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?

1. DrBob222

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

2. Hannah

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

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