### Buffer pH Calculation for Educational Purposes#### Problem Statement:What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 x 10^-4) and 0.200 mol of NaF to which 0.110 mol of NaOH were added?#### Background:A buffer solution is essential in maintaining pH stability, composed typically of a weak acid and its conjugate base. In this scenario, the weak acid is hydrofluoric acid (HF), and its conjugate base is sodium fluoride (NaF). The addition of sodium hydroxide (NaOH), a strong base, will shift the equilibrium and affect the pH of the buffer solution.#### Approach:1. **Initial Concentration Calculations:** - **HF (Weak Acid):** Initially, we have 0.300 mol of HF in 1.0 L, so the initial concentration of HF is 0.300 M. - **NaF (Conjugate Base):** Initially, we have 0.200 mol of NaF in 1.0 L, so the initial concentration of NaF is 0.200 M.2. **Reaction with NaOH:** - Upon adding 0.110 mol of NaOH, a strong base, it reacts completely with HF: \[ \text{HF} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaF} \] - Mole of HF after reaction: 0.300 mol (initial) - 0.110 mol (reacted) = 0.190 mol - Mole of NaF formed from reaction: 0.200 mol (initial) + 0.110 mol (formed) = 0.310 mol3. **New Concentrations:** - **HF:** 0.190 mol / 1.0 L = 0.190 M - **NaF:** 0.310 mol / 1.0 L = 0.310 M4. **Henderson-Hasselbalch Equation:** The pH of the buffer solution can be calculated using the Henderson-Hasselbalch Equation: \[ \text{pH} = \text{p}K_\text{a}

Volume of buffer solution (V) = 1.0 LNumber of moles of HF = 0.300 moleNumber of moles of NaF =…

Answered: What is the pH of a 1.0 L buffer made…

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10-4) and 0.200 mol of NaF to which 0.110 mol of NaOH were added?

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

### Buffer pH Calculation for Educational Purposes

#### Problem Statement:
What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 x 10^-4) and 0.200 mol of NaF to which 0.110 mol of NaOH were added?

#### Background:
A buffer solution is essential in maintaining pH stability, composed typically of a weak acid and its conjugate base. In this scenario, the weak acid is hydrofluoric acid (HF), and its conjugate base is sodium fluoride (NaF). The addition of sodium hydroxide (NaOH), a strong base, will shift the equilibrium and affect the pH of the buffer solution.

#### Approach:
1. **Initial Concentration Calculations:**
- **HF (Weak Acid):** Initially, we have 0.300 mol of HF in 1.0 L, so the initial concentration of HF is 0.300 M.
- **NaF (Conjugate Base):** Initially, we have 0.200 mol of NaF in 1.0 L, so the initial concentration of NaF is 0.200 M.

2. **Reaction with NaOH:**
- Upon adding 0.110 mol of NaOH, a strong base, it reacts completely with HF:
\[
\text{HF} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaF}
\]
- Mole of HF after reaction: 0.300 mol (initial) - 0.110 mol (reacted) = 0.190 mol
- Mole of NaF formed from reaction: 0.200 mol (initial) + 0.110 mol (formed) = 0.310 mol

3. **New Concentrations:**
- **HF:** 0.190 mol / 1.0 L = 0.190 M
- **NaF:** 0.310 mol / 1.0 L = 0.310 M

4. **Henderson-Hasselbalch Equation:**
The pH of the buffer solution can be calculated using the Henderson-Hasselbalch Equation:
\[
\text{pH} = \text{p}K_\text{a}

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