Answer:
Explanation:
a).
conc of Ca²⁺ =0.0025 M
pCa = -log(0.0025) = 2.6
logK,= 10.65 So lc = 4.47 x 10.
Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is [tex]\alpha_{Y^{-4}}[/tex] =0.81
So the Conditional Formation constant= [tex]K_f[/tex] =0.81x 4.47 x10¹⁰
=3.62x10¹⁰
b)
At Equivalence point:
Ca²⁺ forms 1:1 complex with EDTA At equivalence point,
Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol
Number of moles of EDTA= 0.125 mol
Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL
V e= 25.00 mL
At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.
[tex][CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M[/tex]
[tex]{K^'}_f[/tex]
Ca²⁺ + Y⁴ ⇄ CaY²⁻
Initial 0 0 0.001667
change +x +x -x
equilibrium x x 0.001667 - x
[tex]{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\[/tex]
[tex]x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14[/tex]
x = 2.15×10⁻⁷
[Ca+2] = 2.15x10⁻⁷ M
pca = —log(2 15x101= 6.7
