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]

NH4 Cl    +  NaOH    ⟶  NH4OH+NaCl mmol50×0.275×0.1  =10=7.5 mmol10−7.5−7.5 =2.5  ⇒This will result in a basic buffer.⇒pOH=pKb+log⁡[ Salt ][ Base ]⇒pH=14−4.27=9.73