Question

Page 243, Practical Question TABLE 12-1 Values of ye for EDTA at 25°C and -0.10 M pH • Calculate (Ca2+] in 0.10 M CaY2- at pH 8.00, logKf(Ca2+ - EDTA) = 10.65. (answer: 2.3x10-5 M). 1.3 X 10-23 1.4 X 10-18 2.6 X 10-14 2.1 X 10-11 3.0 x 10 2.9 x 10-7 1.8 X 10- 3.8 X 10+ 4.2 x 10 Ca2+ + Y4- Cay2- 0.041 Conditional formation constant [MY-4) K = ayun Ky = m**][**] 0.30 0.81 0.98 1.00 1.00

Answer 1

Consider a complex formation reaction, Ca ²⁺ +Y ^4- $\rightleftharpoons$ CaY ^2-

For above reaction, K'f = [CaY ^2- ] / [Ca ²⁺][Y ^4- ]

At pH = 8 , $\alpha$_Y 4- is 4.2 $\times$ 10 ^-03

$\therefore$ The conditional formation constant K'f = ($\alpha$_Y 4- ) K f = (4.2 $\times$ 10 ^-03 ) ( 10 ^10.65) = 1.876 $\times$ 10 ⁸

Let's use ICE table.

Concentration (M)	Ca 2+	Y 4-	CaY 2-
Initial			0.10
Change	+X	+ X	- X
Equilibrium	X	X	0.10 - X

Therefore, K'f = ( 0.10 - X) / (X )(X) = 1.876 $\times$ 10 ⁸

Value of K'f is very large, hence at equilibrium X is very small in comparison with 0.10. Therefore, we can write 0.10 - X $\simeq$ 0.10.

$\therefore$ 0.10 / X ² = 1.876 $\times$ 10 ⁸

0.10 / 1.876 $\times$ 10 ⁸ = X ²

X ² = 5.330 $\times$ 10 ^-10

X = 2.30 $\times$ 10 ^-05 M = [Ca ²⁺]

ANSWER : [Ca ²⁺] = 2.30 $\times$ 10 ^-05 M