For half titration you should assume EDTA reacted with Ca²⁺ in the solution completely, and plug known concentrations into Kf, the only unknown will be the EDTA concentration.

That's only an approximation (to get the more exact value you should use ICE table), but typically gives acceptable accuracy.

It won't work for equivalence point though. There you can assume you have solution equivalent to the one prepared by dissolving just the proper amount of CaEDTA and calculating how much of it dissociated (hint: concentrations of Ca²⁺ and unreacted EDTA will be identical). Again, use Kf.

Thank you so much for replying, I appreciate it.

I'm not sure about EDTA concentration. Should I say: nEDTA/volume=(0.1mol/L*0.01L*0.0005mol)/0.105L instead?

For equivalent point, it gives Kf=1*10^10. I thought saying: Kf= (CaEDTA2-)/((Ca2+)*(EDTA4-))=1*10^10 then put 0.1M for both Ca2+ and EDTA. But I don't think that's correct.

Would there be any chance of guiding me a bit about Ca-EDTA concentration? If not, completely understandable.
Thank you again SO much for your help.

Electrical impulses are the cause of the heartbeat and arise from a remarkable combination of electrochemistry and the properties of semipermeable cell membranes regarding a number of physiologically important ions (Na+, K+, Ca2+). The concentrations of these ions differ for intracellular fluids (ICF) and extracellular fluids (ECF). In cardiac muscle cells the ICF and ECF concentrations of K+ are approximately 135 and 4 mM, respectively. The difference in ion concentration between the ICF and ECF fluids produces a concentration cell~(...) Despite the fact that ions are present on both sides of the membrane, a potential difference is created, calculated through the Nernst equation, with E°=0. After this brief introduction, show whether the process of entering potassium (K+) ions into heart cells is a spontaneous process. ii) Ascent, descent? Electron flow ?

I've tried: ICF:135MM, ECF 4MM => Q=135/4>1 flow anode->cathode.
Would it be correct to say that it's not spontaneous as a concentration cell is non-spontaneous. What else could I add? Thanks for any help

@OP, Is this a homework assignment, or are you doing this reaction in a laboratory?

That's chitinase test kit.  Sorry, spell checker. It's chitinase3-like 1 human.  Whether it would work, probable not.  But it might point you in the right direction.

I am asking for information on measurements that have already been done, not advertisement for products. And given that the plants did evolve the chitinase, it should have some effects on at least fungi, possibly also small insects. Do you know any information on activity levels?