# EDTA Titrations

I was doing some questions on EDTA titrations and was wondering how you actually derive the second equation here:

\begin{align} α_{\ce{Y^4-}} &= \frac{[\ce{Y^4-}]}{[\ce{H6Y^2+}] + [\ce{H5Y+}] + [\ce{H4Y}] + [\ce{H3Y-}] + [\ce{H2Y^2-}] + [\ce{HY^3-}] + [\ce{Y^4-}]}\\[10pt] &= \scriptsize \frac{K_1K_2K_3K_4K_5K_6}{[\ce{H+}]^6 + [\ce{H+}]^5 K_1 + [\ce{H+}]^4 K_1K_2 + [\ce{H+}]^3 K_1K_2K_3 + [\ce{H+}]^2 K_1K_2K_3K_4 + [\ce{H+}] K_1K_2K_3K_4K_5 + K_1K_2K_3K_4K_5K_6} \end{align}

where $\ce{Y^4-}$ is the fully deprotonated form of EDTA and $K_i$ represents the $i$-th dissociation constant of $\ce{H2Y^2+}$.

• @Joseph.L Was this equation given or did you find it somewhere? Oct 16, 2016 at 17:25

I posted the general solution for the degree of dissociation here: $$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}}$$ with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons. But from the beginning.

If EDTA would be a simple acid that has four protons, this would look like: $$\alpha(\mathrm A^{4-}) = \frac{k_1 k_2 k_3 k_4}{x^4+x^3 k_1+x^2k_1 k_2+x k_1 k_2 k_3 + k_1 k_2 k_3 k_4}$$

This derives from solving a linear equation system based on the mass balance: $$c_0(\ce{H_nA}) = \sum_{i=0}^n c(\ce{H_{n-i}A^{i-}})$$

and all the respective equilibrium/acid constants: $$k_{a,i} = \frac{c(\ce{H3O+})~c(\ce{H_{n-i}A^{i-}})}{c(\ce{H_{n-i+1}A^{(i-1)-})}}$$

What you will get from solving this, is an equation for the concentration of every deprotonated EDTA subspecies, as well as for EDTA itself: $$c(\ce{H_{n-m}A^{m-}}) = \frac{c_0(\ce{H_nA}) \prod_{i=1}^{m} k_i}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}}$$

As the degree of dissociation is defined as "the concentration of the most deprotonated species" devided by "the sum of concentrations of all species of the respective acid in solution", it would be for EDTA with four protons: $$\alpha(\ce{A^{4-}}) = \frac{c(\ce{A^{4-}})}{c_0(\ce{H_nA})} = \frac{c_0(\ce{H_nA}) \prod_{i=1}^{m} k_i}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}} \left(\sum_{i=0}^n c(\ce{H_{n-i}A^{i-}})\right)^{-1}$$

Which then ends up into the equation as stated above: $$\alpha(\mathrm A^{4-}) = \frac{k_1 k_2 k_3 k_4}{x^4+x^3 k_1+x^2k_1 k_2+x k_1 k_2 k_3 + k_1 k_2 k_3 k_4}$$

As in my eyes one can easily see now, is that your equation somehow derives as mine, but with the assumption that EDTA has six protons that can get deprotonated. But following my way, it would look a bit different: $$\alpha(\mathrm A^{6-}) = \frac{k_1 k_2 k_3 k_4 k_5 k_6}{x^6 + x^5 k_1+x^4 k_1k_2+x^3k_1 k_2 k_3+x^2 k_1 k_2 k_3 k_4+x k_1 k_2 k_3 k_4 k_5 + k_1 k_2 k_3 k_4 k_5 k_6}$$

If at all, I think, the proper way of handling EDTA completely would be to take it as an amphoteric compound that can either act as an acid and "lose" four protons or act as an base and "take up" two more protons. The mass balance is nearly the same but the two protonation steps are added: $$\ce{c_0(H_4A)} = \sum_{i=0}^4 c(\ce{H_{4-i}A^{i-}}) + \sum_{i=0}^2 c(\ce{H_{4+i}A^{i+}})$$

And the system of all four equilibrium/acid constants will be expanded with the two equilibrium/base constants for the protonation steps. So for $i=1\ldots4$: $$k_{a,i} = \frac{c(\ce{H3O+})~c(\ce{H_{4-i}A^{i-}})}{c(\ce{H_{4-i+1}A^{(i-1)-})}}$$ and for $i=1\ldots2:$ $$k_{b,i} = \frac{c(\ce{OH-})~c(\ce{H_{4+i}A^{i+}})}{c(\ce{H_{4+i-1}A^{(i-1)+})}} = \frac{k_{\mathrm W}/c(\ce{H3O+})~c(\ce{H_{4+i}A^{i+}})}{c(\ce{H_{4+i-1}A^{(i-1)+})}}$$

In that case, the result would be: $$\alpha(\mathrm A^{4-}) =\\ \frac{x^4 k_\mathrm W^2}{x^6 k_{\mathrm b,1} k_{\mathrm b,2}+x^5 k_{\mathrm b,1} k_{\mathrm W}+k_{\mathrm W}^2 \left(x^4+x^3 k_{\mathrm a,1}+x^2 k_{\mathrm a,1} k_{\mathrm a,2}+x k_{\mathrm a,1} k_{\mathrm a,1} k_{\mathrm a,3}+k_{\mathrm a,1} k_{\mathrm a,1} k_{\mathrm a,3} k_{\mathrm a,4}\right)}$$