Question

A solution is prepared by adding 0.075 mole of K3 [Fe(CN)6] to 0.72 L of 2.2 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. [Fe(CN)6*] = [Fe3+] = Submit Answer Try Another Version 2 item attempts remaining

Answer 1

3K+ + Fe(CN) = K3[Fe(CN)6]

Fe3+ +6CN- = Fe(CN)

The formation constant

[Fe(CNS) [Fe3+] × [CN-16 K = 1.0 x 1042

The initial concentrations are

[Fe(CN)] = [K3Fe(CN)] [Fe(CN):] = 0.72 L - 0.075 mol [Fe(CN)2-1 = 0.104 M

CN = (NaCN CN] = 2.2 M

Let x M of ferric ions are formed at equilibrium. Prepare an ICE table.

	$Fe^{3+}$	$CN^-$	Fe(CN)
Initial concentration (M)	0.00	2.2	0.104
Change in concentration (M)	+x	+6x	-x
Equilibrium concentration (M)	x	2.2 + 6.	0.104 – I

Substitute values in the expression for the formation constant.

[Fe(CN) K= Fe3+] x [CN-16 1.0 x 1042 0.104 - 2 X (2.2 + 6.)

Since, the value of the formation constant is very large, approximate 0.104-x to 0.104 and approximate 2.2+6x to 2.2.

0.104 1.0 x 1042 _ 2 x (2.2) 0.104 = 1.0 x 1042 x (2.2) I = 9.18 x 10-46

[Fe3+]=r r = 9.18 x 10-46 [Fe3+1 = 9.18 x 10-46 M

[Fe(CN) 1 = 0.104 – 2 [Fe(CN): ] = 0.104 – 9.18 x 10-4 Fe(CN) 10.104 M