### Buffer pH Calculation Example#### Problem Statement: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 0.070 mol of HCl were added?#### Explanation:We can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Where:- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)).- \([\text{A}^-]\) is the concentration of the conjugate base (F⁻ from NaF).- \([\text{HA}]\) is the concentration of the weak acid (HF).**Steps to solve the problem:**1. **Calculate \(\text{p}K_a\):**\[ \text{p}K_a = -\log(6.8 \times 10^{-4}) \]2. **Determine the changes in molarity due to the addition of HCl:** - Upon adding 0.070 mol of HCl to the buffer, HCl will react completely with the conjugate base NaF. - \( 0.070 \text{ mol} \) HCl will reduce the amount of \( \text{A}^- (F^-) \) and increase the amount of \(\text{HA (HF)}\). Initial moles before the reaction: - HF: 0.300 mol - NaF: 0.200 mol Change in moles after addition of HCl: - HF: 0.300 mol + 0.070 mol = 0.370 mol - NaF: 0.200 mol - 0.070 mol = 0.130 mol3. **Apply the Henderson-Hasselbalch equation:**\[ \text{pH} = \text{p}K_a + \log \left( \frac{0.130}{0.370} \right) \] #### Diagram Explanation:The image contains a

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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 0.070 mol of HCI were added? %3D

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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 Example

#### Problem Statement:
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 0.070 mol of HCl were added?

#### Explanation:

We can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Where:
- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)).
- \([\text{A}^-]\) is the concentration of the conjugate base (F⁻ from NaF).
- \([\text{HA}]\) is the concentration of the weak acid (HF).

**Steps to solve the problem:**

1. **Calculate \(\text{p}K_a\):**
\[ \text{p}K_a = -\log(6.8 \times 10^{-4}) \]

2. **Determine the changes in molarity due to the addition of HCl:**

- Upon adding 0.070 mol of HCl to the buffer, HCl will react completely with the conjugate base NaF.
- \( 0.070 \text{ mol} \) HCl will reduce the amount of \( \text{A}^- (F^-) \) and increase the amount of \(\text{HA (HF)}\).

Initial moles before the reaction:
- HF: 0.300 mol
- NaF: 0.200 mol

Change in moles after addition of HCl:
- HF: 0.300 mol + 0.070 mol = 0.370 mol
- NaF: 0.200 mol - 0.070 mol = 0.130 mol

3. **Apply the Henderson-Hasselbalch equation:**

\[ \text{pH} = \text{p}K_a + \log \left( \frac{0.130}{0.370} \right) \]

#### Diagram Explanation:
The image contains a

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