NH3+HCl⇌NH4Clmolesat t=050×.2×10−325×.2×10−3−molesat t=t25×.2×10−3025×.2×10−3
\therefore \mathrm{salt} & \mathrm{base} \mathrm{present} \mathrm{in} \mathrm{soluton}, \mathrm{so} \mathrm{it} \mathrm{is} \mathrm{basic} \mathrm{Buffer} \mathrm{solution}.
pOH=pkb(NH3)+logbasesalt=4.75∴pOH=Pkb=4.75
pH=14−pOH=14−4.75=9.25.