To find the pH of the resulting buffer when 500.0 mL of 0.180 M NaOH is mixed with 555 mL of 0.200 M weak acid (with [tex]\( K_{\text{a}} = 7.22 \times 10^{-5} \)[/tex]), we follow these steps:
1. Convert Volumes to Liters:
- Volume of NaOH: [tex]\( 500.0 \, \text{mL} = 0.500 \, \text{L} \)[/tex]
- Volume of Weak Acid (HA): [tex]\( 555 \, \text{mL} = 0.555 \, \text{L} \)[/tex]
2. Calculate Initial Moles:
- Moles of NaOH ([tex]\( \text{OH}^- \)[/tex]): [tex]\( 0.500 \, \text{L} \times 0.180 \, \text{M} = 0.090 \, \text{moles} \)[/tex]
- Moles of Weak Acid ([tex]\( \text{HA} \)[/tex]): [tex]\( 0.555 \, \text{L} \times 0.200 \, \text{M} = 0.1110 \, \text{moles} \)[/tex]
3. Reaction Stoichiometry:
[tex]\[ \text{HA} (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O} (l) + \text{A}^- (aq) \][/tex]
- NaOH completely dissociates and reacts with the weak acid.
- Moles of conjugate base ([tex]\( \text{A}^- \)[/tex]) formed is equal to the moles of [tex]\( \text{OH}^- \)[/tex], which is [tex]\( 0.090 \, \text{moles} \)[/tex].
- Moles of HA remaining: [tex]\( 0.1110 \, \text{moles} - 0.090 \, \text{moles} = 0.0210 \, \text{moles} \)[/tex]
4. Total Volume of the Solution:
[tex]\[ 0.500 \, \text{L} + 0.555 \, \text{L} = 1.055 \, \text{L} \][/tex]
5. Concentrations in the Final Solution:
- Concentration of [tex]\( \text{HA} \)[/tex]: [tex]\( \frac{0.0210 \, \text{moles}}{1.055 \, \text{L}} = 0.0199 \, \text{M} \)[/tex]
- Concentration of [tex]\( \text{A}^- \)[/tex]: [tex]\( \frac{0.090 \, \text{moles}}{1.055 \, \text{L}} = 0.0853 \, \text{M} \)[/tex]
6. Calculate [tex]\( \text{pK}_\text{a} \)[/tex]:
[tex]\[ \text{pK}_\text{a} = -\log_{10}(K_\text{a}) = -\log_{10}(7.22 \times 10^{-5}) = 4.14 \][/tex]
7. Use the Henderson-Hasselbalch Equation:
[tex]\[ \text{pH} = \text{pK}_\text{a} + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Substitute the values:
[tex]\[ \text{pH} = 4.14 + \log_{10} \left( \frac{0.0853}{0.0199} \right) \][/tex]
8. Calculate the pH:
[tex]\[ \log_{10} \left( \frac{0.0853}{0.0199} \right) = \log_{10}(4.29) = 0.633 \][/tex]
[tex]\[ \text{pH} = 4.14 + 0.633 = 4.773 \][/tex]
Thus, the pH of the resulting buffer is [tex]\(\boxed{4.77}\)[/tex].
