A buffer solution is made that is 0.329 M in HF and 0.329 M in KF. If K, for HF is 7.20 x 10 what is the pH of the buffer solution? pH = Write the net ionic equation for the reaction that occurs when 0.093 mol KOH is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter.)
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![**Transcription and Explanation for Educational Purposes**
**Buffer Solution Problem:**
A buffer solution is made that is 0.329 M in HF and 0.329 M in KF.
- If \(K_a\) for HF is \(7.20 \times 10^{-4}\), what is the pH of the buffer solution?
\[ \text{pH} = \boxed{} \]
---
**Reaction Equation:**
Write the net ionic equation for the reaction that occurs when 0.093 mol KOH is added to 1.00 L of the buffer solution.
*(Use the lowest possible coefficients. Omit states of matter.)*
\[ \boxed{} + \boxed{} \rightarrow \boxed{} + \boxed{} \]
---
**Explanation and Guidance:**
1. **Calculating the pH of the Buffer Solution:**
- Use the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{pK}_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
- Where \([\text{A}^-]\) represents the concentration of the conjugate base (KF) and \([\text{HA}]\) represents the concentration of the weak acid (HF).
- Calculate \(\text{pK}_a\) from \(K_a\):
\[
\text{pK}_a = -\log(K_a) = -\log(7.20 \times 10^{-4})
\]
2. **Net Ionic Equation for Reaction with KOH:**
- Identify the reaction components and products. Potassium hydroxide (KOH) will react with the HF in the buffer solution.
- The net ionic equation typically focuses on the ions involved directly in the chemical reaction.
This approach provides a clear understanding of the chemical principles involved in buffer solutions and acid-base reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00585539-79e1-42a7-8a75-95bc6de8d9d5%2Ff7f4bab7-94d8-43d2-8b5c-06914afee251%2Fwym1wgf_processed.jpeg&w=3840&q=75)
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