Using the Henderson-Hasselbalch equation for a basic buffer:
\[
\text{pOH} = pK_b - \log \frac{[\text{base}]}{[\text{salt}]},
\]
and the relationship:
\[
\text{pH} = 14 - \text{pOH}.
\]
Substitute the given values:
\( pK_b = 4.75 \),
\([\text{base}] = 0.1 \, \text{M} \),
\([\text{salt}] = 0.05 \, \text{M} \).
Step 1: Substitute into the equation:
\[
\text{pOH} = 4.75 - \log \frac{0.1}{0.05}.
\]
Step 2: Simplify the logarithmic term:
\[
\frac{[\text{base}]}{[\text{salt}]} = \frac{0.1}{0.05} = 2.
\]
\[
\text{pOH} = 4.75 - \log 2.
\]
Using \( \log 2 \approx 0.301 \):
\[
\text{pOH} = 4.75 - 0.301.
\]
\[
\text{pOH} = 4.449.
\]
Step 3: Calculate the pH:
\[
\text{pH} = 14 - \text{pOH}.
\]
\[
\text{pH} = 14 - 4.449.
\]
\[
\text{pH} = 9.551.
\]
Final Answer:
\[
\boxed{\text{pH} = 9.55}
\]