You are titrating 120.0 mL of 0.080 M Ca2+ with 0.080 M EDTA at pH 9.00....

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Chemistry

You are titrating 120.0 mL of 0.080 M Ca2+ with 0.080 M EDTA atpH 9.00. Log Kf for the Ca2+ -EDTA complex is 10.65, and thefraction of free EDTA in the Y4â€“ form, Î±Y4â€“, is 0.041 at pH9.00.
(a) What is K'f, the conditional formation constant, for Ca2+ atpH 9.00?
(b) What is the equivalence volume, Ve, in milliliters?
(c) Calculate the concentration of Ca2 at V = 1/2 Ve.
(d) Calculate the concentration of Ca2 at V = Ve.
(e) Calculate the concentration of Ca2 at V = 1.1 Ve.

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4.3 Ratings (897 Votes)

a Consider a complex formation reaction Ca 2 EDTA CaY 2For above reaction conditional formation constant is Kf Y 4K fTherefore K f 0041 x 10 1065 183x 10 9b calculation of equivalence volumeFrom above reaction we can write No of moles of Ca2 No of moles of EDTATherefore M ca 2 x V Ca2 MEDTA x V EDTA VEDTA M ca 2 x V Ca2 MEDTA 0080 M x 1200 ml 0080 M 1200 ml Equivalencevolume 1200 mlC Ca 2 at V 12 Vemmol of Ca 2 See Answer

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