Calculate pMn2 at each of the following points in the titration of 24.06 mL of 0

Question

Calculate pMn2 at each of the following points in the titration of 24.06 mL of 0.0560 M EDTA with 0.0280 M MnCl2. The EDTA solution is buffered at a pH of 9.00. The fraction of free EDTA in the Y4– form (?Y4–) can be found here. The formation constant for the Mn2 -EDTA complex is given by log Kf = 13.89.

Link: https://sites.google.com/site/chempendix/fraction-of-free-edta-as-y4

a) 4.812 mL f) 48.12 ml Number Number b) 19.25 mlL g) 48.22 mL Number Number c) 38.50 ml h) 52.93 mL Number Number d) 47.16 mlL i) 57.74 mL Number Number e) 48.02 mlL

Explanation / Answer

logKf = 13.89

Kf = 7.76 x 10^13

Kf' = Kf x alfa[Y4-] = 7.76 x 10^13 x 0.041 = 3.2 x 10^12

a) moles of EDTA = 0.056 M x 24.06 ml = 1.35 mmol

moles of MnCl2 = 0.028 M x 4.812 ml = 0.135 mmol

excess [EDTA] = 1.215/28.872 = 0.0421 M

[MnY2-] = 0.135/28.872 = 0.0047 M

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

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

[Mn2+] = 3.50 x 10^-14 M

pMn2 = -log[Mn2+] = 13.46

b) moles of EDTA = 0.056 M x 24.06 ml = 1.35 mmol

moles of MnCl2 = 0.028 M x 19.25 ml = 0.539 mmol

excess [EDTA] = 0.811/43.31 = 0.019 M

[MnY2-] = 0.539/43.31 = 0.0124 M

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

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

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

pMn2 = -log[Mn2+] = 12.69

c) moles of EDTA = 0.056 M x 24.06 ml = 1.35 mmol

moles of MnCl2 = 0.028 M x 38.50 ml = 1.078 mmol

excess [EDTA] = 0.272/62.56 = 0.004 M

[MnY2-] = 1.078/62.56 = 0.017 M

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

3.2 x 10^12 = 0.017/0.004[Mn2+]

[Mn2+] = 1.33 x 10^-12 M

pMn2 = -log[Mn2+] = 11.88

d) moles of EDTA = 0.056 M x 24.06 ml = 1.35 mmol

moles of MnCl2 = 0.028 M x 47.16 ml = 1.3205 mmol

excess [EDTA] = 0.03/71.22 = 0.0004 M

[MnY2-] = 1.3205/71.22 = 0.0047 M

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

3.2 x 10^12 = 0.0047/0.0004[Mn2+]

[Mn2+] = 3.67 x 10^-12 M

pMn2 = -log[Mn2+] = 11.43

e) moles of EDTA = 0.056 M x 24.06 ml = 1.35 mmol

moles of MnCl2 = 0.028 M x 48.02 ml = 1.34456 mmol

excess [EDTA] = 0.00544/72.08 = 0.000075 M

[MnY2-] = 1.34456/72.08 = 0.0186 M

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

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

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

pMn2 = -log[Mn2+] = 10.11

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