Posted by jain on Monday, October 17, 2011 at 6:31pm.
The secret to these titration problems is to recognize where you are on the titration curve.
a)at the beginning of the titration you have pure triethylamine.
..........R3N + H2O ==> R3NH+^ + OH^-
initial...0.1M............0........0
change.....-x.............x.........x
equil....0.1-x............x..........x
Kb = (R3NH^+)(OH^-)/(R3N)
Substitute into Kb expression and solve for x, then convert to pH.
b), c), and d
mL x M = millimoles
mmoles R3N = 20.00 x 0.1000 = 2.00 mmols.
mmoles HCl = 10.00 x 0.1000 = 1.00 mmols.
...........R3N + HCl ==>R3NH^+ + Cl^-
initial....2.00
add...............1.00
change......-1.00..-1.00..1.00..1.00
equil.....1.00......0......1.00..1.00
Use Kb expression from a) and substitute as in this ICE table OR use the Henderson-Hasselbalch equation and solve for H^+ and/or pH.
c and d are worked the same as b.
e. At 20.00 mL acid you are at the equivalence point. The pH is determined by the hydrolysis of the salt. The salt will be 2.00 mmoles and the volume will be 40 mL; therefore, (R3NH^+) = 20/40 = 0.05M
........R3NH^+ + H2O ==> H3O^+ + R3N
initiial...0.05...........0.......0
change....-x..............x........x
equil....0.05-x............x........x
Ka = (Kw/Kb) = (H3O^+)(R3N)/(RcNH^+)
Substitute 1E-14 for Kw, Kb for the R3N which is listed in the problem, x and x for the numerator and 0.05-x for the denominator. Solve for x and convert to pH.
f)You have used all of the triethylamine (at 20.00 mL) so this last 5.00 mL is simply excess HCl. Since that is a strong acid, calculate the concn of HCl at that point and convert to pH.