Solubility of a tribasic weak acid I have a question that reads:


  $\ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
  
  $\ce{H3A &lt;=&gt;[pKa1 = 2.9] H2A- &lt;=&gt;[pKa2 = 4.2] HA^{2-} &lt;=&gt;[pKa3 = 5.2] A^{3-}}$
  
  What is the predominant species in a 0.10M solution of $\ce{H3A}$ at pH 5?
  
  a) $\ce{H3A}$
  
  b) $\ce{H2A-}$
  
  c) $\ce{HA^{2-}}$
  
  d) $\ce{A^{3-}}$


I worked this out as

$\ce{Ka1 = \frac{[H+][H2A-]}{[H3A]}}, \ce{Ka2 = \frac{[H+][HA^{2-}]}{[H2A-]}}, \ce{Ka3 = \frac{[H+][A^{3-}]}{[HA^{2-}]}}$

At pH 5, $\ce{[H+] = 10^{-5}M}$

So, 

$\ce{[H+] = 10^{-5} = Ka1\frac{[H2A-]}{[H3A]} = 10^{-2.9} \frac{[H2A-]}{[H3A]}}\ce{\implies \frac{[H2A-]}{[H3A]} = 10^{2.1} }$

and similarly,

$\ce{\frac{[HA2-]}{[H2A-]} = 10^{0.8}}$

$\ce{\frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$

Since $\ce{[HA^{2-}] &gt; [H3A], [HA^{2-}]&gt;[H2A-] and [A^{3-}]&lt;[HA^{2-}]}$, the predominant species is $\ce{HA^{2-}}$, so c).

This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).
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.