(Solved): EDTA is a hexaprotic system with the pKa values: pKa1=0.00,pKa2=1.50,pKa3=2.00,pKa4= ...

EDTA is a hexaprotic system with the $\mathrm{p}{K}_{\mathrm{a}}$ values: $\begin{array}{l}\mathrm{p}{K}_{\mathrm{a}1}=0.00,\mathrm{p}{K}_{\mathrm{a}2}=1.50,\mathrm{p}{K}_{\mathrm{a}3}=2.00,\mathrm{p}{K}_{\mathrm{a}4}=2.69,\\ \mathrm{p}{K}_{\mathrm{a}5}=6.13\text{, and }\mathrm{p}{K}_{\mathrm{a}6}=10.37\end{array}$ The distribution of the various protonated forms of EDTA will therefore vary with $\mathrm{pH}$. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, ${\mathrm{Y}}^{4-}$. This fraction is designated ${\alpha }_{{\mathrm{Y}}^{4-}}$. Calculate ${\alpha }_{{\mathrm{Y}}^{4-}}$ at two $\mathrm{pH}$ values. $\begin{array}{ll}\mathrm{pH}=3.55& {\alpha }_{{\mathrm{Y}}^{4-}}=\\ \mathrm{pH}=10.70{\alpha }_{{\mathrm{Y}}^{4-}}=& \end{array}$