Question

A beaker containing 50.0 mL of 0.300 M Ca2+ at pH 9 is titrated with 0.150 M EDTA. The pCa2+ at the equivalence point is

Answer 1

The reaction between Ca2+ and EDTA is a 1 mole to 1 mole reaction

50 mL of 0.30 M Ca²⁺ solution will have 50 x 0.3/1000 = 0.015 moles of calcium

To get 0.015 moles of EDTA from a solution with concentration 0.150 M (0.150 moles/L or 0.150 moles/1000mL)

0.015moles/0.15moles/1000mL = 100 mL

So the final volume if 50 + 100 mL = 150 mL

at equivalence point there will be 0.015 moles of CaEDTA^2- so its concentration will be (0.015/150ml)x1000 = 0.1M

Now the formation constant for EDTA with Ca2+ is reported to be 10.65

At its most basic level, Kf can be explained as the following, where M is a metal ion, L is the ligand EDTA, ML is the complex

M + L ML

Kf=[ML]/[M][L]

Kf=forward/reverse

If we have to calculate pCa2+ at equivalence point we will need to take a reverse equilibrium

ML M + L

Where M = Ca²⁺, L=EDTA and ML = CaEDTA^2-

K=reverse/forward

K=1/Kf = 1/10.65 = 0.0938

To find concentrations we will set up a ICE table

	CaEDTA2-	Ca2+	EDTA
Initial	0.1M	0	0
Change	-X	+X	+X
Equilibrium	0.1-x	x	x

K for this process is

K = [Ca²⁺][EDTA]/[CaEDTA^2-]

0.0938 = x²/0.1-x

x² - 0.00938 + 0.0938x = 0

This is a quadratic equation with roots 0.0607 and -0.1545; negative root can be ignored

so x that is Ca²⁺ concentration will be 0.0607

pCa²⁺ = -log[Ca²⁺]

pCa²⁺ = - log(0.0607)

pCa²⁺ = 1.21