Calculation of pH

Refer to: http://www.jesuitnola.org/upload/clark/refs/aqueous.htm
Ka1 of H2CO3 = 4.3 x 10-7 M
Ka2 of H2CO3 = 5.6 x 10-11 M
Ksp of Mg(OH)2 = 1.6 x 10-12 M3
Kw of water = 1 x 10-14 M2

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Calculate the pH of a solution of 2.5 M Na2CO3.

Consider the reactions that affect the pH of the solution:
CO32-(aq) + H2O(l) ≒ HCO3-(aq) + OH-(aq) **Kh1 = Kw/Ka2 = 1.79 x 10-4 M
HCO3-(aq) + H2O(l) ≒ H2CO3­(aq) + OH-(aq) **Kh2 = Kw/Ka1 = 2.33 x 10-8 M
HCO3-(aq) + H2O(l) ≒ CO32-­(aq) + H3O+(aq) **Ka2 = 5.6 x 10-11 M
Since Kh1 >> Kh2 >> Ka2, thus only the first reaction is considered.

CO32-(aq) + H2O(l) ≒ HCO3-(aq) + OH-(aq) **Kh1 = 1.79 x 10-4 M
[CO32-(aq)]o = 2.5 M
Let the concentration of OH- at equilibrium be y M.
At equilibrium:
[HCO3-] = [OH-] = y M
[CO32-] = (2.5 - y) M

Kh1 = y2/(2.5 - y) = 1.79 x 10-4
y2 = (4.48 x 10-4) - (1.79 x 10-4)y
y2 + (1.79 x 10-4)y - (4.48 x 10-4) = 0
y = 0.021
[OH-] = 0.021 M

[H3O+] = Kw/[OH-] = (1 x 10-14)/(0.021) = 4.76 x 10-13 M
pH = -log[H3O+] = -log(4.76 x 10-13) = 12.3

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Calculate the pH of 3.3 M of Mg(OH)2.

3.3 M of Mg(OH)2 does not exist, because Mg(OH)2 is only sparingly soluble in water.
If 3.3. M Mg(OH)2 exists:
[OH-] = 3.3 x 2 = 6.6 M
[H3O+] = Kw/[OH-] = (1 x 10-14)/(6.6) = 1.52 x 10-15
pH = 14.8

Calculate the pH of saturated Mg(OH)2:
Mg(OH)2(s) ≒ Mg2+(aq) + 2OH-(aq)
At equilibrium:
Let [Mg2+] be y M.
Then, [OH-] = 2y M

Ksp = y(2y)2 = 1.6 x 10-12
4y3 = 1.6 x 10-12
y = 7.37 x 10-5
[OH-] = 7.37 x 10-5 M
[H3O+] = Kw/[OH-] = (1 x 10-14)/(7.37 x 10-5) = 1.36 x 10-10 M
pH = 9.87
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