Calculate pMn at each of the following points in the titration of 24.06 mL of 0.

Question

Calculate pMn at each of the following points in the titration of 24.06 mL of 0.0560 M EDTA with 0.0280 M Map MnCI2. The EDTA solution is buffered at a pH of 9.00. The fraction of free EDTA in the YA form (ay can be found here. The formation constant for the Mn? -EDTA complex is given by log Kf 13.89. f 48.12 mL a) 4.812 mL Number Number 13.46 b) 19.25 mL g) 48.22 mL Number Number 12.69 c) 38.50 mL h) 52.93 mL Number Number 11.88 Scroll down to view the entire question. i) 57.74 mL d) 47.16 mL Number Number

Explanation / Answer

Kf = 7.67 x 10^13

Kf' = Kf.alpha[Y4-] = 7.67 x 10^13 x 0.041 = 3.2 x 10^12

(a) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 4.812 ml = 0.135 mmol

excess [EDTA] = (1.35 - 0.135) mmol/28.872 ml = 0.0421 M

[MnY2-] formed = 0.135 mmol/28.872 ml = 0.0047 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.0047/[Mn2+](0.0421)

[Mn2+] = 3.5 x 10^-14 M

pMn2+ = -log[Mn2+] = 13.456

(b) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 19.25 ml = 0.539 mmol

excess [EDTA] = (1.35 - 0.539) mmol/43.31 ml = 0.019 M

[MnY2-] formed = 0.539 mmol/43.31 ml = 0.0124 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.0124/[Mn2+](0.019)

[Mn2+] = 2.04 x 10^-13 M

pMn2+ = -log[Mn2+] = 12.69

(c) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 38.50 ml = 1.078 mmol

excess [EDTA] = (1.35 - 1.078) mmol/62.56 ml = 0.0043 M

[MnY2-] formed = 1.078 mmol/62.56 ml = 0.0172 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.0172/[Mn2+](0.0043)

[Mn2+] = 1.25 x 10^-12 M

pMn2+ = -log[Mn2+] = 11.90

(d) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 47.16 ml = 1.3208 mmol

excess [EDTA] = (1.35 - 1.3208) mmol/71.22 ml = 0.00041 M

[MnY2-] formed = 1.3208 mmol/71.22 ml = 0.0185 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.0185/[Mn2+](0.00041)

[Mn2+] = 1.41 x 10^-11 M

pMn2+ = -log[Mn2+] = 10.851

(e) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 48.02 ml = 1.34456 mmol

excess [EDTA] = (1.35 - 1.34456) mmol/72.08 ml = 0.000075 M

[MnY2-] formed = 1.34456 mmol/72.08 ml = 0.0186 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.0186/[Mn2+](0.000075)

[Mn2+] = 7.75 x 10^-11 M

pMn2+ = -log[Mn2+] = 10.11

(f) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 48.12 ml = 1.35mmol

Equivalence point

[MnY2-] formed = 1.35 mmol/72.18 ml = 0.019 M

Kf' = [MnY2-]/[Mn2+][EDTA]

3.2 x 10^12 = 0.019/x^2

[Mn2+] = 7.71 x 10^-8 M

pMn2+ = -log[Mn2+] = 7.113

(g) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 48.22 ml = 1.35016 mmol

excess [Mn2+] = 0.00016 mmol/72.28 ml = 2.24 x 10^-6 M

pMn2+ = -log[Mn2+] = 5.65

(h) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 52.93 ml = 1.48204 mmol

excess [Mn2+] = 0.13204 mmol/76.99 ml = 0.0017 M

pMn2+ = -log[Mn2+] = 2.77

(i) initial EDTA = 0.0560 M x 24.06 ml = 1.35 mmol

MnCl2 added = 0.028 M x 57.74 ml = 1.62 mmol

excess [Mn2+] = 0.27 mmol/81.8 ml = 0.0033 M

pMn2+ = -log[Mn2+] = 2.481

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