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Hi does anyone know how to solve this question?

A tricarboxylic acid $\ce{H3A}$ has three different types of $\text{p}K_\text{a}$ $(\text{p}K_\text{a, 1} =2.0,$ $\text{p}K_\text{a, 2}=5.0,$ $\text{p}K_\text{a, 3}=8.0)$.

  1. Identify the ratio of the three charged components($\ce{H2A-}$,$\ce{HA^{2-}}$ and $\ce{A^{3-}}$) in $0.2 ~\text{M}$ solution of tricarboxylic acid $\ce{H3A}$.

  2. $0.55~\text{M}$ of the strong base $\ce{NaOH}$ is added into a solution of $0.20~\text{M}$ tricarboxylic acid $(\ce{H3A})$. Calculate the final pH. (Assume that the addition of $\ce{NaOH}$ will not influence the volume of the solution.)

I managed to get $0.04~\mathrm{M / L}$ for the $\ce{H2A-}$ species but I do not know how to find the concentration of the acids.

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  • $\begingroup$ Hi and welcome to Chem.SE! I've taken the liberty of editing your question to make it more clear. If I messed something up, feel free to make the edit to correct it :) $\endgroup$ – tschoppi Mar 31 '14 at 12:30
  • $\begingroup$ For question 1, what is the pH, or are we to assume that it was added to water and we are looking for the equilibrium when it dissolves? $\endgroup$ – LDC3 Apr 1 '14 at 0:56
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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.

$$\ce{C6H8O7 + H_2O <=> C6H7O7- + H3O+ <=> C6H6O7^{2-} + H3O^+ <=> C6H5O7^{3-} + H3O+}$$

Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive $pK_a$ values and concentrations.

Primary proton transfer equilibrium:

$$\ce{C6H8O7 + H_2O <=> C6H7O7- + H3O+} $$

\begin{array} {|c|c|c|c|c|} \hline \text{Initial conc.} & 0.20 & - & 0 & 0 \\ \hline \text{Change conc.} & -x & - & +x & +x\\ \hline \text{End conc.} & 0.20 - x & - & x & x \\ \hline \end{array}

$$K_{a1} = .01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$

$$ K_{a1} = .01 = \frac{x^2}{0.20-x} $$

Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:

$$ x = 0.04 \ \text{or} \ x = -0.05$$

We disregard our negative root. From $[\ce{C6H7O7}]^- = [\ce{H3O+}]$:

$$[\ce{C6H7O7}]^- = 0.04 \ \text{mol} \cdot \text{L}^{-1}$$

Now we use our second $pK_a$ to find the concentration of $\ce{C6H6O7^{2-}}$. Since $Ka_2 << Ka_1$, we can assume that the hydronium ion concentration is unchanged by the second deportonation.

$$\ce{C6H7O7- + H2O <=> C6H6O7^{2-} + H3O^+} $$

\begin{array} {|c|c|c|c|c|} \hline \text{Initial conc.} & 0.04 & - & 0 & 0.04 \\ \hline \text{Change conc.} & -x & - & +x & +x\\ \hline \text{End conc.} & 0.04 - x & - & x & 0.04 + x \\ \hline \end{array}

$$ K_{a2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$

$$ x = -0.04002 \ \text{or} \ x = 9.995*10^{-6} $$

Ignoring the negative root:

$$ [\ce{C6H6O7}]^{2-} = 9.995*10^{-6} \ \text{mol} \cdot \text{L}^{-1}$$

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 \text{Initial conc.} & 9.995*10^{-6} & - & 0 & 0.04 \\ \hline \text{Change conc.} & -x & - & +x & +x\\ \hline \text{End conc.} & 9.995*10^{-6} - x & - & x & 0.04 + x \\ \hline \end{array}

$$ K_{a3} = .00000001 = \frac{x(0.04 + x)}{9.995*10^{-6} -x} $$

$$ x = -0.04002 \ \text{or} \ x = 2.49875*10^{-12}$$

$$ [\ce{C6H5O7}]^{3-} = 2.49875*10^{-12} \ \text{mol} \cdot \text{L}^{-1}$$

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