pH of a solution made by mixing 50 mL of 0.2 M NH4Cl and 75 mL of 0.1 M NaOH is [pKb of NH3(aq) = 4.74. log 3 = 0.47]

N+H4            +      NaOH     ⟶      NH4OH+NaCl mmol             50 × 0.2             75 × 0.1                          =10                       =75 mmol           10-7.5                         -                7.5 =2.5⇒This will result in a basic buffer.⇒pOH=pKb+log⁡[ Salt ][ Base ]=4.74+log⁡2.57.5=4.27⇒pH=14−4.27=9.73