A buffer solution is made that is 0.402 M in HF and 0.402 M in KF. If Ka for HF is 7.20 x 10-4 what is the pH of the buffer solution? pH = Write the net ionic equation for the reaction that occurs when 0.107 mol HNO3 is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter. Use H3O* instead of H+)

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### Buffer Solution Problem

#### Buffer Composition:
A buffer solution is composed of:
- 0.402 M in HF
- 0.402 M in KF

#### Problem Statement:
If the ionization constant (\(K_a\)) for HF is \(7.20 \times 10^{-4}\), determine the pH of the buffer solution.

\[ \text{pH} = \_\_\_\_\_\_ \]

#### Equation:
Write the net ionic equation for the reaction when 0.107 mol of \(HNO_3\) is added to 1.00 L of the buffer solution.
(Use the lowest possible coefficients and omit states of matter. Use \( H_3O^+ \) instead of \( H^+ \)).

\[ \_\_\_\_\_\_ + \_\_\_\_\_\_ \rightarrow \_\_\_\_\_\_ + \_\_\_\_\_\_ \]

#### Additional Instructions:
- Ensure to use the correct coefficient.
- Ignore states of matter.

### Explanation:

A buffer solution typically contains a weak acid and its conjugate base. In this case, HF (a weak acid) and KF (which provides \(K^+\) that doesn't react, leaving \(F^-\) as the conjugate base) make up the buffer system.

To find the pH, use the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{Conjugate Base}]}{[\text{Acid}]} \right) \]

Given:
- \( [HF] = 0.402 \, \text{M} \)
- \( [F^-] = 0.402 \, \text{M} \)
- \( K_a = 7.20 \times 10^{-4} \)

First, find \( pK_a \):
\[ pK_a = -\log(K_a) \]
\[ pK_a = -\log(7.20 \times 10^{-4}) \]
\[ pK_a \approx 3.14 \]

Substitute the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 3.14 + \log \left( \frac{0.402}{0.402} \right) \]
\[ \text{pH} =
Transcribed Image Text:### Buffer Solution Problem #### Buffer Composition: A buffer solution is composed of: - 0.402 M in HF - 0.402 M in KF #### Problem Statement: If the ionization constant (\(K_a\)) for HF is \(7.20 \times 10^{-4}\), determine the pH of the buffer solution. \[ \text{pH} = \_\_\_\_\_\_ \] #### Equation: Write the net ionic equation for the reaction when 0.107 mol of \(HNO_3\) is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients and omit states of matter. Use \( H_3O^+ \) instead of \( H^+ \)). \[ \_\_\_\_\_\_ + \_\_\_\_\_\_ \rightarrow \_\_\_\_\_\_ + \_\_\_\_\_\_ \] #### Additional Instructions: - Ensure to use the correct coefficient. - Ignore states of matter. ### Explanation: A buffer solution typically contains a weak acid and its conjugate base. In this case, HF (a weak acid) and KF (which provides \(K^+\) that doesn't react, leaving \(F^-\) as the conjugate base) make up the buffer system. To find the pH, use the Henderson-Hasselbalch equation: \[ \text{pH} = pK_a + \log \left( \frac{[\text{Conjugate Base}]}{[\text{Acid}]} \right) \] Given: - \( [HF] = 0.402 \, \text{M} \) - \( [F^-] = 0.402 \, \text{M} \) - \( K_a = 7.20 \times 10^{-4} \) First, find \( pK_a \): \[ pK_a = -\log(K_a) \] \[ pK_a = -\log(7.20 \times 10^{-4}) \] \[ pK_a \approx 3.14 \] Substitute the values into the Henderson-Hasselbalch equation: \[ \text{pH} = 3.14 + \log \left( \frac{0.402}{0.402} \right) \] \[ \text{pH} =
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