frac{M \times V
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