posted by Hannah

Calculate the pH of an aqueous solution of 0.15 M potassium carbonate.

I know that pH = -log(H30+) but I am not sure how to start this problem.

Hydrolyze the CO3^2-.
CO3^- + HOH ==> HCO3^- + OH^-
Do an ICE chart, Kb = Kw/k2 for H2CO3.

so for kb I do 1e-14/.15?

No. Kb = 1E-14/k2 for H2CO3.

How do I set up an ice table when I only have .15M?

1. bobpursley

.15M initial concentration of what? Or is it final concentration?

2. DrBob222

.........CO3^2- + HOH ==> HCO3^- + OH^-
initial..0.15...............0......0
change....-x................x......x
equil....0.15-x.............x......x

Kb = (Kw/K2) = (HCO3^-)(OH^-)/(CO3^2-)

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