An analytical chemist is titrating 73.6 mL of a 0.8200M solution of benzoic acid (HC H, CO₂) with a 0.3800M solution of KOH. The p K of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. D Round your answer to 2 decimal places. olo Ar pH = 0 X B

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**Titration Challenge: Determining pH after Addition of KOH**

An analytical chemist is titrating 73.6 mL of a 0.8200 M solution of benzoic acid (HC6H5CO2) with a 0.3800 M solution of KOH. The \( pK_a \) of benzoic acid is 4.20. Determine the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it.

**Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.

**Instructions**: Round your answer to 2 decimal places.

In the input field provided, enter the calculated pH value.

### Example Calculation:
1. **Determine initial moles of benzoic acid:**
   \[ \text{Volume of acid} \times \text{Concentration of acid} = 0.0736 \, \text{L} \times 0.8200 \, \text{mol/L} = 0.060352 \, \text{mol} \]

2. **Determine moles of KOH added:**
   \[ \text{Volume of KOH} \times \text{Concentration of KOH} = 0.177 \, \text{L} \times 0.3800 \, \text{mol/L} = 0.06726 \, \text{mol} \]

3. **Perform stoichiometry to find moles of acid remaining and moles of salt formed:**
   \[ 0.060352 \, \text{mol acid} - 0.06726 \, \text{mol KOH} = -0.006908 \, \text{mol acid remaining (indicating excess base)} \]

4. **Calculate the concentration of excess OH\(^-\) ions in the final solution:**
   \[ 0.006908 \, \text{mol OH}^-/ (0.0736 \, \text{L} + 0.177 \, \text{L}) = 0.0290088 \, \text{mol/L} \]

5. **Determine the pOH of the solution:**
   \[ \text{pOH} = -\log(
Transcribed Image Text:**Titration Challenge: Determining pH after Addition of KOH** An analytical chemist is titrating 73.6 mL of a 0.8200 M solution of benzoic acid (HC6H5CO2) with a 0.3800 M solution of KOH. The \( pK_a \) of benzoic acid is 4.20. Determine the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it. **Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. **Instructions**: Round your answer to 2 decimal places. In the input field provided, enter the calculated pH value. ### Example Calculation: 1. **Determine initial moles of benzoic acid:** \[ \text{Volume of acid} \times \text{Concentration of acid} = 0.0736 \, \text{L} \times 0.8200 \, \text{mol/L} = 0.060352 \, \text{mol} \] 2. **Determine moles of KOH added:** \[ \text{Volume of KOH} \times \text{Concentration of KOH} = 0.177 \, \text{L} \times 0.3800 \, \text{mol/L} = 0.06726 \, \text{mol} \] 3. **Perform stoichiometry to find moles of acid remaining and moles of salt formed:** \[ 0.060352 \, \text{mol acid} - 0.06726 \, \text{mol KOH} = -0.006908 \, \text{mol acid remaining (indicating excess base)} \] 4. **Calculate the concentration of excess OH\(^-\) ions in the final solution:** \[ 0.006908 \, \text{mol OH}^-/ (0.0736 \, \text{L} + 0.177 \, \text{L}) = 0.0290088 \, \text{mol/L} \] 5. **Determine the pOH of the solution:** \[ \text{pOH} = -\log(
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