Calculate pMn 2+ at each of the following points in the titration of 26.65 mL of

Question

Calculate pMn2+ at each of the following points in the titration of 26.65 mL of 0.0212 M EDTA with 0.0106 M MnCl2. The EDTA solution is buffered at a pH of 10.00. The fraction of free EDTA in the Y4- (ay4) can be form found here. The formation constant for the Mn -EDTA complex is given by log Kf 13.89

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

A) 53.19 mL

B) 53.30 mL

C)53.41 mL

D) 58.63 mL

E) 63.96 mL

Calculate pMn at each of the following points in the titration of 26.65 mL of 0.0212 M EDTA with 0.0106 M Mncl2. The EDTA solution is buffered at a pH of 10.00. The fraction of free EDTA in the Y4 form (ay4 can be found here. The formation constant for the Mn2+-EDTA complex is given by log Kf 13.89

Explanation / Answer

This titration is given by the following equation:

Mn2+ + EDTA4- ------>MnY2-

Using M1V1 = M2V2, we can find the volume of Mn2+ solution to reach the equivalence point.

V1 = M2V2 / M1

= 26.65 mL X 0.0212 M / 0.0106 M

= 53.3 mL

Therefore to reach the equivalence point B

One volume of Mn2+ = 2 X volumes of EDTA

[MnY2-] at the equivalence point = Concentration of Mn2+ / total volume

(0.0106) /3

= 0.0035 M

If x is the concentration of Mn2+ and EDTA4- at equivalence point, then the concentration of MnY2- will be 0.0035-x

ay4- Kf = [MnY2-] / [Mn2+] [EDTA4-]

We know that at pH 10, ay4- = 0.3

Therefore,

0.3 X 13.89 = 0.0035 –x / x2

4.017 = 0.0035 –x / x2

4.017x2 = 0.0035 –x

4.017x2 + x -0.0035 = 0

For a quadratic equation ax2 +bx +c=0

x1,2 = [-b ± (b2-4ac)] / 2a

In our case a = 4.017, b =1, c = -0.0035

x = [-1 ± (12+4•4.017•0.0035)] / 2•4.017

= [-1 ± (1+0.056238)] / 8.034

= [-1 ± (1.056238)] / 8.034

x = [-1 + 1.0277] / 8.034 or x = [-1 – 1.0277] / 8.034

x = 0.0277 / 8.034 or x = -2.0277 / 8.034

x= 0.00345 M or x = -0.2489 M

Concentration cannot be negative, therefore x = 0.00345 M

[Mn2+] = 0.00345 M

pMn2+ = -log [Mn2+]

= -log 0.00345

pMn2+ = 2.46 at equivalence point B

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