Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The value of Ka for HClO is 2.9 × 10⁻⁸. Determine the moles of the ractant and product after the reaction of the acid and base. (Similar to ICE format but is instead Before (mol), Change (mol), and After (mol) Determine the ICE table for HClO (aq) + H2O - H3O+ + ClO- (aq) Fill in Ka= ? = 2.9 * 10-8
Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The value of Ka for HClO is 2.9 × 10⁻⁸. Determine the moles of the ractant and product after the reaction of the acid and base. (Similar to ICE format but is instead Before (mol), Change (mol), and After (mol) Determine the ICE table for HClO (aq) + H2O - H3O+ + ClO- (aq) Fill in Ka= ? = 2.9 * 10-8
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The value of Ka for HClO is 2.9 × 10⁻⁸.
- Determine the moles of the ractant and product after the reaction of the acid and base. (Similar to ICE format but is instead Before (mol), Change (mol), and After (mol)
- Determine the ICE table for HClO (aq) + H2O - H3O+ + ClO- (aq)
- Fill in Ka= ? = 2.9 * 10-8
- Calculate pH
Transcribed Image Text:**Determining the Resulting pH of a Buffer Solution**
This section provides guidance for a chemistry problem focused on calculating the pH of a buffer solution after a reaction. The problem states:
Determine the resulting pH when 0.003 mol of solid NaOH is added to 100.0 mL of a buffer containing 0.13 M HClO and 0.37 M NaClO. The dissociation constant (Ka) for HClO is given as \(2.9 \times 10^{-8}\).
**Instructions:**
Use the table below to determine the moles of reactant and product after the reaction of acid and base. The amount of liquid water in the reaction can be ignored.
**Reaction Table:**
| | HClO(aq) | + | OH⁻(aq) | → | H₂O(l) | + | ClO⁻(aq) |
|----------------|----------|---|---------|---|--------|---|----------|
| Before (mol) | | | | | | | |
| Change (mol) | | | | | | | |
| After (mol) | | | | | | | |
**Interactive Inputs:**
- The table allows for inputs in terms of the changes in moles for reactants and products.
- Below the table are buttons to input specific mole values, such as 0, 0.13, 0.37, \(+x\), \(-x\), etc.
- The "RESET" button clears any changes made.
The goal is to use this information to determine how the addition of NaOH affects the buffer system and calculate the resulting pH.
![**Lesson 9: Acid-Base Reactions and Buffer Systems**
**Objective:** Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The acid dissociation constant (Ka) for HClO is 2.9 × 10⁻⁸.
---
**Step 1: Understand the Buffer System**
You are given a buffer solution composed of Hypochlorous acid (HClO) and its conjugate base, sodium hypochlorite (NaClO).
**Step 2: Setting Up the ICE Table**
To solve this problem using the ICE (Initial, Change, Equilibrium) table, consider the following reaction:
\[ \text{HClO(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{ClO}^-(aq) \]
- **Initial concentrations (M):**
- HClO: 0.13 M
- ClO⁻: 0.37 M
- H₃O⁺: Determined via reaction with NaOH
- **Change (M):**
- Calculate the change in molarity as NaOH dissociates and reacts, affecting the equilibrium concentrations of the HClO and ClO⁻.
- **Equilibrium (M):**
- Analyze concentrations once the reaction re-establishes equilibrium.
**Step 3: Calculating pH**
Use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]
Calculate the pH considering the changes in concentrations.
---
**Interactive Features:**
- **ICE Table Inputs:** Fields to fill in initial, change, and equilibrium molarities.
- **Calculate Button:** Automates the solving of equilibrium expressions and pH.
- **Reset:** Clears all entries for a new attempt.
Learning to set up an ICE table and using the Henderson-Hasselbalch equation are key skills in understanding buffer systems and acid-base equilibria.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8f1d932-7acd-490f-810d-23c78bee5bcd%2Fc4d0a191-897a-402d-81ea-3d5369f3977c%2Fx1m1l5t_processed.png&w=3840&q=75)
Transcribed Image Text:**Lesson 9: Acid-Base Reactions and Buffer Systems**
**Objective:** Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The acid dissociation constant (Ka) for HClO is 2.9 × 10⁻⁸.
---
**Step 1: Understand the Buffer System**
You are given a buffer solution composed of Hypochlorous acid (HClO) and its conjugate base, sodium hypochlorite (NaClO).
**Step 2: Setting Up the ICE Table**
To solve this problem using the ICE (Initial, Change, Equilibrium) table, consider the following reaction:
\[ \text{HClO(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{ClO}^-(aq) \]
- **Initial concentrations (M):**
- HClO: 0.13 M
- ClO⁻: 0.37 M
- H₃O⁺: Determined via reaction with NaOH
- **Change (M):**
- Calculate the change in molarity as NaOH dissociates and reacts, affecting the equilibrium concentrations of the HClO and ClO⁻.
- **Equilibrium (M):**
- Analyze concentrations once the reaction re-establishes equilibrium.
**Step 3: Calculating pH**
Use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]
Calculate the pH considering the changes in concentrations.
---
**Interactive Features:**
- **ICE Table Inputs:** Fields to fill in initial, change, and equilibrium molarities.
- **Calculate Button:** Automates the solving of equilibrium expressions and pH.
- **Reset:** Clears all entries for a new attempt.
Learning to set up an ICE table and using the Henderson-Hasselbalch equation are key skills in understanding buffer systems and acid-base equilibria.
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