Question

25.4 mL of 0.12 M methylamine (CH3NH2) is titrated with 0.15 M HCI. (a) What is the pH after 14.2 mL of the HCI has been added? (b) What is the pH at the equivalence point in the titration? K(CH3NH2) = 4.4 x 10-4

Answer 1

Notice that methylamine CH3NH2 is a weak base while HCl is a strong acid. And, when they are mixed, a neutralisation reaction will occur which can be written as CH3NH2(aq) + H+ (aq) CH2NH3(aq) Initial amount of CH3NH2 = 25.4ml * 0.12M = 3.048mmol In part (a), the amount of HCl added = 14.2ml 0.15M = 2.13mmol Doing stoichiometry, CH3NH2(aq) + (aq) = CH3NH3(aq) 3.048mmol 2.13mmol 0 Initial Final 0.918mmol 0 2.13mmol Now using the pOH equation for basic buffer. [salt] pOH = pkp + log10 base] 2.13 pOH = -log10 (4.4 x 100+) + log10918 pOH = 3.36+0.36 pOH = 3.72 Therefore pH = 14 -3.72 = 10.28

In part (b) amount of HCI added will be same as initial amount of CH3NH, because equivalence point is to be reached. Volume of HCl required = 3.048mmol/0.15M = 20.32ml Total volume of solution = 20.32ml+25.4ml = 45.72ml Doing stoichiometry, CH3NH2(aq) + H+ (aq) CH3NH3(aq) 3.048mmol 2.13mmol Ommol Initial Final Ommol Ommol 3.048mmol In this case we are left with 3.048mmol CH3NH3(aq) in 45.72ml solution. Now, the hydrolysis of this ion becomes important. Initial molarity of CH3NH3(aq) = 3.048mmol/45.72ml = 0.066M

The hydrolysis reaction will be CH NH3(aq) CH3NH2(aq) + H+ (aq) The equilibrium constant for this reaction will be Kw/Kb, which will be 10-14/4.4x104 = 2.27x10-11. Doing stoichiometry, CH3NH3(aq) CH3NH2(aq) + H+ (aq) K = 2.27x10-11. Initial conc. 0.066M OM OM Final conc. 0.066(1-a)M 0.066M 0.0660M

where a is degree of hydrolysis which will be very small compared to 1, looking at the K value of reaction On solving, 0.066a x 0.066a -= 2.27 x 10-11 0.066(1 - a) a = 1.85 x 10-5 Therefore [H] = 0.066x1.85x10-5 = 1.22x10-6 Hence pH = -log10(1.22*10*)=5.91