Okay, let's try to tackle this problem. Hmm, the question is about finding the concentration of calcium ions in a solution where there's a Ca-EDTA complex. The given information is that the solution is 0.02 M stochiometric Ca-EDTA, the formation constant (Kf) is 5x10^10, and the temperature is 25°C. The options are A through D. Alright, let's start breaking this down.

First, I remember that EDTA is a hexadentate ligand, meaning it can bind to a metal ion like calcium in multiple places. The formation of the Ca-EDTA complex is a precipitation reaction where EDTA reacts with calcium ions. The formation constant, Kf, is the equilibrium constant for this reaction. So the reaction would look something like Ca^2+ + EDTA^4- ⇌ CaEDTA^2−. 

The problem states that the solution is 0.02 M stochiometric Ca-EDTA complex. Wait, stochiometric means that the concentrations of Ca^2+ and EDTA are in the stoichiometric ratio required by the reaction. Since EDTA is a hexadentate ligand, each EDTA can bind with one Ca^2+. So the stoichiometric ratio is 1:1. Therefore, if the concentration of the complex is 0.02 M, then the concentrations of Ca^2+ and EDTA should each be 0.02 M. But wait, that might not be the case because the complex dissociates in solution. So maybe the initial concentration of the complex is 0.02 M, and we need to find the equilibrium concentrations of the free Ca^2+ and EDTA, as well as the complex.

Wait, no. The problem says "stochiometric Ca-EDTA complex". Maybe that means that the solution is prepared by mixing Ca^2+ and EDTA such that their concentrations are stoichiometrically equivalent. So if you start with 0.02 M Ca^2+ and 0.02 M EDTA, but they react to form the complex. However, the problem states that the solution contains the complex, so maybe the initial concentration of the complex is 0.02 M. But if so, then the initial concentrations of Ca^2+ and EDTA are each 0.02 M, and they react to form the complex. But when the complex forms, some of the Ca^2+ and EDTA are consumed, and some remains as free ions. 

Alternatively, perhaps the solution is prepared by dissolving the complex, so the total concentration of the complex is 0.02 M. But then the actual concentrations of free Ca^2+ and EDTA would be lower because some of them are tied up in the complex. Wait, but the problem says "stochiometric Ca-EDTA complex". That might mean that the solution is prepared such that the ratio of Ca^2+ to EDTA is 1:1, so when the complex forms, the concentrations of free Ca^2+ and EDTA are equal. Let me think.

Let me try to write the equilibrium expression. The formation constant Kf is given as 5x10^10. The reaction is Ca^2+ + EDTA^4− ⇌ CaEDTA^2−. The equilibrium expression is Kf = [CaEDTA^2−]/([Ca^2+][EDTA^4−]). 

If the solution contains the Ca-EDTA complex at a concentration of 0.02 M, then the total concentration of the complex is 0.02 M. Let's denote the concentration of the complex as C = 0.02 M. Let the concentration of free Ca^2+ be x. Then, the concentration of free EDTA^4− would also be x, because for every Ca^2+ that dissociates, one EDTA^4− dissociates as well. The total concentration of the complex would be C = [CaEDTA^2−] + [Ca^2+] + [EDTA^4−]. Wait, but if the initial solution is prepared with the complex, then the total concentration of the complex is 0.02 M. But as the complex dissociates, the free ions are formed. So the total concentration of the complex, EDTA, and Ca^2+ would be related. Hmm, maybe I need to approach this using the common ion effect.

Alternatively, since the Kf is very large (5x10^10), which is much greater than 1, the reaction strongly favors the formation of the complex. Therefore, the concentration of free Ca^2+ and EDTA^4− should be very low. So, if we start with 0.02 M Ca^2+ and 0.02 M EDTA, most of them would form the complex, leaving very little free Ca^2+ and EDTA. But since the problem states that the solution contains the complex, perhaps we need to find the free concentration of Ca^2+ given that the complex is 0.02 M.

Wait, maybe the problem is more straightforward. If the solution is 0.02 M in the Ca-EDTA complex, then the concentration of the complex is 0.02 M. Let's denote [CaEDTA^2−] = 0.02 M. Let the concentration of free Ca^2+ be x. Then, the concentration of free EDTA^4− would also be x, because the stoichiometry is 1:1. The total concentration of the complex would be 0.02 M, but the total EDTA in solution would be the complex plus the free EDTA. Similarly for Ca^2+. Wait, but the problem says "stochiometric Ca-EDTA complex", which might mean that the total concentration of Ca and EDTA is stoichiometrically equivalent. Hmm, this is a bit confusing.

Alternatively, perhaps the problem is simpler. Since the Kf is given, and the concentration of the complex is 0.02 M, we can use the Kf expression to find the free concentrations. Let me set up the equation. Let [Ca^2+] = x, [EDTA^4−] = x, then [CaEDTA^2−] = 0.02 M. Wait, but the total concentration of EDTA would be x (free) + 0.02 (bound in complex). Similarly for Ca^2+. However, the problem states that the solution contains the complex, so the total concentration of the complex is 0.02 M. But the total concentration of Ca^2+ would be x (free) + 0.02 (bound). Similarly for EDTA. But if the solution is prepared such that the concentration of the complex is 0.02 M, then the total concentration of Ca^2+ and EDTA would be 0.02 M each? Wait, that can't be right because if the complex is 0.02 M, then the free ions would be negligible.

Alternatively, perhaps the solution is 0.02 M in the complex, and we need to find the free Ca^2+ concentration. Let's assume that the total concentration of the complex is 0.02 M. Let the concentration of free Ca^2+ be x, then the concentration of free EDTA would also be x. The total concentration of the complex would be 0.02 M, which equals [CaEDTA^2−] + [Ca^2+] + [EDTA^4−]. But if [CaEDTA^2−] is 0.02 M, then [Ca^2+] + [EDTA^4−] must be negligible. Wait, that doesn't make sense. Alternatively, maybe the total concentration of the complex is 0.02 M, so [CaEDTA^2−] = 0.02 M. Then, the free Ca^2+ and EDTA are each x, so [Ca^2+] = x, [EDTA^4−] = x. Therefore, the total concentration of Ca^2+ in solution is x (free) + 0.02 (bound in complex). Similarly for EDTA. But the problem doesn't specify the total concentration of Ca^2+ or EDTA, only that the complex is 0.02 M. Hmm, this is getting a bit tangled.

Wait, perhaps the problem is simpler. Since the solution contains the complex at 0.02 M, and the Kf is 5x10^10, which is very large, the dissociation of the complex is negligible. Therefore, the concentration of free Ca^2+ and EDTA can be approximated as zero. But that would mean that the answer is zero, which isn't one of the options. Wait, but the options don't have zero. Hmm. Maybe I'm missing something here.

Alternatively, perhaps the problem is about the formation of the complex from free Ca^2+ and EDTA. If the solution contains 0.02 M of the complex, then the total concentration of free Ca^2+ and EDTA can be calculated. Let me set up the equation again. Let the concentration of free Ca^2+ be x. Then, the concentration of free EDTA would also be x (since the stoichiometry is 1:1). The total concentration of the complex would be 0.02 M, so [CaEDTA^2−] = 0.02 M. The total concentration of Ca in solution is x + 0.02 M. Similarly, the total concentration of EDTA is x + 0.02 M. But since the problem states that the solution contains the complex, perhaps the total concentration of the complex is 0.02 M, and the total concentrations of Ca and EDTA are each 0.02 M. But that would mean that x is negligible, which can't be right. Alternatively, maybe the solution is prepared by adding 0.02 M Ca^2+ and 0.02 M EDTA, and then the complex forms. In that case, the initial concentrations are 0.02 M each, but after complexation, some of each is consumed. Let's denote the amount of complex formed as s. Then, [Ca^2+] = 0.02 - s, [EDTA^4−] = 0.02 - s, and [CaEDTA^2−] = s. The Kf expression would be Kf = s / [(0.02 - s)(0.02 - s)]. Since Kf is very large (5x10^10), s would be approximately 0.02 M. Therefore, [Ca^2+] ≈ 0.02 - 0.02 = 0. But again, that leads to zero, which isn't an option. Hmm.

Wait, but maybe the problem is asking for the concentration of free Ca^2+ in a solution where the complex is 0.02 M, and the solution is in equilibrium. So in that case, the total concentration of Ca in solution is the sum of free Ca^2+ and the complex. Let's denote the total concentration of Ca as C_total. Similarly for EDTA. But the problem doesn't give the total concentration of Ca or EDTA. Hmm, this is confusing. Wait, the problem states "the solution containing 0.02 M stochiometric Ca-EDTA complex". Maybe that means that the concentration of the complex is 0.02 M, and we need to find the concentration of free Ca^2+.

In that case, using the Kf expression: Kf = [CaEDTA^2−]/([Ca^2+][EDTA^4−]). Let [Ca^2+] = x, [EDTA^4−] = x (since stoichiometry is 1:1), and [CaEDTA^2−] = 0.02 M. Then, Kf = 0.02 / (x * x) = 5x10^10. Solving for x: x^2 = 0.02 / (5x10^10) = 4x10^-13. Therefore, x = sqrt(4x10^-13) = 2x10^-6.5. Wait, sqrt(4e-13) is 2e-6.5? Wait, 1e-13 is (1e-6.5)^2. So sqrt(4e-13) is 2e-6.5, which is 2e-6.5 = 2 * 10^-6.5 = 2 * 3.16e-7 ≈ 6.3e-7. So x ≈ 6.3e-7 M. Which is option D. So the concentration of free Ca^2+ is 6.3e-7 M. 

But wait, let me check the calculation again. Kf = 5e10 = 0.02 / (x^2). So x^2 = 0.02 / 5e10 = 2e-3 / 5e10 = 4e-13. Then x = sqrt(4e-13) = 2e-6.5. 10^-6.5 is 3.16e-7. So 2 * 3.16e-7 ≈ 6.3e-7. Yes, that's correct. So the concentration of free Ca^2+ is approximately 6.3e-7 M, which is option D. 

But wait, let me think again. If the complex is 0.02 M, and the free Ca^2+ is 6.3e-7 M, then the concentration of the complex would be [CaEDTA^2−] = 0.02 M, and the free calcium is negligible compared to the complex. So the answer would be D. 

Alternatively, maybe the problem is about the total concentration of EDTA being 0.02 M, but that doesn't make sense because the complex is 0.02 M. Hmm. Let me make sure I didn't misinterpret the problem. The question says "What is the concentration of calcium ions in a solution containing 0.02 M stochiometric Ca-EDTA complex". So the solution contains the complex at 0.02 M. The formation constant is given. So using the Kf expression, we can find the free Ca^2+. So the answer is option D, 6.3e-7 M.
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