chemistry equil. ~Highhand 請入

Data needed are shown below, which may be varied from different sources.

http://www.ktf-split.hr/periodni/en/abc/kpt.html
Ksp[Mg(OH)2(s)] = 5.61 x 10^-12 M^3
(This data is needed in method 3 only.)

http://www.scribd.com/doc/94270/Acids-and-Bases
Kb[NH3(aq)] = 1.8 x 10^-5 M
Hence, Ka[NH4^+(aq)] = Kw/Kb[NH3] = (1 x 10^-14)/(1.8 x 10^-5) = 5.56 x 10^-10 M
(This data is needed in method 2 only.)

=====
Method 1 :
NH3(aq) + H2O(l) = NH4^+(aq) + OH^-(aq) .. Kb= 1.8 x 10^-5 M

The reaction starts from the backward reaction. At the start of the reaction :
[NH4^+]o = 1 M
[NH3]o = 0 M

At eqm :
Let [NH3] = p M
Then [NH4^+] = (1 - p) M
pOH = 14 - pH = 14 - 9 = 5
[OH^-(aq)] = 10^-5 = 1 x 10^-5 M

Kb = [NH4^+][OH^-]/[NH3]
(1 - p)(1 x 10^-5)/p = 1.8 x 10^-5
(1 - p)/p = 1.8
1 - p = 1.8p
2.8p = 1
p = 0.357 M
[NH3] = 0.357 M
[NH4^+] = 1 - 0.357 = 0.643 M

=====
Method 2 :
NH4^+(aq) = NH3(aq) + H^+(aq) .. Ka = 5.56 x 10^-10 M

At the start of the reaction :
[NH4^+]o = 1 M
[NH3]o = 0 M

At eqm :
Let [NH3] = w M
Then [NH4^+] = (1 - w) M
[H^+] = 10^-9 = 1 x 10^-9 M

Ka = [NH3][H^+]/[NH4^+]
w(1 x 10^-9)/(1 - w) = 5.56 x 10^-10
w/(1 - w) = 0.556
w = 556 - 0.556w
1.556w = 0.556
w = 0.357
[NH3] = 0.357 M
[NH4^+] = 1 - 0.357 = 0.643 M

=====Method 3 :

Consider the solubility eqm of Mg(OH)2:
Mg(OH)2(aq) = Mg^2+(aq) + 2OH^-(aq) .. Ksp[Mg(OH)2(s)]

At eqm :
pOH = 14 - pH = 14 - 9 = 5
[OH^-(aq)] = 10^-5 = 1 x 10^-5 M
Ksp = [Mg^2+](1 x 10^-5)^2 = 5.61 x 10^-12
Hence, [Mg^2+] = 0.0561 M

Consider both the solubility eqm of Mg(OH)2 and the dissociation of NH3 :
Mg(OH)2(aq) = Mg^2+(aq) + 2OH^-(aq) .. Ksp[Mg(OH)2(s)]
2NH4^+(aq) + 2OH^-(aq) = 2NH3(aq) + 2H2O(l) .. 1/Kb[NH3(aq)]^2

Add the above two equations:
Mg(OH)2­(s) + 2NH4^+(aq) = Mg^2+(aq) + 2NH3(aq) + 2H2O(l)
Eqm constant K
= Ksp[Mg(OH)2(s)]/Kb[NH3(aq)]^2
= (5.61 x 10^-12)/(1.8 x 10^-5)^2
= 0.0173 M^-1

At eqm :
Let [NH3] = y M
Then [NH4^+] = (1 - y) M
[Mg^2+] = 0.0561 M

K = [Mg^2+][NH3]^2/[NH4^+]^2
(0.0561)y^2/(1 -y)^2 = 0.0173
y/(1 - y) = 1/1.8
1 - y = 1.8y
2.8y = 1
y = 0.357
[NH3] = 0. 357 M
[NH4^+] = 1 - 0.357 = 0.643 M