Answer: 8.78
pKa(NH4+)=9.26" role="presentation" style="position: relative;">pKa(NH+4)=9.26pKa(NH4+)=9.26pK_a(NH^+_4) = 9.26
∴pKb(NH3)=pKw−pKa(NH4+)" role="presentation" style="position: relative;">∴pKb(NH3)=pKw−pKa(NH+4)∴pKb(NH3)=pKw−pKa(NH4+)\therefore pK_b (NH_3) = pK_w - pK_a(NH^+_4) = 14 - 9.26 = 4.74
On adding 15 m mol (50 mL of 0.3 M) HCl to 20 m mol (50 mL of 0.4 M) NH3" role="presentation" style="position: relative;">NH3NH3NH_3 the resulting solution contains 15 m mol (in 100 mL) of NH4Cl" role="presentation" style="position: relative;">NH4ClNH4ClNH_4Cl and 5 m mol (in 100 mL) of NH3" role="presentation" style="position: relative;">NH3NH3NH_3.
∴[Salt]=15×10−2M,[Base]=5×10−2M" role="presentation" style="position: relative;">∴[Salt]=15×10−2M,[Base]=5×10−2M∴[Salt]=15×10−2M,[Base]=5×10−2M\therefore [Salt] = 15 \times 10^{-2} M , [Base] = 5 \times 10^{-2} M
pOH=pKb+log[Salt][Base]=4.74+log15×10−25×10−2=4.74+log3=4,74+0.48=5.22" role="presentation" style="position: relative;">pOH=pKb+log[Salt][Base]=4.74+log15×10−25×10−2=4.74+log3=4,74+0.48=5.22pOH=pKb+log[Salt][Base]=4.74+log15×10−25×10−2=4.74+log3=4,74+0.48=5.22pOH = pK_b + log \frac{[Salt]}{[Base]} = 4.74 + log \frac{15 \times 10^{-2}}{5 \times 10^{-2}} = 4.74 + log 3 = 4,74 + 0.48 = 5.22
pH = 14 - pOH = 14 - 5.22 = 8.78