apling Learning cmillan learning You are titrating 120.0 mL of o.080 M Ca2\" wit

Question

apling Learning cmillan learning You are titrating 120.0 mL of o.080 M Ca2" with o 080 M EDTA at pH 9.00. Log K for the Ca "-EDTA complex is 10.65, and the fraction of free EDTA in the Y4 form, ay4 is 0.041 at pH 9.00 (a) what is Kf, the conditional formation constant, for Ca at pH 9.00? Number (pH 9.00 b) What is the equivalence volume, Ve, in milliliters? Number m L c) Calculate the concentration of Ca at V 1/2 ve. Number (d) Calculate the concentration of Ca at V Ve. Number M a Previous Give Un & View Solution Check Answer a Next

Explanation / Answer

Titration,

(a) conditional formation constant Kf' = Kf x alpha[Y^4-]

log Kf = 10.65

Kf = 4.47 x 10^10

So,

Kf' = 4.47 x 10^10 x 0.041 = 1.83 x 10^9

(b) Equivalence Ve volume

= 0.08 M x 120 ml/0.08 M

= 120 ml

(c) concentration Ca2+ at V = 1/2Ve

1/2Ve = 60 ml

[Ca2+] = 0.08 M x 120 ml/(120 + 60)ml = 0.053 M

(d) concentration of Ca2+ at V = Ve

All of Ca2+ has formed CaEDTA^2- complex

[CaEDTA^2-] = 0.08 M x 120 ml/240 ml = 0.04 M

Ca2+ + EDTA <==> CaEDTA^2-

let x amount of complex has dissolved

Kf' = [CaEDTA^2-]/[Ca2+][EDTA]

1.83 x 10^9 = 0.04/x^2

x = [Ca2+] = 4.66 x 10^-6 M

(e) concentration of Ca2+ at V = 1.1Ve

1.1Ve = 120 + 0.1 x 120 = 132 ml

[CaEDTA^2-] formed = 0.08 M x 120 ml/252 ml = 0.0381 M

[EDTA] excess = 0.08 M x 12 ml/252 ml = 0.00381 M

Kf' = 1.83 x 10^9 = 0.0381/[Ca2+] x 0.00381

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

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