Question

You are titrating 120.0 mL of 0.080 M Ca2+ with 0.080 M EDTA at pH 9.00. Log Kf for the Ca2+ -EDTA complex is 10.65, and the fraction of free EDTA in the Y4– form, αY4–, is 0.041 at pH 9.00.

(a) What is K'f, the conditional formation constant, for Ca2+ at pH 9.00?

(b) What is the equivalence volume, Ve, in milliliters?

(c) Calculate the concentration of Ca2 at V = 1/2 Ve.

(d) Calculate the concentration of Ca2 at V = Ve.

(e) Calculate the concentration of Ca2 at V = 1.1 Ve.

Answer 1

a) 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

b) calculation of equivalence volume

From above reaction we can write , No. of moles of Ca ²⁺ = No. of moles of EDTA

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

V _EDTA = M _ca 2+ x V _Ca2+ / M _EDTA = 0.080 M x 120.0 ml / 0.080 M = 120.0 ml

Equivalence volume = 120.0 ml

C) [Ca ²⁺] at V = 1/2 Ve

mmol of Ca ²⁺ = concentration x volume = 0.080 x 120 = 9.6 mmol

mmol of EDTA = concentration x volume = 0.080 x 60.0 =4.8 mmol

mmol of excess Ca ²⁺ = mmol of Ca ²⁺ - mmol of EDTA = 9.6 mmol - 4.8 mmol = 4.8 mmol

Total volume of solution = Volume of Ca ²⁺ + volume of EDTA= 120.0 + 60.0 = 180.0 ml

[Ca ²⁺] = No. of moles of Ca ²⁺ / Volume of solution in L

[Ca ²⁺] = [ 4.8 / 1000] / [ 180.0 /1000 ] = 0.0267 M

D) [Ca ²⁺] at V = Ve

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-.

mmol of Ca ²⁺ = concentration x volume = 0.080 x 120 = 9.6 mmol

mmol of EDTA = concentration x volume = 0.080 x 120 =9.6 mmol

mmol of [CaY] ^2- produced = mmol of Ca ²⁺ = mmol of EDTA = 9.6 mmol

Total volume of solution = Volume of Ca ²⁺ + volume of EDTA= 120.0 +120 = 240 ml

[ [CaY] ^2- ] = No. of moles of  [CaY] ^2- / Volume of solution in L

   [ [CaY] ^2- ] =  [9.6/ 1000] / [ 240 /1000 ] = 0.04 M

Lets use ICE table.

Concentration	Ca 2+	EDTA	[CaY] 2-
Initial			0.04
Change	x	x	-x
equilibrium	x	x	0.04-x

We have, K' _f = [CaY^2-] / [Ca ²⁺] [ EDTA] = 1.83 x 10 ⁹

Therefore, 0.04 -x / (x)(x) = 1.83 x 10 ⁹

0.04 -x / x ² = 1.83 x 10 ⁹

Assume x is negligible as compared 0.04. Then we can write 0.04 / x ² = 1.83 x 10 ⁹

x ² = 0.04 / 1.83 x 10 ⁹ = 2.186 x 10 -11

x = 4.68 x 10 ^{- 6} M

[Ca ²⁺] = 4.68 x 10 ^{- 6} M =[EDTA]

E) [Ca ²⁺] at 1.1 Ve ( 1.1 x 120 = 132 ml)

After equivalence point , there is excess EDTA its concentration is calculated as shown below.

[EDTA] = original concentration of EDTA x Dilution factor.

[EDTA] = 0.080 x (12 ml / 252 ml) = 3.81 x 10 ^-03 M

Here 12 ml is volume of EDTA added after equivalence point.

Total volume of solution = Volume of calcium + volume of EDTA = 120 + 132 = 252 ml

[CaY ^2-] = original concentration of EDTA x Dilution factor.

[CaY ^2-] = 0.080 x ( 120 ml / 252 ml) = 3.80 x 10 ^{- 02} M

We have,  K' _f = [CaY^2-] / [Ca ²⁺] [ EDTA] = 1.83 x 10 ⁹

( 3.80 x 10 ^{- 02}) / [Ca ²⁺] (3.81 x 10 ^-03 ) = 1.83 x 10 ⁹

[Ca ²⁺] (3.81 x 10 ^-03 ) = ( 3.80 x 10 ^{- 02}) / 1.83 x 10 ⁹ = 2.08 x 10 ^-11

  [Ca ²⁺] = 2.08 x 10 ^-11/ (3.81 x 10 ^-03 ) = 5.5 x 10 ^-09 M

[Ca ²⁺] = 5.5 x 10 ^-09 M