# What is the pOH of a 0.0100 F solution of dipotassium phtalate?

What is the pOH of a 0.0100 F solution of dipotassium phtalate? pKa1 = 2.950 and pKa2 = 5.408 for phtalic acid.

I did it for phthalic acid by calculating pH and then just doing pOH = 14.00 - pH, which gave me pOH = 11.60. Not sure if I did it correctly, I'm only just (re)learning this. I just need a clear and basic explanation, if possible. The two pKa's are what confuse me... which one do I use? I've been searching on google and through my textbook for 2 days now but I'm just confusing myself even more at this point. I know it's probably really easy stuff, but I haven't done chem in 4 years... trying to refresh my knowledge.

• Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. – Philipp Nov 6 '14 at 22:17

So, we know that dipotassium phtalate is a conjugate base of a diprotic acid, so the addition of it into water will remove some $\ce{[H+]}$s from $\ce{[H2O]}$ and increase the level of $\ce{[OH^{-}]}$ with means a lower pOH.
We calculate the pOH by first converting to $\mathrm{K_a}$ $$\mathrm{pK_{a2}=10^{-5.408}=3.908*10^{-06}}$$ $$\mathrm{K_{b1}=\frac{K_w}{K_{a2}}= \frac{1*10^{-7}}{3.908*10^{-06}}=2.559*10^{-2}}$$ $$\ce{\frac{[HA][OH^-]}{[A^-]}=\mathrm{K_{b1}}}$$ $$\mathrm{\frac{x^2}{F-x}=2.559*10^{-2}\rightarrow 1.60*10^{-2}}$$ We did that last one assuming that x was negligibly small. We now know $$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$ $$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$ $$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$