An ammonia-ammonium chloride buffer, pH = 9. Concentration of ammonia = 0.25M. By how much the pH will change if 75ml 0.1M KOH be added to 200ml buffer solution. Kb = 2 x 10^-5

For Ammonium chloride and ammonia buffer:
so pOH = pK_b + log [NH₄Cl]/[NH₃] 
pH = 14-pOH
​Since the initial pH is given we will directly calculate change in pH after adding NaOH.
After adding NaOH following reaction takes place:
NH₄Cl + NaOH -------> NH₃ + NaCl + H₂O 
Concentration of NH₄Cl will decrease and concentration of NH₃ will increases.
No. of moles of NH₄Cl in 200 ml of buffer solution = 0.25/1000 * 200 = 0.05 moles
No. of moles of NH₃ in 200 ml of buffer solution = 0.25/1000 * 200 = 0.05 moles
No of moles of KOH added = 0.1/1000* 75 = 0.0075
No. of moles of NH₄Cl after adding KOH = 0.05-0.0075 = 0.0425 moles
No. of moles of NH₃ after adding KOH = 0.05 + 0.0075 = 0.0575 moles
Pkb = -log (kb) = - log(2 x 10^-5)= 4.7
putting the values in above equation to calculate pOH:
pOH = 4.7 + log [0.0425]/log [0.0575]
pOH = 4.7 + log [0.0425]- log [0.0575]
pOH = 4.7 - 0.13 = 4.57
pH = 14 - 4.57 = 9.43
change in pH = 9.43 - 9 = 0.43