Question

Like to a hydrogen-ion buffer, a metal-ion buffer tends to maintain a particular metal-ion concentration in solution.
By rough analogy between the equations,
HA(aq) ⇌ H⁺(aq) + A⁻(aq)    K = K_a
and
CaY²⁻ ⇌ Ca²⁺ + "EDTA"    K = K'_f^-1
determine the total mass of Na₂EDTA·2H₂O (FM 372.24) which must be dissolved, along with 2.15 g of Ca(NO₃)₂·2H₂O (FM 200.13), in 500.0 mL of solution (acid-buffered at pH 9.24) to give a calcium-ion buffer with pCa²⁺ = 8.70?

For the formation equation

Y⁴⁻ + Ca²⁺ ⇌ CaY²⁻, logK_f   = 10.65

Answer 1

Given data set:

The reaction equations are:

HA(aq) H=(aq) + A-(aq) K= Ka

CaY2- Ca2+ + EDTA K = K'f-1

Mass of Ca(NO3)2.H2O = 2.15 g

Molar mass of Ca(NO3)2.H2O = 200.13 g/mol

Volume of solution = 500 ml = 0.5 L

pH = 9.24

Let us calculate the concentration of CaY2- using the relation:

Molarity = moles/Volume

Moles= Given mass/Molar mass

Moles of Ca(NO3)2.H2O = 2.15 g/200.13 g/mol = 0.0107 moles

So, Molarity of Ca(NO3)2.H2O = 0.0107 moles/0.5 L = 0.0214 M

i.e [CaY2-] = 0.0214 M

Now,

Y⁴⁻ + Ca²⁺ ⇌ CaY²⁻ logK_f   = 10.65

Kf = 10^10.65 = 4.46*10^10

Now for pH = 9.24, Y4- = 0.041

Also, Kf' = (Y4-)*(Kf)

= 0.041*4.46*10^10

= 1.83*10^9

So, Kf' = [CaY2-]/[EDTA][Ca2+]

Substituting given,

1.83*10^9 = 0.0214/(x*10^-9.24)

1.83*10^9 = 0.0214/(x*(5.75*10^-10))

x = 0.0109