EDTA titrations A solution of 0.0100 moles of pure Na2H2EDTA is prepared in 1.00

Question

EDTA titrations A solution of 0.0100 moles of pure Na2H2EDTA is prepared in 1.00 Lof aqueous solution. Note: EDTA is abbreviated Y in equilibrium expressions. A. (2 pts) Calculate the molarity of this solution. B. (3 pts) Suppose EDTA is of unknown purity, and this solution needs to be standardized. In one or two complete sentences, maximum 50 words, describe a titration procedure (naming specific reagent(s)) that could be used to establish the concentration of the solution. C. (5 pts) Calculate the conditional formation constant for the titration of TI (aq) at pH 250; use data from Tables in Harris, and you may interpolate a Table for ouY4. D. (5 pts) write an appropriate algebraic expression that could be used for calculating ay4. at pH 2.50, defining all terms clearly; include the appropriate value of IH30 1.

Explanation / Answer

A. molarity of solution = moles/L

                                    = 0.01 mol/1 L

                                    = 0.01 M

B. To titrate excess of EDTA in solution, we add another anion which gives absorbance at different lambda-max value. The moles of anion solution used tells us about the amount of excess EDTA present in solution.

C. conditional formation constant for Tl3+

logKf = 21.3

at pH 2.50

Kf' = Kf x alpha[Y4-]

     = 2 x 10^21 x 1.3 x 10^-11

     = 2.6 x 10^10

D. pH = -log[H+] = 2.50

[H+] = 3.16 x 10^-3 M

Expression for alpha[Y4-]

from K1, K2, K3, K4, K5, K6 for EDTA

alpha[Y4-] = [H+]^6/([H+]^6 + K1[H+]^5 + K1K2[H+]^4 + K1K2K2[H+]^3 + K1K2K3K4[H+]^2 + K1K2K3K4K5[H+] + K1K2K3K4K5K6)

Feeding all the values we can easily calculate alpha[Y4-]

E. Titration

F. 12.5 ml EDTA added

this is half equivalence point

[Tl3+] = 1/Kf' = 1/2.6 x 10^10 = 3.85 x 10^-11 M

pTl = -log[Tl3+] = 10.41

G. Equivalence point

[TlY2-] formed = 0.01 M x 25 ml/50 ml = 0.005 M

[Tl3+] = sq.rt.(0.005/2.6 x 10^-10) = 4.38 x 10^-7 M

pTl = 6.36

H. 37.5 ml of EDTA added

excess [EDTA] = 0.01 M x 12.5 ml/62.5 ml = 0.002 M

[TlY2-] formed = 0.01 M x 25 ml/62.5 ml = 0.004 M

[Tl3+] = 0.004/2.6 x 10^10 x 0.002 = 7.69 x 10^-11 M

pTl = 10.11

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