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

Question

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.

Answer 1

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 $$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$ $$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$ $$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$