Solubility of a tribasic weak acid
You can also use the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

rearranges to:

$$\mathrm{pH} - \mathrm{p}K_{\mathrm{a}} = \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

$\log \frac{\ce{[A-]}}{\ce{[HA]}} &gt; 0$ for conjugate base dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} &lt; 0$ for acid dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} = 0$ if they're present in the same concentrations.

The equilibrium of interest is the third dissociation.

Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($\ce{HA^{2-}}$) dominates over the conjugate base ($\ce{A^{3-}}$). You should be able to extend this for each equilibrium.