Lost track of the hydrogen in step 3 of the Krebs cycle

I'm trying to follow the Krebs cycle step by step, accounting for the number of hydrogens in
each molecule before and after reactions.


I've run into a problem starting from step 3.


[![Citric Acid Cycle - Step 3](https://i.stack.imgur.com/31ax1.png)](https://i.stack.imgur.com/31ax1.png)


Isocitrate has five hydrogens.
It undergoes oxidisation to Alpha-Ketoglutarate, with NAD+ being reduced to NADH/H+.


I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+.


What I don't understand (and this lack of understanding cascades through steps 4 and 5) is that Alpha Ketoglutarate has four hydrogens.


My arithmetic says that there should be only three.


**Where does the fourth hydrogen come from?**


**Note:**
If I start from Step 6, my accounting method works, all the way round to Step 3.


Step 6: Succinate (four hydrogens) is oxidised to Fumarate (two hydrogens), reducing FAD to FADH2.


Step 7: Fumarate (two hydrogens) is reduced to Malate (four hydrogens), by the addition of H2O, gaining
two hydrogens.


Step 8: Malate (four hydrogens) is oxidised to Oxaloacetate (two hydrogens), reducing NAD+ to NADH/H.


Etc.


I came across a similar question at <https://biology.stackexchange.com/questions/93775/what-exactly-happens-to-hydrogen-atoms-in-step-4-of-citric-acid-cycle> but as this was about step 4, and for me the problem occurs one step earlier.




Your problem stems from the statement


"I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+."


This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.


If you start with *isocitric acid*, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.


The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that "NADH/H+" is indicated as a product even though only NADH has actually been produced in the reaction shown.


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