After Addition of 0.10 M HCI Tube 1A Tube 2A Tube 3A Tube 4A 0.50 mL 1.50 mL 3.00 mL 4.50 mL . Measured pH of Buffer 6.55 5.79 4.25 3.20 After Addition of 0.10 M NAOH Tube 1B Tube 2B Tube 3B Tube 4B 0.50 mL 1.50 mL 3.00 mL 4.50 mL E. Measured pH of Buffer 7.05 7.39 8.29 9.99 E. mmoles of phosphate (HPO,?- + H2PO.) in 5.00 mL of 0.10 M buffer solution 0.50 mmoles [HPO,] . Ratio [H,PO, ] Calculations for Line 5 & Line 6 Below Using Ratio (Show All Work Here OR on the Back Side!) (From Part A) 0.5011 . mmoles of H2PO, in 5.00 mL of 0.10 M buffer solution 5. mmoles of HPO,2- in 5.00 mL of 0.10 M buffer solution

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter16: Reactions Between Acids And Bases
Section: Chapter Questions
Problem 16.7QE
icon
Related questions
Question
100%
### Buffer and pH Analysis

#### Table 1: pH Measurements After Addition of HCl and NaOH

|                | Tube 1A (0.50 mL) | Tube 2A (1.50 mL) | Tube 3A (3.00 mL) | Tube 4A (4.50 mL) |
|----------------|-------------------|-------------------|-------------------|-------------------|
| **After Addition of 0.10 M HCl** |                   |                   |                   |                   |
| Measured pH    | 6.55              | 5.79              | 4.25              | 3.20              |
| **After Addition of 0.10 M NaOH** |                   |                   |                   |                   |
| Measured pH    | 7.05              | 7.39              | 8.29              | 9.99              |

#### Calculations Section

1. **Moles of Phosphate Combination:**  
   Total mmoles of phosphate (HPO₄²⁻ + H₂PO₄⁻) in 5.00 mL of 0.10 M buffer solution:  
   **0.50 mmoles**

2. **Ratio of Phosphate Ions:**  
   \[
   \text{Ratio} \left( \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} \right) \text{ (From Part A)}
   \]  
   **0.5011**

3. **Additional Calculations:**  
   Calculations for Line 5 & Line 6 below using ratio. (Show all work here or on the back side!)

4. **Specific Moles Calculated:**

   - **mmoles of H₂PO₄⁻ in 5.00 mL of 0.10 M buffer solution:** (To be calculated)
   - **mmoles of HPO₄²⁻ in 5.00 mL of 0.10 M buffer solution:** (To be calculated)

### Explanation:
This analysis involves measuring the buffer capacity by adding known concentrations of HCl and NaOH to phosphate buffers and observing the pH changes. The information helps in understanding the effectiveness of the buffer system across different concentrations of acidic or basic additions.
Transcribed Image Text:### Buffer and pH Analysis #### Table 1: pH Measurements After Addition of HCl and NaOH | | Tube 1A (0.50 mL) | Tube 2A (1.50 mL) | Tube 3A (3.00 mL) | Tube 4A (4.50 mL) | |----------------|-------------------|-------------------|-------------------|-------------------| | **After Addition of 0.10 M HCl** | | | | | | Measured pH | 6.55 | 5.79 | 4.25 | 3.20 | | **After Addition of 0.10 M NaOH** | | | | | | Measured pH | 7.05 | 7.39 | 8.29 | 9.99 | #### Calculations Section 1. **Moles of Phosphate Combination:** Total mmoles of phosphate (HPO₄²⁻ + H₂PO₄⁻) in 5.00 mL of 0.10 M buffer solution: **0.50 mmoles** 2. **Ratio of Phosphate Ions:** \[ \text{Ratio} \left( \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} \right) \text{ (From Part A)} \] **0.5011** 3. **Additional Calculations:** Calculations for Line 5 & Line 6 below using ratio. (Show all work here or on the back side!) 4. **Specific Moles Calculated:** - **mmoles of H₂PO₄⁻ in 5.00 mL of 0.10 M buffer solution:** (To be calculated) - **mmoles of HPO₄²⁻ in 5.00 mL of 0.10 M buffer solution:** (To be calculated) ### Explanation: This analysis involves measuring the buffer capacity by adding known concentrations of HCl and NaOH to phosphate buffers and observing the pH changes. The information helps in understanding the effectiveness of the buffer system across different concentrations of acidic or basic additions.
Expert Solution
Step 1

Given:

mmoles of H2PO4-+mmoles of HPO4-2=0.50 mmoles[HPO4-2][H2PO4-]=0.5011

 

 

steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
Acid-Base Titrations
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry: Principles and Practice
Chemistry: Principles and Practice
Chemistry
ISBN:
9780534420123
Author:
Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:
Cengage Learning
Chemical Principles in the Laboratory
Chemical Principles in the Laboratory
Chemistry
ISBN:
9781305264434
Author:
Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Publisher:
Brooks Cole