Question

Find the conditional formation constant for Ba(EDTA)2− at pH 10.00, where log⁡Kf is 7.88 and αY4− is 0.30.

Kf′=

Find the concentration of free Ba2+ in 0.040 M Na2[Ba(EDTA)] at pH 10.00.

Answer 1

The conditional formation constant can be calculated as:

${K_{f}}'=K_{f}\cdot \alpha _{Y^{4-}}=10^{7.8}\cdot 0.3=1.89x10^{7}$

And for the equilibrium:

$Ba^{2+}+EDTA^{4-}\leftrightharpoons [Ba(EDTA)^{2-}]$

it represents:

${K_{f}}'=\frac{[Ba(EDTA)^{2-}]}{[Ba^{2+}]+C_{EDTA}}$

Were C_EDTA is the total concentration of all EDTA species.

We can build an ICE chart to determine the desired Ba(II) concentration:

	Ba(II)	EDTA	Ba(EDTA)2-
initial	0	0	0.040 M
change	+x	+x	-x
equilibrium	x	x	0.040 - x

Se we can write the expression of the constant as:

${K_{f}}'=1.89x10^{7}=\frac{0.040-x}{x^{2}}$

This is a quadratic equation, which can be solved to yield:

$x_{1}=4.59x10^{-5}\, M\, ;\, x_{2}=-4.60x10^{-5}\, M$

x₁ is the only one whic makes chemical sense (negative concentrations don't exist), so the concentration of Ba²⁺ is 4.59 x 10^-5.