Find the concentration of free Na + in 0.15 M Li 3 [Na(EDTA)] at pH = 10.00.

Now hopefully by following the previous examples, you're able to determine the amount of free sodium ions in solution, that is in equilibrium with our EDTA complex. Here when this breaks up, it breaks up into lithium ions and then NAEDTA 3 minus. And it's that EDTA complex that we're focused on. It's reacting with sodium. So it creates this complex here. Now we need to first determine what our conditional formation constant will be. So we're gonna say k prime sub f equals our fraction of EDTA in the basic form times the formation or stability constant, kf. So at 10 a pH of 10, we find out that value is 0.30. Then if we look on our chart, we see that log Kf for our sodium ion equals 1.86 which means that the kf of sodium equals 10 to 1.86. So that gives me 21.73 as the conditional formation constant. Now we have our initial molar. These, initially, are 0. We're gonna lose some of our EDTA complex here, and we're gonna lose some of our EDTA complex here. We're gonna lose some of our EDTA complex here. We're gonna lose some of our EDTA complex here. We're gonna lose some of our EDTA complex here. We're gonna lose We're gonna lose some of our EDTA complex here. We have the formation of these free ions. So bring down everything. So our k f is 21.73. It equals our products over reactant, so that's 0.15 minuteus x divided by x squared. In order to figure out the concentration of our sodium ions, we now need to isolate our x. So multiply both sides here by x squared. So we'll find that 21.73x squared equals 0.15 minuteus x. We're gonna bring all the terms over to this side here. So that gives me 21.73xsquaredplusxminus.15. Alright. So now, we use the quadratic formula. So negative one, plus or minus, 1 squared minus 4 times a times times c divided by 2 times a. So that's gonna give me x equals negative 1 plus or minus 3.745 divided by 43.46. Again, we'll get 2 answers here, one that's positive and one that's negative. At equilibrium, our final concentrations cannot be negative values. So we're going to ignore the negative answer and focus only on the x that gives me a positive concentration, which comes out to 0.06 3 molar. Here now, that tells me that this is the concentration of sodium. Now, realize the pattern here. We have a smaller charge of sodium, therefore our our conditional formation constant is not as large as it was in the previous examples. So not as much of this EDTA complex is being produced which explains why my sodium concentration at the end is Tin two ion in the previous example. So just remember, the larger your k f or conditional formation constants are, the more of your complex you're forming and the less available free metal ions that are circulating within your solution. Here in this example, because our conditional formation constant was not as large as in our previous example, which was 8.18 times 10 to 16. Very large number. Because it's not as large, we don't make as much of this EDTA complex here, which means that there's more of this metal ion that's free floating around within our solution. Understand the whole idea of formation constants and their relationship to charge and how the greater the KF is, the more product will form.

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1. Chemical Measurements1h 50m
2. Tools of the Trade1h 17m
3. Experimental Error1h 52m
4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
6. Chemical Equilibrium3h 41m
7. Activity and the Systematic Treatment of Equilibrium1h 0m
8. Monoprotic Acid-Base Equilibria1h 53m
9. Polyprotic Acid-Base Equilibria2h 17m
10. Acid-Base Titrations2h 37m
11. EDTA Titrations1h 34m
12. Advanced Topics in Equilibrium1h 16m
13. Fundamentals of Electrochemistry2h 19m
14. Electrodes and Potentiometry41m
15. Redox Titrations1h 14m
16. Electroanalytical Techniques57m
17. Fundamentals of Spectrophotometry50m

11. EDTA Titrations

EDTA

Analytical Chemistry

11. EDTA Titrations

EDTA

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Multiple Choice

Find the concentration of free Na ⁺ in 0.15 M Li ₃[Na(EDTA)] at pH = 10.00.

0.024 M

0.13 M

0.063 M

0.22 M

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Verified step by step guidance

Identify the chemical species involved: Li3[Na(EDTA)] is a complex where Na+ is bound to EDTA. At pH 10, consider the dissociation of this complex to release free Na+ ions.

Understand the equilibrium: The dissociation of Na(EDTA)3- can be represented as Na(EDTA)3- ⇌ Na+ + EDTA4-. The equilibrium constant for this reaction, Kd, will help determine the concentration of free Na+.

Consider the effect of pH: At pH 10, EDTA is mostly in the form of EDTA4-. This affects the equilibrium position and the concentration of free Na+.

Set up the equilibrium expression: Use the equilibrium constant expression Kd = [Na+][EDTA4-]/[Na(EDTA)3-] to relate the concentrations of the species at equilibrium.

Solve for [Na+]: Substitute the known values and solve the equilibrium expression for the concentration of free Na+ in the solution.