chem acid base equilibrium (III)

Q1
NH3 + H2O <===> NH4+ + OH-
Kb = [NH4+][OH-] / [NH3]
no. of mole of NH3 = 0.1X(150/1000) = 0.015 M
no. of mole of NH4+ = 0.1X(100/1000) = 0.01
Since they are in the same volume of solution;
[NH4+] / [NH3] = 0.01 / 0.015
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = [OH-](0.01 / 0.015) ; [OH-] = 2.7x10^-5
pOH = 4.57 ; pH = 9.43
Q2
(a) before the addition of NaOH
HA <===> H+ + A-
[HA] = 0.125 - x
[H+] = x
[A-] = x
Ka = 1.8x10^-5 = [H+][A-] / [HA] = x^2 / (0.125-x)
x^2 = 2.56X10^-6 - 1.8x10^-5x
x^2 + 1.8x10^-5x - 2.56X10^-6 = 0
x = 1.589x10^-3 or -1.61x10^-3(rejected)
pH = -log(1.589x10^-3) = 2.80
(b)after the addition of NaOH
no. of mole of acid in 0.125M 1dm3 acid = 0.125
no. of mole of NaOH added = 0.1
no. of mole of A- prodeuced = 0.1
no. of mole of HA remain = 0.125-0.1 = 0.025
pH = p Ka + log([A-]/[HA]) = 4.74 + log(0.1/0.025) = 5.34

Q1

Consider dilution in mixing two solutions.
In the solution, [NH3]o = 0.1 x (150/250) = 0.06 M
and [NH4^+]o = 0.1 x (100/250) = 0.04 M

Assume that the changes in [NH3] and [NH4+] are negligible.
.............. NH3 + H2O = NH4^+ . + . OH^-
At eqm .. 0.06 ................ 0.04 ............ y ............ (M)

Kb = 0.04y / 0.06 = 1.8 x 10^-5
[OH^-] = y = 2.7 x 10^-5 mol dm^-3
pOH = -log[OH^-] = 4.57
pH = 14 – 4.57 = 9.43

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Q2(a)

.............. CH3COOH .. = .. H^+ .. + ..CH3COO^-
At eqm .... 0.125-z............... z ................. z ............. M

Ka = z^2/(0.125-z) = 1.8 x 10^-5
[H^+] = z = 1.49 x 10^-3 mol dm^-3
pH = -log[H^+] = 2.83

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Q2(b)

Assume that volume change is negligible.

Part of CH3COOH reacts with NaOH to give CH3COO^-.
[CH3COOH]o = 0.125 – 0.1 = 0.025 M
[CH3COO^-]o = 0.1 M

Assume that the changes in [CH3COOH] and [CH3COO^-] are negligible.
.............. CH3COOH .. = .. H^+ .. + ..CH3COO^-
At eqm .... ~0.025............... w .................~0.1 ............. M

Ka = 0.1w / 0.025 = 1.8 x 10^-5
[H^+] = w = 4.5 x 10^-6 mol dm^-3
pH = -log[H^+] = 5.35