What is the pOH of a 0.0100 F solution of dipotassium phtalate?
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
$$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$
$$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$
$$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$