Calculate pMn2 at each of the following points in the titration of 20.31 mL of 0

Question

Calculate pMn2 at each of the following points in the titration of 20.31 mL of 0.0472 M EDTA with 0.0236 M MnCl2. The EDTA solution is buffered at a pH of 10.00. 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.

a) 4.062 mL

Values of Y4-for EDTA at 25°C and =0.10M

b) 16.25 mL

c) 32.50 mL

d) 39.81 mL

e) 40.54 mL

f) 40.62 mL

g) 40.70 mL

h) 44.68 mL

i) 48.74 mL

pH Y4- 0 1.3 x 10-23 1 1.4 x 10-18 2 2.6 x 10-14 3 2.1 x 10-11 4 3.0 x 10-9 5 2.9 x 10-7 6 1.8 x 10-5 7 3.8 x 10-4 8 4.2 x 10-3 9 0.041 10 0.30 11 0.81 12 0.98 13 1.00 14 1.00

Explanation / Answer

Step 1: Formula:

EDTA + Y4- = EDTA-Mn

Step 2:Equilibration:

EDTA + Mn- = EDTA-Mn

I: 0 + 0 =13.89

C: +x +Y-4+x =-x

E=x +Y-4+x =13.89 -x

Kf= [EDTA-Mn]/[Mn2] + [EDTA]

[Mn2] + [EDTA] * Kf=[EDTA-Mn]

[Mn]= [EDTA-Mn]/[EDTA] * Kf

[Mn]= [13.89]/[0.0472]*[0.30]

[Mn]=88

pMn=-log[Mn]=4.5664X1088

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