5.73 is the answer
Explanation:
Here, moles of ethyl amine = 0.240 L x 0.7318 M = 0.175 moles
To reach equivalence point, 0.175 moles of HNO3 must be added
Therefore,
The volume of HNO3 that must be added = 0.177 moles / 0.3280 M = 535 mL
Total volume = 240 + 535 = 775 mL
Concentration of salt = 0.175 moles / 0.775 L = 0.225 M
We have,
pOH = [tex]=(1 / 2)[\mathrm{p} \mathrm{E} \mathrm{w}+\mathrm{p} \mathrm{K} \mathrm{b}+\log \mathrm{C}]=(1 / 2)[14+3.19+\log 0.225]=8.27[/tex]
pH = 14 – pOH = 14-8.27 = 5.73
pH = 5.73
