Question

A 120.0 mL sample of 0.040 M Ca2+ is titrated with 0.040 M EDTA at pH 9.00. The value of 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.

What is Kf′, the conditional formation constant, for Ca2+ at pH 9.00?

Kf′=

What is the equivalence point volume, Ve, in milliliters?

Ve=    mL

Calculate the concentration of Ca2+ at V=12Ve.

[Ca2+]= M

Calculate the concentration of Ca2+ at V=Ve.

[Ca2+]= M

Calculate the concentration of Ca2+ at V=1.1Ve.

[Ca2+]= M

Answer 1

Consider a complex formation reaction: Ca ²⁺ + EDTA [CaY] ^2- ------------------->(1)

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

Calculation of volume of EDTA required to reach equivalence point

From above reaction (1), we have 1 mol Ca ²⁺$\equiv$ 1 mol  EDTA

Therefore, M _ca 2+ $\times$ V _Ca2+ = M _EDTA$\times$ V _EDTA

V _EDTA = M _ca 2+ $\times$ V _Ca2+ / M _EDTA =

V _EDTA = 0.040 M $\times$ 120.0 ml / 0.040 M = 120.0 ml

Equivalence point volume = 120.0 ml

[Ca ²⁺] at V = 1/2 Ve

We know that, Molarity = No .of moles of solute / volume of solution in L

Therefore, No .of moles of solute = Molarity $\times$ volume of solution in L

No. of moles of Ca ²⁺ present initially in the solution = 0.040 mol / L $\times$ 0.1200 L = 0.0048 mol

No. of moles of EDTA added in the solution = 0.040 mol / L $\times$ 0.0600 L = 0.0024 mol

Excess moles of Ca ²⁺ left in the solution after reaction with EDTA = 0.0048 - 0.0024 = 0.0024 mol

Volume of solution at this stage = Volume of Ca ²⁺ solution + volume of EDTA solution = 120.0 + 60.00 ml = 180.0 ml

[Ca ²⁺ ] = 0.0024 mol / 0.1800 L = 0.0133 M

[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-.

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

No. of moles of EDTA added in the solution = 0.040 mol / L $\times$ 0.1200 L = 0.0048 mol

Volume of solution at this stage = Volume of Ca ²⁺ solution + volume of EDTA solution = 120.0 + 120.0 ml = 240.0 ml

[CaY] ^2-. = 0.0048 mol / 0.2400 L = 0.0200 M

Let’s use ICE table.

Concentration	Ca 2+	EDTA	[CaY] 2-
Initial			0.0200
Change	+X	+X	-X
Equilibrium	X	X	0.0200 - X

Therefore, K' _f = [CaY^2-] / [Ca ²⁺] [EDTA] = 0.0200 -X / (X)(X) = 1.83$\times$10 ⁹

0.0200 -X / (X)(X) = 1.83 $\times$ 10 ⁹

0.0200  -X / X ² = 1.83 $\times$ 10 ⁹

At equilibrium X is negligible as compared 0.0200. Then we can write 0.0200 / x ² = 1.83 x 10 ⁹

X ² = 0.0200 / 1.83 $\times$ 10 ⁹ =1.093 $\times$ 10 ^-11

X = 3.306 x 10 ^{- 6} M

[Ca ²⁺] = 3.306 x 10 ^{- 6} M

[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

No. of moles of Ca ²⁺ present initially in the solution = 0.040 mol / L $\times$ 0.1200 L = 0.0048 mol

No. of moles of EDTA added in the solution = 0.040 mol / L $\times$ 0.1320 L = 0.00528 mol

Excess moles of EDTA = 0.00528 - 0.0048 = 0.00048 mol

No. of moles [CaY ^2-] produced = No. of moles of Ca ²⁺ present initially in the solution = 0.0048 mol

Volume of solution at this stage = Volume of Ca ²⁺ solution + volume of EDTA solution = 120.0 + 132.0 ml = 252.0 ml

[EDTA] = 0.00048 mol / 0.2520 L = 1.905 $\times$ 10 ^-03 M

[CaY ^2-] = 0.0048 mol / 0.2520 L = 0.01905 M

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

(0.01905 ) / [Ca ²⁺] (1.905 $\times$ 10 ^-03) = 1.83 $\times$ 10 ⁹

[Ca ²⁺] (1.905 $\times$ 10 ^-03) = (0.0195) / 1.83$\times$ 10 ⁹ = 1.06 $\times$ 10 ^-11

  [Ca ²⁺] = 1.06 x 10 ^-11/ (1.905 $\times$ 10 ^-03) = 5.6$\times$ 10 ^-09 M

[Ca ²⁺] = 5.6 x 10 ^-09 M