Answer to Question #107974 in Chemistry for tony

Question

You have 75 mL of a solution with [Zn2+] = 0.031 M. The pH of this solution = 8. You titrate this solution with 0.055M EDTA. Remember that pMn+ = -log(Mn+)
a) What is true at the equivalence volume for any titration?
b) Calculate the Ve (the equivalence volume) for this titration?
c) Calculate the pZn2+ after addition of 20 mL EDTA
d) Calculate the pZn2+ at the equivalence volume
e) Calculate the pZn2+ after addition of 50 mL EDTA

Answer 1

a) At the equivalence point of any titration the number of equivalents (in moles) of the titrant is equival to the number of equivalents (in moles) of the titrated substance.

b) The titration is based on complex formation:

Zn²⁺ + EDTA^4- = ZnEDTA^2-

From the definition of equivalence point, [EDTA^4-]*Ve = V_Zn*[Zn²⁺]

Ve = 75*0.031/0.055 = 42.27 mL

c) after the reaction with 20 mL of EDTA solution, the rest amount of Zn²⁺ is

0.075*0.031 - 0.020*0.055 = 0.001225 mole,

the final volume of solution is 0.075 + 0.020 = 0.095 L,

concentration of [Zn²⁺] = 0.001225/0.095 = 0.013 M,

pZn²⁺ = 1.89

d) at the equivalence point the concentration of Zn²⁺ is only due to the ZnEDTA^2- dissociation equilibrium, no excess concentration of Zn²⁺ and EDTA^2- are in the solution. The logarithm of formation constant of the complex is lgK_f = 16.5.

K_f = [ZnEDTA^2-]/([Zn²⁺]*[EDTA^4-])

Concentration of [ZnEDTA^2-] equals to that of initial [Zn²⁺] concentration (or the added [EDTA^4-] concentration) with an account of dilution:

[ZnEDTA^2-] = 0.031*75/(75+42.27) = 0.020 M

Equilibrium concentrations of [Zn²⁺] and [EDTA^4-] are equal, hence:

[Zn²⁺] = sqrt([ZnEDTA^2-]/K_f) = sqrt(0.020/10^16.5) = 10^-9 M,

pZn²⁺ = 9

e) after addition of 50 mL EDTA (above the equivalence volume) the concentration of Zn²⁺ is due to complex ZnEDTA^2- dissociation in presence of excess of EDTA^4-. The amoun of free EDTA^4- ligand is 0.050*0.055 - 0.04227*0.055 = 0.000425 mole, the final volume is 0.075 + 0.050 = 0.125 L, the concentration of excess [EDTA^4-] = 0.000425/0.125 = 0.0034 M.

The concentration of [ZnEDTA^2-] equals to that that of initial [Zn²⁺] with an account of dilution:

[ZnEDTA^2-] = 0.031*75/(75+50) = 0.0186 M.

From the expression for formation constant we find that:

[Zn²⁺] = [ZnEDTA^2-]/(K_f*[EDTA^4-]) = 0.0186/(10^16.5*0.0034) = 10^-15.76

pZn²⁺ = 15.76