HI!!
I don't understand how to calculate pH and [Ag
+] of a solution of:
NH
3 0,10M
AgNO
3 10
-2M
(pk
b NH3 =4.7 , logK
1=3.32, logK
2 =3.89 )
K
1 -----> Ag
++ NH
3 >>> [AgNH
3]
+K
2 -----> Ag
+ + AgNH3
+ >>> Ag(NH
3)
2 +I should use the two "mass" balance of the ligand NH
3 and [Ag
+] :
C
NH3= [NH
3] + [NH4
+] + [AgNH
3]
+ + 2[ Ag(NH
3)
2 +]
C
Ag+ = [Ag
+] + [AgNH
3+] + [ Ag(NH
3)
2 +]
Then i know this formula:
β' = β*α
NH3 acid/base[NH
3']= [NH
3] + [NH4
+] (the not-complexated ligand)
[Ag
+] = C
Ag+ * 1/(1+β'[NH3'] + β'[NH3']
2) ≈C
Ag+ * 1/( β'[NH3']
2)
But my problem now is to calculate the value of [NH3] because it's a funciotn of pH,but also of the complex equilibrium (i don't have the pH fixed/set)
Usually i use this relation (when i don't have competitive equilibrium):
C
ligand - [L] = nC
metal - n[M]
And put it in:
[M]= C
M* 1/ β[L]
n (i suppose that β
n >> β
i ...with i=1,2...n-1)
And i solve this for [M]^2...
But now i have also an acid base equilibrium and i think it's not corret...: (i have β' and [L'] and not only β, [L] )
Some help??
Thanks