Combining acid dissociation constants to determine pH of diprotic acid
This should be broken into:
\begin{align}
\ce{H2A + H2O &amp;-&gt; HA- + H3O+}\\
\ce{HA- + H2O &amp;-&gt; A^{2-} + H3O+}
\end{align}

With equilibrium equations:
\begin{align}
K_\mathrm{a1}&amp;=\frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}\tag1\label{Ka1}\\
K_\mathrm{a2}&amp;=\frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]}{[\ce{HA-}]}\tag2\label{Ka2}\\
\end{align}

Let us solve for a general case. From logic and general knowledge we know the water dissociation constant, that the masses must balance, and that the charges must balance. These are shown here:
\begin{align}
K_\mathrm{w}&amp;=[\ce{H3O+}]\cdot[\ce{OH-}]\tag3\label{Kw}\\
I&amp;=[\ce{H2A}]+[\ce{HA-}]+[\ce{A^{2-}}]\tag4\label{Mass Balance}\\
[\ce{H3O+}]&amp;=[\ce{OH-}]+[\ce{HA-}]+2[\ce{A^{2-}}]\\
&amp;=\frac{K_\mathrm{w}}{[\ce{H3O+}]}+[\ce{HA-}]+2[\ce{A^{2-}}]\tag5\label{Charge Balance}\\
\end{align}

Using equations \eqref{Ka1} and \eqref{Ka2}, we can solve for $[\ce{HA-}]$ and $[\ce{A^{2-}}]$:
\begin{align}
K_\mathrm{a1} &amp;= \frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]} 
&amp;\Rightarrow &amp;&amp;
[\ce{HA-}] &amp;= \frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}\tag6\label{HA-}\\
K_\mathrm{a1} \cdot K_\mathrm{a2} &amp;= \frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]^2}{[\ce{H2A}]} 
&amp;\Rightarrow &amp;&amp;
[\ce{A^{2-}}] &amp;= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\tag7\label{A2-}\\
\end{align}

Substituting into \eqref{Mass Balance}:
\begin{align}
I &amp;= [\ce{H2A}]+\frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}   
     +\frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\\
  &amp;= [\ce{H2A}] \cdot \frac{[\ce{H3O+}]^2 + [\ce{H3O+}] 
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H3O+}]^2}\tag9\label{Sub Mass Balance}
\end{align}

Solving \eqref{Sub Mass Balance} for $[\ce{H2A}]$:

$$[\ce{H2A}]=\frac{I \cdot [\ce{H3O+}]^2}{[\ce{H3O+}]^2 + [\ce{H3O+}] 
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{10}\label{H2A}$$

Substituting \eqref{H2A} into \eqref{HA-} and \eqref{A2-}:
\begin{align}
[\ce{HA-}] &amp;= \frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
\tag{11}\label{Sub HA-}\\
[\ce{A^{2-}}] &amp;= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{12}\label{Sub A2-}
\end{align}

Substituting equations \eqref{Sub HA-} and \eqref{Sub A2-} into equation \eqref{Charge Balance}:
\begin{align}
&amp;&amp;
[\ce{H3O+}] &amp;= \frac{K_\mathrm{w}}{[\ce{H3O+}]}+\frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]+2K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\\
\Rightarrow &amp;&amp;
[\ce{H3O+}]^2 &amp;= K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot ([\ce{H3O+}]^2+2K_\mathrm{a2}\cdot[\ce{H3O+}])}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{13}\label{Final Eq}
\end{align}

The above equation exactly calculates the $[\ce{H3O+}]$ for a solution of any diprotic acid. The same method used to solve for this equation can be applied to polyprotic acids as well. To make this into a more manageable equation, however, you may choose to assume $K_\mathrm{a1} \cdot K_\mathrm{a2}=0$ (but only in the case of weak acids).



As for the case you proposed, using $K_\mathrm{w}=1 \times 10^{-14}$, $I=1\ \mathrm{M}$, $K_\mathrm{a1}=5.9 \times 10^{−2}$ and $K_\mathrm{a2}=6.4 \times 10^{−5}$ in the equation $y=K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot (x^2+2K_\mathrm{a2})}{x^2 + x \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}-x^2$, and using a graphic calculator to find where the equation is equal to 0, we find $[\ce{H3O+}]=0.215508\ \mathrm{M}$ and $\mathrm{pH}=0.666537$.