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Equilibrium II 57
3. Acid/base reaction: CH3C02H+NaOH-,CH3CO; + H20+Na+
(Na+ is a spectator ion, i.e. an ion not actively involved in the
equilibrium expression).
4. Amount of CH3C02H (expressed in moles) = (35 x 0.1)/1000
= 3.5 x Amount of NaOH (expressed in moles) = (15 x
o.i)/iooo = 1.5 x 10-3.
5. CH3C02H + NaOH 4 CH3C0, + H20 + Nat
Initial 3.5 x~O-~ 1.5~10-~ 0 0
End (3.5-1.5)~ 1 0-3
=2.0~10-~ o 1.5 x~O-~ 1.5 x1OV3

6. Molarity of CH3C02H = [amount (in moles) x 1000]/[total
volume (in cm3)]. The total volume is now (35 + 15) cm3 = 50
cm3. Therefore, molarity of CH3C02H = (2 x x 1000)/50
= 0.04 M and the molarity of CH3C0, = (1.5 x x lOOO)/
50 = 0.03 M.
7. Although the concentration of acid, [CH3C02H] = 0.04 M, is
only just greater than the concentration of anion, [CH3C0~] =
0.03 M, acid dissociation is the dominant reaction which occurs!
8. CH3C02H + H20 + CH3C02 + H30+. This is the equili-
brium reaction, where Ka = ([H30+][CH3CO,]}/[CH3CO2H].
9. CH3C02H + H20 * CH3CO: + H30+

Initial 0.04 0.03 0
Change -x +X +X
Final (0.04 - x) (0.03 + x) x

10. Ka = {[CH3CO,][H30'])/[CH,C02H] = ((0.03 + x)(x))/
(0.04-x) = 1.8 x loa5. Assume x << 0.04 =+ (0.03x))/0.04 =
1.8 x i.e. x = 2.4 x
Therefore, (2.4 x 10-5)/(0.03)% = 0.08% and (2.4 x low5)/
(0.04)% = 0.06%. Both are < 5%, hence the assumption made
was justified.
11. W3O+] = x = 2.4 x M. pH =-logIo[H30+] =-loglo
(2.4 x = 4.62.

Answer:pH = 4.62

Example No. 2: In a titration of 15 cm3 of 0.35 M (CH3)3N with
0.25 M HCl, what is the pH of the solution after the addition of
15 cm3 of 0.25 M HCI, given that &(CH3)3N = 7.4 x