Question

Question 6. In titration of 500 ml 0.2 Mn2+ with 0.8 M EDTA (pH = 4), when 200 ml EDTA is added (3 pts) What is the fraction of EDTA in totally unprotonated form? (use the table in lecture notes no calculation is necessary) (3 pts) What is the conditional formation constant? (3 pts) Indicate and calculate the excess and limiting species? (3 pts) What is pMn2+? (3 pts) What are the sources of Mn2+ at beforethe equivalence point, after the equivalence point and the equivalence point?

Answer 1

Solution:

(A) At any pH a mass balance requires that the

total concentration of unbound EDTA equal the

combined concentrations of each of its forms.

CEDTA = [H6Y2+] + [H5Y+] + [H4Y] + [H3Y–] +

[H2Y2–] + [HY3–] + [Y4–]

at pH = 4 :

@Y4–  =   [Y4–]/  CEDTA = 3.8*10^-9 ( from literature value )

note:(@ denotes alpha symbol)

(b) What is the conditional formation constant?

at pH = 4

log Kf = 13.89

Kf' = Kf*@Y4–    = 10^13.89* 3.8*10^-9=

2.95*10^5

(c) Indicate and calculate the excess and limiting species?

this is 1:1 reaction :

moles of EDTA = 0.200 L * 0.8 mol/L = 0.160

moles (excess species)

moles of Mn2+ = 0.500 L * 0.2 mol/L = 0.100

moles (limiting species)

(d) What is pMn2+?

this is after equivalence point :
Kf' = [MnY2-] / [Mn2+] [EDTA] = 2.95*10^5

[EDTA] = 0.160 mol/0.700 L = 0.2286 M

since there is completion of reaction :

[MnY2-]eq = (0.100 mole/ 0.700 L) = 0.01429 M - x

[Mn2+]eq = x

Kf' = (0.01429 - x ) / (x)* 0.2286   = 2.95*10^5

[Mn2+] eq = 2.12*10^-7 M

pMn2+ = 6.67

If you have any doubts please comment. I will help you.