You are titrating 130.0 mL of 0.080 M Ca2 with 0.080 M EDTA at pH 9.00. Log Kf f

Question

You are titrating 130.0 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.

I just need help with E. Please!

Map ost Numlbe K. (pH 9.00) 183 x 1010 (b) What is the equivalence volume, Ve, in milliliters? Number mL (c) Calculate the concentration of Ca2+ at V-1/2 Ve Number Ca2+1= 11.02667 (d) Calculate the concentration of Ca2 at V-Ve. Incorrect. Number [ca 1-114.68 x 10-1 M (e) Calculate the concentration of Ca2+ at V= 1.1 V. Number

Explanation / Answer

(E) at V = 1.1Ve

Volume of EDTA added = 130 x 0.1 + 130 = 143 ml

moles of Ca2+ = 0.08 M x 130 ml = 10.4 mmol

moles of EDTA = 0.08 M x 143 ml = 11.44 mmol

[CaY2-] formed = 10.4 mmol/273 ml = 0.0381 M

[EDTA] remained = 1.04 mmol/273 ml = 0.004 M

Kf = 4.47 x 10^10

Kf' = alpha[Y4-] x Kf = 0.041 x 4.47 x 10^10 = 1.83 x 10^9

Kf' = [CaY2-]/[Ca2+][EDTA]

1.83 x 10^9 = 0.0381/[Ca2+](0.004)

[Ca2+] concentration = 5.46 x 10^-9 M

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