First, calculate the moles of acetic acid (\(\mathrm{HCH_3CO_2}\)) and KOH.
\[
\text{Moles of acetic acid} = 0.2084 \, \text{M} \times 0.1800 \, \text{L} = 0.03751 \, \text{moles}
\]
\[
\text{Moles of KOH} = 0.5401 \, \text{M} \times V_{\text{KOH}} \, \text{L}
\]
At equivalence point, moles of KOH added will be equal to the moles of acetic acid.
\[
0.03751 \, \text{moles} = 0.5401 \, \text{M} \times V_{\text{KOH}} \, \text{L}
\]
Solving for \(V_{\text{KOH}}\):
\[
V_{\text{KOH}} = \frac{0.03751 \, \text{moles}}{0.5401 \, \text{M}} = 0.06944 \, \text{L} = 69.44 \, \text{mL}
\]
The total volume of the solution at equivalence point is the sum of the initial volume of acetic acid solution and the volume of KOH added.
\[
V_{\text{total}} = 180.0 \, \text{mL} + 69.44 \, \text{mL} = 249.44 \, \text{mL} = 0.2494 \, \text{L}
\]
At equivalence, all acetic acid has been converted to acetate ion (\(\mathrm{CH_3CO_2^-}\)).
\[
\text{Concentration of } \mathrm{CH_3CO_2^-} = \frac{\text{moles of } \mathrm{CH_3CO_2^-}}{V_{\text{total}}} = \frac{0.03751 \, \text{moles}}{0.2494 \, \text{L}} = 0.1504 \, \text{M}
\]
The acetate ion hydrolyzes in water:
\[
\mathrm{CH_3CO_2^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COOH} + \mathrm{OH^-}
\]
The equilibrium expression for this reaction is:
\[
K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.76}} = 1.74 \times 10^{-10}
\]
Using the ICE table and assuming \(x\) is the concentration of \(\mathrm{OH^-}\):
\[
K_b = \frac{x^2}{0.1504 - x} \approx \frac{x^2}{0.1504}
\]
\[
1.74 \times 10^{-10} = \frac{x^2}{0.1504}
\]
Solving for \(x\):
\[
x^2 = 1.74 \times 10^{-10} \times 0.1504 = 2.617 \times 10^{-11}
\]
\[
x = \sqrt{2.617 \times 10^{-11}} = 1.617 \times 10^{-6} \, \text{M}
\]
\[
\text{pOH} = -\log(1.617 \times 10^{-6}) = 5.79
\]
\[
\text{pH} = 14 - \text{pOH} = 14 - 5.79 = 8.21
\]
\(\boxed{\text{pH} = 8.21}\)
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