Question

constract the titration curve of pCa versus Volume of EDTA for 50.00 ml of 0.00500 M Ca+ titrated with 0.0100 M EDTA in a solution buffered to ph 10.0

a) in the pre-equivalent point regim at 5.00 ml

b) at the equivalent point 25.00 ml

c) after equivalent point 26.00 ml

Answer 1

Initial moles of Ca2+ = MxV = 0.00500 Mx 0.05000 L = 2.5x10^-4 mol

(a) When 5.00 ml( = 0.005 L) of 0.0100 M ETDA is added,

Moles of EDTA added = MxV = 0.0100M x 0.005L =  5x10^-5 mol

The chemical reaction is

----------------Ca²⁺ + EDTA^4-  ----- > CaEDTA^2-

Init.mol:   2.5x10^-4 mol, 5x10^-5 mol ------- 0

Eqm.mol:2.0x10^-4 mol, 0   mol ------- 5x10^-5 mol

Total volume, Vt = 50.0 mL + 5 mL = 55 mL = 0.055 L

=> [Ca²⁺] = 2.0x10^-4 mol /  0.055 L = 0.003636

pCa²⁺ = - log[Ca²⁺] = 2.439 (answer)

(b):

When 25.00 ml( = 0.005 L) of 0.0100 M ETDA is added,

Moles of EDTA added = MxV = 0.0100M x 0.025L = 2.5x10^-4 mol

The chemical reaction is

----------------Ca²⁺ + EDTA^4-  ----- > CaEDTA^2-

Init.mol:   2.5x10^-4 mol, 2.5x10^-4 mol ------- 0

Eqm.mol: 0 mol, 0   mol ------- 2.5x10^-4 mol

Total volume, Vt = 50.0 mL + 25 mL = 75 mL = 0.075 L

=> [Ca²⁺] 0