The acid dissociation K, of benzoic acid (C6H5CO,H) is 6.3 × 10 –³. Calculate the pH of a 1.5 × 10¯°M aqueous solution of benzoic acid. Round your answer to 2 decimal places.
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![### Calculating the pH of a Benzoic Acid Solution
**Problem Statement:**
The acid dissociation constant, \( K_a \), of benzoic acid \( \left( C_6H_5CO_2H \right) \) is \( 6.3 \times 10^{-5} \).
Calculate the pH of a \( 1.5 \times 10^{-3} \, M \) aqueous solution of benzoic acid. Round your answer to 2 decimal places.
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**Instructions:**
1. **Equation for \( K_a \):**
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
where \( [H^+] \) is the concentration of hydrogen ions, \( [A^-] \) is the concentration of the conjugate base, and \( [HA] \) is the concentration of the acid.
2. **Initial concentration:**
- The initial concentration of benzoic acid, \( [C_6H_5CO_2H] \), is \( 1.5 \times 10^{-3} \, M \).
3. **Assumption and approximation:**
- Assume that the concentration of \( H^+ \) is equal to that of \( A^- \) due to the dissociation.
- Let \( x \) represent the concentration of \( H^+ \).
4. **Simplified expression:**
\[
K_a \approx \frac{[H^+]^2}{[HA - x]} \approx \frac{x^2}{[HA]}
\]
5. **Substituting known values:**
\[
6.3 \times 10^{-5} = \frac{x^2}{1.5 \times 10^{-3} - x}
\]
Assuming \( x \) is much smaller than \( 1.5 \times 10^{-3} \), simplify:
\[
6.3 \times 10^{-5} \approx \frac{x^2}{1.5 \times 10^{-3}}
\]
6. **Solving for \( x \) or \( [H^+] \):**
\[
x^2 = 6.3 \times 10^{-5} \times 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feadc81de-613f-46f4-aced-a703abfea63b%2Fda01b38a-7acf-4dd7-a5cb-96d3a50345c7%2F1qxsn3.png&w=3840&q=75)
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