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.
Chemistry(Please help) - DrBob222, Saturday, April 14, 2012 at 11:09pm
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?
Chemistry(Please help) - DrBob222, Tuesday, April 17, 2012 at 3:27pm
No. Kb = 1E-14/k2 for H2CO3.
How do I set up an ice table when I only have .15M?
2 answers
.15M initial concentration of what? Or is it final concentration?
.........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-)
initial..0.15...............0......0
change....-x................x......x
equil....0.15-x.............x......x
Kb = (Kw/K2) = (HCO3^-)(OH^-)/(CO3^2-)
2 answers