Question

For 290.0 mL of a buffer solution that is 0.280 M in CH3CH2NH2 and 0.250 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Answer 1

CH3CH2NH2 , pKb = 4.3 x 10^-4

pKb = 3.37

pOH = pKb + log [CH3CH2NH3Cl]/[CH3CH2NH2]

pOH = 3.37 + log (0.250 / 0.280)

pOH = 3.32

pH + pOH = 14

pH = 10.68

initial pH = 10.68

moles of base = 0.280 x 290 / 1000 = 0.0812

moles of salt = 0.250 x 290 / 1000 = 0.0725

on addition of C moles of strong base to the basic buffer base moles increases and salt moles decreases

now new pOH = pKb + log [salt -C]/[base + C]

pOH = 3.37 + log [0.0725 -0.01] / [0.0812 + 0.01]

pOH = 3.21

pH + pOH = 14

pH = 10.79

final pH = 10.79