50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :

Number of millimoles of NH3 = 50mL × 0.2M = 10 mmol
Number of millimoles of HCl = 25mL × 0.2M = 5 mmol

The reaction between ammonia (NH3) and hydrochloric acid (HCl) is:
NH3 + HCl → NH4Cl

Since HCl is a strong acid, it will completely react with NH3. After the reaction:
Millimoles of NH3 remaining = 10 mmol - 5 mmol = 5 mmol
Millimoles of NH4Cl formed = 5 mmol

This is a buffer solution containing a weak base (NH3) and its conjugate acid (NH4Cl). We can use the Henderson-Hasselbalch equation to calculate the pOH:
pOH = pKb + log([NH4+]/[NH3])

Since the volumes are the same, we can use millimoles directly:
pOH = 4.75 + log(5 mmol / 5 mmol)
pOH = 4.75 + log(1)
pOH = 4.75 + 0
pOH = 4.75

Now we can calculate the pH:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 4.75
pH = 9.25

Therefore, the pH of the mixture will be 9.25