Chemistry concentration

Question

Calculate the concentration of free Ca2+ in a solution in which 25.00 mL of 0.1500 M Ca2+ has been titrated with 25.00 mL of 0.1500 M EDTA4-. The Kr for the formation of the calcium/EDTA complex is 4.5x10^10

Answer 1

At the equivalence point you have added exactly enough EDTA to convert all of the calcium into the CaY2- complex, so the only free Ca2+ left is what leaks back out when that complex dissociates. Kf tells you how little that is.

Step 1: how much complex forms, and in what volume.
moles Ca2+ = (0.02500 L)(0.1500 M) = 3.750e-3 mol
moles EDTA4- = (0.02500 L)(0.1500 M) = 3.750e-3 mol

They combine 1:1, so 3.750e-3 mol of CaY2- forms and neither reactant is left over. That is exactly what makes this the equivalence point. The volumes add: 25.00 + 25.00 = 50.00 mL.

[CaY2-] = 3.750e-3 mol / 0.05000 L = 0.07500 M

Notice that is exactly half of 0.1500 M. Mixing equal volumes halves every concentration, and forgetting that dilution is the most common way this problem goes wrong.

Step 2: let the complex fall back apart.
CaY2- <==> Ca2+ + Y4-

That is the reverse of the formation reaction, so its equilibrium constant is 1/Kf:
1 / (4.5e10) = 2.2e-11

Let x = [Ca2+]. The complex releases one Ca2+ and one Y4- together, so [Y4-] = x as well, and [CaY2-] = 0.07500 - x:

x^2 / (0.07500 - x) = 2.2e-11

Kf is enormous, so x will be minuscule next to 0.07500 and you can drop it from the denominator:

x^2 = (0.07500)(2.2e-11) = 1.67e-12
x = 1.3e-6 M free Ca2+, or pCa = 5.89

Check the approximation: 1.3e-6 is about 0.002% of 0.07500, so dropping it changed nothing.

Two things worth carrying forward. First, the "Kr" in your problem is Kf, the formation constant, and the whole trick here is seeing that the titration reaction IS the formation reaction, so the equivalence-point calculation is just that same reaction run backwards. Second, this answer assumed all of the uncomplexed EDTA is present as Y4-, which is only true at high pH, roughly 12 and above. At any lower pH some free EDTA is protonated as HY3-, H2Y2- and so on, and you use the conditional constant K'f = alpha(Y4-) x Kf instead. Ca/EDTA titrations are normally run at pH 10 in an ammonia buffer, where alpha(Y4-) = 0.30, giving K'f = 1.4e10 and [Ca2+] = 2.4e-6 M. Drop to pH 8 and alpha falls to 5.4e-3, K'f = 2.4e8, and the free calcium climbs to 1.8e-5 M. If your problem set supplies a pH and a table of alpha values, that is the version they want.

The shape of the answer is the point of a complexometric titration: 0.075 M calcium has been driven down to about one part per million of that, which is why EDTA gives such a sharp endpoint.