Question

A beaker containing 50.0 ml of 0.300M Ca2+ at ph 9 is titrated with 0.150M EDTA. Calculate the pCa2+ at the equivalence point

Answer 1

Consider a complex formation reaction: Ca ²⁺ + EDTA [ CaY ] ^2-

For above reaction, conditional formation constant is K' _f = (_Y 4-) K _f

Therefore, K' _f = 0.041 x 10 ^10.65 = 1.83 x 10 ⁹

First , we Calculate  of volume of EDTA required to reach equivalence point

From above reaction , we have 1 mol Ca ²⁺ 1 mol  EDTA

Therefore, M _ca 2+ V _Ca2+ = M _EDTA V _EDTA

V _EDTA = M _ca 2+ V _Ca2+ / M _EDTA =

V _EDTA = 0.300 M 50.0 ml / 0.150 M

V _EDTA = 100 ml

volume of EDTA required to reach equivalence point = 100 ml

At the equivalence point, all the calcium ions are consumed by added EDTA. Hence, at this stage concentration of Ca ²⁺ will be due to dissociation of [CaY] ^2-.

Moles of [ CaY ] ^2- = moles of EDTA added to the solution

No. of moles of EDTA added in the solution = 0.150 mol / L 0.100 L = 0.015 mol

Volume of solution at equivalence point = Volume of Ca ²⁺ solution + volume of EDTA solution

Volume of solution at equivalence point = 50.0 + 100 ml =150.0 ml = 0.150 L

[CaY] ^2-. = 0.015 mol / 0.150 L = 0.100 M

Let’s use ICE table.

Concentration ( M )	Ca 2+	EDTA	[CaY] 2-
I			0.100
C	+X	+X	-X
E	X	X	0.100 - X

We have , K' _f = [CaY^2-] / [Ca ²⁺] [EDTA]

Therefore,K' _f = 0.100 -X / (X)(X) = 1.8310 ⁹

0.100 -X / (X)(X) = 1.83 10 ⁹

0.100  -X / X ² = 1.83 10 ⁹

At equilibrium X is negligible as compared 0.100. Then we can write 0.100 / x ² = 1.83 x 10 ⁹

X ² = 0.100 / 1.83 10 ⁹ =5.464 10 ^-11

X = 7.392 10 ^{- 6} M

[Ca ²⁺] = 7.932 10 ^{- 6} M

We have relation, pCa ²⁺ = - log [Ca ²⁺]

pCa ²⁺ = - log 7.932 10 ^{- 6} = 5.13

ANSWER : pCa ²⁺ at equivalence point = 5.13