Ratio of conjugate bases of a tricarboxylic acid
Let's use citric acid as an example. Although the $pK_a$ values of citric acid might not necessarily be the same as those above, it will be useful for demonstration.
$$\ce{C6H8O7 + H2O &lt;=&gt; C6H7O7- + H3O+ &lt;=&gt; C6H6O7^2- + H3O+ &lt;=&gt; C6H5O7^3- + H3O+}$$
Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive $\mathrm pK_\mathrm a$ values and concentrations.
Primary proton transfer equilibrium:
$$\ce{C6H8O7 + H2O &lt;=&gt; C6H7O7- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} &amp; 0.20 &amp; - &amp; 0 &amp; 0 \\
\hline
\Delta C &amp; -x &amp; - &amp; +x &amp; +x\\
\hline
C_\mathrm{final}&amp; 0.20 - x &amp; - &amp; x &amp; x \\
\hline
\end{array}
$$K_\mathrm {a_1} = 0.01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$
$$ K_\mathrm{a1} = 0.01 = \frac{x^2}{0.20-x} $$
Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:
$$ x = 0.04 \ \mathrm{or} \ x = -0.05$$
We disregard our negative root. From $[\ce{C6H7O7}]^- = [\ce{H3O+}]$:
$$[\ce{C6H7O7}-] = \pu{0.04 mol L-1}$$
Now we use our second $\mathrm pK_\mathrm a$ to find the concentration of $\ce{C6H6O7^{2-}}$. Since $K_\mathrm {a_2} &lt;&lt; K_\mathrm {a_1}$, we can assume that the hydronium ion concentration is unchanged by the second deportonation.
$$\ce{C6H7O7- + H2O &lt;=&gt; C6H6O7^2- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} &amp; 0.04 &amp; - &amp; 0 &amp; 0.04 \\
\hline
\Delta C &amp; -x &amp; - &amp; +x &amp; +x\\
\hline
C_\mathrm{final} &amp; 0.04 - x &amp; - &amp; x &amp; 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 9.995 \times 10^{-6} $$
Ignoring the negative root:
$$ [\ce{C6H6O7^2-}] = \pu{9.995 \times 10^{-6} mol L-1}$$
Finally, assuming that the concentration of hydronium ions and the conjugate base of citric acid after the 2nd deprotonation is unaffected by additional deprotonation, we can write up our equilibrium table again:
\begin{array} {|c|c|c|c|c|}
\hline\
C_\mathrm{initial}&amp; 9.995*10^{-6} &amp; - &amp; 0 &amp; 0.04 \\
\hline
\Delta C &amp; -x &amp; - &amp; +x &amp; +x\\
\hline
C_\mathrm{final} &amp; 9.995*10^{-6} - x &amp; - &amp; x &amp; 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_3} = .00000001 = \frac{x(0.04 + x)}{9.995 \cdot 10^{-6} -x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 2.49875 \times 10^{-12}$$
$$ [\ce{C6H5O7^3-}] = \pu{2.49875 \times 10^{-12} mol L-1}$$