At the one side, you can always calculate the percentages of the acid or it's ions in solution:
$$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{a,j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{a,j}}$$
with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons.
But on the other hand there are some special points during titrations. One of those points did you choose, which is where $\mathrm{pH} = \mathrm{pK}_{a,i}$.
For $\ce{H3PO4}$ there exist three different acidity constants, which describe the three possible dissociation steps. For the first step, the reaction is as you described: $$\ce{H3PO4 + H2O \xrightleftharpoons{k_{a,1}} H2PO4- + H3O+}$$
A little work on it gives you the Henderson-Hasselbalch-equation, which usually is good approximation to buffer regions: $$\mathrm{pH} = \mathrm{pK}_a - \lg\left(\frac{[HA]}{[A^-]}\right)$$
If you enter the same values for pH and pKa you have to solve $$-\lg(x)=0$$ which is true for $x=1$. So you know that you have a 1:1 ratio at this point.
This means for phosphoric acid:
At $\mathrm{pH=pK}_{a1}$, you've got exactly $50~\%~\ce{H3PO4}$ and $50~\%~\ce{H2PO4-}$ (blue, yellow). The same procedure can be applied to the other dissociation steps. Therefor at $\mathrm{pH=pK}_{a2}$ you've got $50~\%~\ce{H2PO4-}$ and $50~\%~\ce{HPO4^2-}$ (yellow, green) and at the last possible $\mathrm{pH=pK}_{a3}$ there are $50~\%~\ce{HPO4^2-}$ and $50~\%~\ce{PO4^3-}$ (green, red).
This can be applied to most acid-base dissociation steps but you have to pay attention for close successive $\mathrm{pK}_a$ values, as if the gap gets too small, the 1:1 ratio doesn't hold.
