To calculate the pH of a buffer solution containing [tex]$12.3 \text{ grams}$[/tex] of sodium benzoate [tex]$(\text{NaC}_6\text{H}_5\text{COO})$[/tex] and [tex]$500.0 \text{ mL}$[/tex] of [tex]$0.175 \text{ M}$[/tex] benzoic acid, follow these steps:
1. Calculate the moles of sodium benzoate:
- Given data:
- Mass of sodium benzoate = [tex]$12.3 \text{ grams}$[/tex]
- Molar mass of sodium benzoate [tex]$(\text{NaC}_6\text{H}_5\text{COO})$[/tex] = [tex]$144.11 \text{ g/mol}$[/tex]
- Calculation:
[tex]\[
\text{Moles of sodium benzoate} = \frac{12.3 \text{ grams}}{144.11 \text{ g/mol}} = 0.085 \text{ mol}
\][/tex]
2. Calculate the moles of benzoic acid:
- Given data:
- Volume of benzoic acid solution = [tex]$500.0 \text{ mL} = 0.500 \text{ L}$[/tex]
- Concentration of benzoic acid = [tex]$0.175 \text{ M}$[/tex]
- Calculation:
[tex]\[
\text{Moles of benzoic acid} = 0.175 \text{ M} \times 0.500 \text{ L} = 0.0875 \text{ mol}
\][/tex]
3. Determine the pKa of benzoic acid:
- Given data:
- Dissociation constant (Ka) of benzoic acid = [tex]$6.46 \times 10^{-5}$[/tex]
- Calculation:
[tex]\[
\text{pKa} = -\log_{10}(6.46 \times 10^{-5}) = 4.19
\][/tex]
4. Use the Henderson-Hasselbalch equation to calculate the pH:
- Henderson-Hasselbalch equation:
[tex]\[
\text{pH} = \text{pKa} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\][/tex]
where [tex]$[\text{A}^-]$[/tex] is the concentration of the conjugate base (sodium benzoate) and [tex]$[\text{HA}]$[/tex] is the concentration of the acid (benzoic acid).
- Calculation:
[tex]\[
\frac{[\text{A}^-]}{[\text{HA}]} = \frac{\text{Moles of sodium benzoate}}{\text{Moles of benzoic acid}} = \frac{0.085 \text{ mol}}{0.0875 \text{ mol}} = 0.97
\][/tex]
[tex]\[
\text{pH} = 4.19 + \log_{10}(0.97) \approx 4.18
\][/tex]
Hence, the pH of the buffer solution is approximately [tex]\( 4.18 \)[/tex].
