Question

A 28.5 mL sample of 0.253 M diethylamine, (C2H5)2NH, is titrated with 0.247 M hydrobromic acid. After adding 13.0 mL of hydrobromic acid, the pH is_____

Answer 1

(C2H5)2NH + HBr ----> (C2H5)2NH·HBr

no of moles of (C2H5)2NH = 0.253M x 0.0285L = 0.00721 moles

no of moles of HBr = 0.247 M x 0.013L = 0.003211 moles

total volume = 28.5 + 13 = 41.5 mL = 0.0415L

from the balanced equation

onemole of it is clear that one mole of (C2H5)2NH can react with one mole of HBr

here HBr is limiting agent so

0.003211 moles of salt will form

concentration of salt = 0.003211 moles / 0.0415 L = 0.0774 M

no of moles remaining diethylamine = 0.00721 moles - 0.003211 = 0.004 moles

concentration of diethylamine = 0.004 moles / 0.0415 L = 0.0964 M

pKb of triethyl amine = 11.09

pKa = 14-11.09 = 2.91

now use the handerson equation

pH = pKa + log(salt/acid)

pH = 2.91 + log(0.0774 / 0.0964)

pH = 2.91 + (-0.095)

pH = 2.81