In a buffer solution, the pH is given by the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)
\]
Given that the pK\(_a\) is related to pK\(_b\) by the equation pK\(_a\) + pK\(_b\) = 14, we first find pK\(_a\) by substituting the given pK\(_b\):
\[
\text{pK}_a = 14 - \text{pK}_b = 14 - 5 = 9
\]
The concentrations of the base (NH\(_3\)OH) and the acid (NH\(_4\)Cl) are both 0.1 M and 0.01 M.
Thus, we apply the equation:
\[
\text{pH} = 9 + \log \left( \frac{0.1}{0.01} \right) = 9 + \log(10) = 9 + 1 = 10
\]
Thus, the pH is 10.