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: \[\boxed{\text{pH} = 9.55}\]