Determine the pH at the point in the titration of 40.0 mL of 0.200 M H,NNH, with 0.100 M HNO, after 10.0 mL of the strong acid has been added. The value of Kb for H,NNH, is 3.0 x 10°. 1 2 3 4 NEXT> Use the table below to determine the moles of reactant and product after the reaction of the acid and base. H,NNH,(aq) + H*(aq) H,NNH, (aq) Before (mol) Change |(mol) After (mol) 100 00
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H,NNH, with 0.100 M HNO, after 10.0 mL of the strong acid has been added. The value of Kb for H,NNH, is 3.0 x 10°. 1 2 3 4 NEXT> Use the table below to determine the moles of reactant and product after the reaction of the acid and base. H,NNH,(aq) + H*(aq) H,NNH, (aq) Before (mol) Change |(mol) After (mol) 100 00
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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![**Question 9 of 9**
**Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 10.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.**
**Use the table below to determine the moles of reactant and product after the reaction of the acid and base.**
**Reaction:**
\[ \text{H}_2\text{NNH}_2 (\text{aq}) + \text{H}^+ (\text{aq}) \rightarrow \text{H}_2\text{NNH}_3^+ (\text{aq}) \]
| | H₂NNH₂ (aq) | H⁺ (aq) | H₂NNH₃⁺ (aq) |
|----------------------|-------------|---------|--------------|
| **Before (mol)** | | | |
| **Change (mol)** | | | |
| **After (mol)** | | | |
**Options for Filling the Table:**
- \[ -6.00 \times 10^{-3} \]
- \[ 6.00 \times 10^{-3} \]
- \[ 7.00 \times 10^{-3} \]
- \[ -7.00 \times 10^{-3} \]
- \[ 8.00 \times 10^{-3} \]
- \[ -8.00 \times 10^{-3} \]
- \[ 1.00 \times 10^{-3} \]
- \[ -1.00 \times 10^{-3} \]
- \[ 2.00 \times 10^{-3} \]
- \[ -2.00 \times 10^{-3} \]
- 0
- 0.200
- 0.100
**RESET Button Available**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc93c4c2c-87bf-4a2e-b2f9-fa896ab5e7e6%2Fb4860731-5077-46ee-afc3-a65cbd595e1b%2Fwujghi_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 9 of 9**
**Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 10.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.**
**Use the table below to determine the moles of reactant and product after the reaction of the acid and base.**
**Reaction:**
\[ \text{H}_2\text{NNH}_2 (\text{aq}) + \text{H}^+ (\text{aq}) \rightarrow \text{H}_2\text{NNH}_3^+ (\text{aq}) \]
| | H₂NNH₂ (aq) | H⁺ (aq) | H₂NNH₃⁺ (aq) |
|----------------------|-------------|---------|--------------|
| **Before (mol)** | | | |
| **Change (mol)** | | | |
| **After (mol)** | | | |
**Options for Filling the Table:**
- \[ -6.00 \times 10^{-3} \]
- \[ 6.00 \times 10^{-3} \]
- \[ 7.00 \times 10^{-3} \]
- \[ -7.00 \times 10^{-3} \]
- \[ 8.00 \times 10^{-3} \]
- \[ -8.00 \times 10^{-3} \]
- \[ 1.00 \times 10^{-3} \]
- \[ -1.00 \times 10^{-3} \]
- \[ 2.00 \times 10^{-3} \]
- \[ -2.00 \times 10^{-3} \]
- 0
- 0.200
- 0.100
**RESET Button Available**
![**Educational Text Transcription and Explanation**
**Problem Statement:**
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 10.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.
**Process:**
Based on the result of the acid-base reaction, set up the ICE (Initial, Change, Equilibrium) table in order to determine the unknown.
**ICE Table:**
For the reaction:
\[ \text{H}_2\text{NNH}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{OH}^-(\text{aq}) + \text{H}_2\text{NNH}_3^+(\text{aq}) \]
- **Initial (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
- **Change (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
- **Equilibrium (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
**Interactive Buttons:**
- Values you can select include:
- Initial concentrations: 0, 0.200, 0.0200, 0.100, 0.140, 0.175
- Changes and equilibrium values: -x, 0.200 + x, 0.200 - x, 0.0200 + x, 0.0200 - x, 0.100 + x, 0.100 - x, 0.140 + x, 0.140 - x, 0.175 + x
- Option to RESET the values.
This setup is used to calculate the p](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc93c4c2c-87bf-4a2e-b2f9-fa896ab5e7e6%2Fb4860731-5077-46ee-afc3-a65cbd595e1b%2Fsjng3uu_processed.png&w=3840&q=75)
Transcribed Image Text:**Educational Text Transcription and Explanation**
**Problem Statement:**
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 10.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.
**Process:**
Based on the result of the acid-base reaction, set up the ICE (Initial, Change, Equilibrium) table in order to determine the unknown.
**ICE Table:**
For the reaction:
\[ \text{H}_2\text{NNH}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{OH}^-(\text{aq}) + \text{H}_2\text{NNH}_3^+(\text{aq}) \]
- **Initial (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
- **Change (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
- **Equilibrium (M):**
- H₂NNH₂: [ ]
- H₂O: [ ]
- OH⁻: [ ]
- H₂NNH₃⁺: [ ]
**Interactive Buttons:**
- Values you can select include:
- Initial concentrations: 0, 0.200, 0.0200, 0.100, 0.140, 0.175
- Changes and equilibrium values: -x, 0.200 + x, 0.200 - x, 0.0200 + x, 0.0200 - x, 0.100 + x, 0.100 - x, 0.140 + x, 0.140 - x, 0.175 + x
- Option to RESET the values.
This setup is used to calculate the p
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