Calculate αY4- at the pH value 3.70 and pH value 10.15

Question

EDTA is a hexaprotic system with the pKa values:

pKa1= 0.00, pKa2= 1.50, pKa3= 2.00, pKa4=2.69. pKa5= 6.13, pKa6=10.37.

The distribution of the various protinated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4-. This fraction is designated αY4-

Calculate αY4- at pH 3.70 = ?

Calculate αY4- at pH 10.15 = ?

Answer 1

Hi Lizzi B.,

this looks tough, but it isn't really.

You jump right in with successive approximations, using the pKa's which bracket the stated condition, first. Assume 1M for EDTA for convenience; you'll only figure relative proportions anyway. Call this "Quick-ICE", if you will.

So, that 4th proton has pKa 2.69, vs. the stated condition of pH 3.70. The stated is a whole pH unit above the pKa, so about 1/11 of the (4th) protons are still on the EDTA, and 10/11 of the 4th protons are off -- that's about a 10:1 ratio, which is one pH unit on the Henderson Hasselbach equation, or whatever it's called -- you should however express it (antilog(1.01)) more exactly, aim for 4 sig figs, since your other data is given to 3 sig figs, and you'll be rounding to 3 sig figs AT THE END OF YOUR CALCULATIONS. So, that's a starting point. Next, you tweak those two relative values ([y4-] and [Y3-]) using the next further pKa's. So the [y4-] "feels the effect of" the 6.13, which tweaks it down by antilog(6.13-3.70) (to Y5-) (that's a quite small perturbation!); the [Y3-] "feels the effect of" the 2.00, which tweaks it down proportionately by antilog(2.69-2.00) to Y2- (that's a small perturbation, but enough to see on your final value). The 1.50 pKa pKa then pulls down Y2- to Y1- (by antilog (2.00-1.50) proportionately), but again, within 1/1000 adjustment, so include it too. All the other pKas are so far away from 3.70 that you can neglect them for that portion of the problem. That leaves you with 5 fractions of your original 1M, namely Y1-, Y2-, Y3-, Y4-, and Y5-. Of which you need the Y4-, BUT you needed to figure the rest to get the precision you needed. Round to 3 sig figs, and you're done. That's round 1 of "quick-ICE". I'm thinking that you wouldn't even need a second round of ICE, but you might do it anyway just to make sure (Neglect Y0 and Y6- completely, they're each negligible.)

Hope this has helped you a bit. -- Mr. d.