Question

A buffer solution contains 0.437 M CH_3NH_3CI and 0.213 M CH_3NH_2 (methylamine). Determine the pH change when 0.045 mol KOH is added to 1.00 L of the buffer. pH after addition - pH before addition = pH change =

Answer 1

pH = pKa + log [ salt / acid]

or pOH = pKb + log [ salt /base]

pKb of methylamine = 3.38

Intial pH :

pOH = 3.38 + log [ 0.437 / 0.213]

pOH = 3.69

pH = 14 - 3.69

pH = 10.31

After addition of 0.045 mol KOH

(CH3NH3Cl = ) CH3NH2.HCl + KOH ----> CH3NH2 + KCl + H2O

addition of KOH increases moles of CH3NH2 and decreases CH3NH3Cl

So after addition CH3NH2 = 0.213 + 0.045 = 0.258 mol / L = 0.258 M

CH3NH3Cl = 0.437 - 0.045 = 0.392 mol / L = 0.392 M

pOH = 3.38 + log [ 0.392 / 0.258]

pOH = 3.56

pH = 14 - 3.56 = 10.44

Change in pH = 10.44 - 10.31 = 0.13