Question

the formation constant for the complex formed on reaction of calcium ion with EDTA, CaY^₂-, is 4.47 x 10^_-11. calculate the concentration of free ca²⁺ in a solution of 0.1 M CaY^₂ at pH 10 and at pH 6. ionic strength effects may be ignored

Answer 1

From EDTA titrations,

Ca²⁺+EDTA <----------->CaY^2- ,K_f=4.47×10^-11~10^10.65

^{we know that ,} Formation constant K_f^'=α_Y^4-×K_f

*At P^H=10.00, (α_Y^4-=0.30)

K_f^'=α_Y^4-×K_f =>(0.30)(10^10.65)=1.3×10¹⁰

*At P^H=6.00, (α_Y⁴=1.8×10^-5)

K_f^'=α_Y⁴×K_f=>(1.8×10^-5)(10^10.65)=8.0×10⁵

Ca²⁺+EDTA <---------------------->CaY^2-

Initial concentration 0 + 0 0.1

Final concentration x + x 0.1-x

Formation constantK_f^'=[CaY^2-]/[Ca²⁺][EDTA]

=0.1-X /X²

*If P^H=10 then K_f^'=0.1-X/X²=1.3×10¹⁰ (* aX²+bX+c=0)

[Ca²⁺]=X=2.773×10^-6M

*If P^H=6 then K_f^'=0.1-X/X²=8.0×10⁵

[Ca²⁺]=X=3.5×10^-4M