(N/A) Given $pH = 4.50$ and initial concentration $C = 0.1 \ M$.
$1$. Calculate $[H^{+}]$:
$[H^{+}] = 10^{-pH} = 10^{-4.50} = 3.16 \times 10^{-5} \ M$.
$2$. Determine $[A^{-}]$ and $[HA]$ at equilibrium:
Since the acid is monobasic $(HA \rightleftharpoons H^{+} + A^{-})$,$[H^{+}] = [A^{-}] = 3.16 \times 10^{-5} \ M$.
$[HA]_{eq} = C - [H^{+}] = 0.1 - 3.16 \times 10^{-5} \approx 0.1 \ M$.
$3$. Calculate $K_{a}$:
$K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(3.16 \times 10^{-5})^{2}}{0.1} = \frac{9.9856 \times 10^{-10}}{0.1} \approx 1.0 \times 10^{-8}$.
$4$. Calculate $pK_{a}$:
$pK_{a} = -\log(K_{a}) = -\log(1.0 \times 10^{-8}) = 8.0$.