You are titrating 150.0 mL of 0.080 M Ca2 with 0.080 M EDTA at pH 9.00. Log Kf f

Question

You are titrating 150.0 mL of 0.080 M Ca2 with 0.080 M EDTA at pH 9.00. Log Kf for the Ca2 -EDTA complex is 10.65, and the fraction of free EDTA in the Y4– form, ?Y4–, is 0.041 at pH 9.00.

Please answer parts A-E, thank you so much

(a) What is K, the conditional formation constant, for Ca2+ at pH 9.00? Number K, (pH 9.00)= ID (b) What is the equivalence volume, Ve, in milliliters? Number mL (c) Calculate the concentration of Ca2+ at V= 1 /2 V. Number 2+ (d) Calculate the concentration of Ca2+ at V= Ve. Number [Ca2+ ]- (e) Calculate the concentration of Ca2+ at V= 1.1 V.

Explanation / Answer

(a) Kf = 10^10.65 = 4.47*10^10

Kf' = alpha * Kf = 0.041 * 4.47*10^10 = 0.183*10^10

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M1 V1 = M2V2 . where M1 and m2 is the concentration of Ca^2+ solution and EDTA solution respectively. V1 and V2 is the volume of Ca^2+ solution and EDTA solution.

V2 = M1V1/M2 = 150mL*0.08 M/0.08 M = 150 mL

equivalence volume if 150 mL

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volume at equivalence point = 150mL = 0.15 L, V= 1/2Ve = 0.075 L

inital concentration od ca^2+ = 0.08 M *0.15L = 0.012

moles of EDTA added = 0.08 * 0.075 L = 0.006 moles

[Ca^2+] = 0.012-0.006 = 0.006 moles

Molarity of Ca^2+ = 0.006moles/0.0.75 = 0.08 M

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At equivalence point a equilibrium will exist and Ca_EDTA comples will break down to form Ca^2+ and EDTA.

Ca-EDTA <===>Ca^2+ + EDTA

Kf' =[Ca-EDTA]/[Ca^2+][EDTA]

or, 0.183*10^10 = 0.012-x/x^2

or, x =2.56*10^-6

[Ca^2+] = 2.56*10^-6 M

Ca-EDTA EDTA ca^2+ inital 0.012 0 0 change -x +x +x equilibrium 0.012-x x x

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