Question

a) Calculate the pH during the titration of 30.00 mL of 0.1000 M trimethylamine, (CH₃)₃N(aq), with 0.1000 M HCl(aq) after 24 mL of the acid have been added. K_b of trimethylamine = 6.5 x 10^-5.

b)Calculate the pH during the titration of 20.00 mL of 0.1000 M methylamine, CH₃NH₂(aq), with 0.1000 M HNO₃(aq) after 19.9 mL of the acid have been added. K_b of methylamine = 3.6 x 10^-4.

Answer 1

a) mols of (CH3)3N = molarity x volume = 0.1 x 0.03 = 0.003 mols

moles of HCl = 0.1 x 0.024 = 0.0024 mols

reamining moles of (CH3)3N = 0.003 - 0.0024 = 0.0006 mols

moles of (CH3)3NHCl = 0.0024 mols

Total volume = 30 + 24 = 54 ml = 0.054 L

molarity of (CH3)3N = 0.0006/0.054 = 0.011 M

molarity of (CH3)3NHCl = 0.0024/0.054 = 0.044 M

Ka = Kw/Kb = 1 x 10^-14/6.5 x 10^-5 = 1.54 x 10^-10

pKa = -log[Ka] = 9.81

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

     = 9.81 + log(0.011/0.044)

     = 9.21

b) moles of CH3NH2 = 0.1 x 0.02 = 0.002 mols

moles of HNO3 - 0.1 x 0.0199 = 1.99 x 10^-3 mols

remaining moles of CH3NH2 = 0.002 - 0.00199 = 1 x 10^-5 mols

moles of CH3NH3NO3 = 0.00199 mols

total volume = 20 + 19.9 = 39.9 ml = 0.0399 L

molarity of CH3NH2 = 0.00001/0.0399 = 2.51 x 10^-4 M

molarity of CH3NH3NO3 = 0.0199/0.0399 = 0.05 M

Ka = 1 x 10^-14/3.6 x 10^-4 = 2.78 x 10^-11

pKa = 10.56

Using Hendersen-Haselbalck eqution,

pH = pKa + log([base]/[acid])

     = 10.56 + log(2.51 x 10^-4/0.05)

     = 8.26