Page 74 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

58 Chapter 5
Solution:

1. Acid/base reaction.
2. (CH3)3N-weak base; HCl-strong acid.
3. Acid/base reaction: (CH3)3N + HCl --$ (CH3)3NH' + C1-.
4. Amount of (CH3)3N (expressed in moles) = (15 x 0.35)/1000
= 5.25 x amount of HC1 (in moles) = (15 x 0.25)/1000
= 3.75 10-3.
5. (CH3)3N + HC1 4 (CH3)3NH'+ C1-

Initial 5.25 x 3.75 x 0 0
End (5.25 - 3.75) x
= 1.5 x 10-3 o 3.75 10-3 3.75 x 10-3

6. Molarity of (CH3)3N = [amount of (CH3)3N (in moles) x
1000]/total volume (in cm3)]. The total volume is now (15 + 15)
cm3 = 30 cm3. Therefore, the molarity of (CH3)3N = (1.5 x
x 1000)/30 = 0.05 M. Molarity of (CH3)3NHf = (3.75 x
x 1000)/30 = 0.125 M.
7. [(CH3)3NH'] = 0.125 M > [(CH3)3Nl = 0.05 M. Therefore,
cation hydrolysis occurs!
8. (CH3)3NH' + H20 * (CH3)3N + H30+.
This is the equilibrium reaction,
where Kh = ([(CH3)3~[H3o'l}/[(CH3)3NH'l.
9. (CH3)3NHf + H20 + (CH3)3N + H30f
Initial 0.125 0.05 0
Change -x +X +X
Final (0.125 - x) (0.05 + x) x
10. Kh = ([(CH3)3~[H30'l}/[(CH,),NH '1 = ((Omo5 + x>(x)>/
(0.12%~).
11. K, = Kb Kh Kh = KW/& = 10-14/(7.4 X w5) = 1.351 X
10-l0.
12. Assume x << 0.05 + (0.05 x)/0.125 = 1.351 x 1O-Io, i.e. x =
3.378 x lo-'' M.
Therefore, (3.378 x 10-1°)/(0.05)% = 6.8 x % and also,
(3.378 x 10-1°)/(0.125)% = 2.7 x %. Both are <5%,
hence the assumption made was justified.
13. [H30'] = x = 3.378 x lo-'' M.
pH = -log1o[H3Of] = -10g10(3.378 x lo-'') = 9.47.
Answer:pH = 9.47