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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Illa Tristia Illa Tristia
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11 years ago
Calculate the pH of 0.32 M K2CO3. What are the concentrations of HCO3-, and H2CO3 in the solution? H2CO3 is a diprotic acid with Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11.
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#1
11 years ago
(CO3)2- + H2O <==> HCO3- + OH-

Equilibrium constant,
Kc = Kw/Ka2 = 1.00 x 10^-14 / 4.7 x 10^-11= 2.13x10^-4

Initial concentration:
[CO32-] = 0.32
[HCO3-] = [OH-] = 0


Final concentration at equilibrium:
[CO32-] = (0.32 - X)
[HCO3-] = [OH-] = X

Kc = [HCO3-][OH-] / [CO32-]
2.13 x 10^-4 = X^2 / (0.32 - X)

Solve for X:
X = 8.15x10^-3 M , X = -8.36x10^-3 M

Since concentration cannot have negative value, X = 8.15x10^-3 M

pOH = - log[OH-] = -log(8.15x10^-3) = 2.09
pH = 14 - 2.09=11.9
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