[xshadow]

HI!!

I don't understand how to calculate pH and [Ag⁺] of a solution of:

NH₃ 0,10M
AgNO₃  10^-2M
(pk_b NH3 =4.7 , logK₁=3.32, logK₂ =3.89 )

K₁ ----->     Ag⁺+ NH₃   >>>   [AgNH₃]⁺
K₂ ----->     Ag⁺ + AgNH3⁺  >>>  Ag(NH₃)₂ ⁺

I should use the  two "mass" balance of the ligand NH₃ and [Ag⁺] :

C_NH3= [NH₃] + [NH4⁺]  +  [AgNH₃]⁺ + 2[ Ag(NH₃)₂ ⁺]

C_Ag+ = [Ag⁺] + [AgNH₃⁺] + [ Ag(NH₃)₂ ⁺]


Then i know this formula:

β' = β*α_{NH3 acid/base}
[NH₃']=  [NH₃] + [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

[AWK]

Assume that complex does not dissociate and the rest of ammonia treat simply as water solution.

[xshadow]

Assume that complex does not dissociate and the rest of ammonia treat simply as water solution.

So I 've to assume that:

2L  + M >>> ML2 

(the reaction is quantitative, "there isn't" an equlibrium)

But in order to calculate pH how can i do??
Can I calculate the  pH of this solution (AgNO3 and NH3) only considering a solution of NH3?? (i.e. without considering the fact that some [NH3] is "taken" away from the acid-base equilibrium by the reaction: 2L  + M >>> ML2 )

Is this correct??
Or to calculate pH I have also to consider that some of [NH3] will become Ag(NH3)_2??


pH of a solution 0,10M NH₃ should be about 11-11.5  (i extimate this  value using pH vs Concentration diagram)...but this is the pH of NH3 in pure water...and not with Ag+ (and the complexation equilibrium)
Thanks.

[AWK]

Calculate stoichiometry of complexation, forget about ammonia released from complex, and calculate concentration of ammonia not involved in complex (0.1 - 2 x 0.01 - is it so difficult?). Complex is not involved in dissociation equilibrium of ammonia. Ammonia released in the process of complex dissociation is order of10^-8 M. Is it worthy to take it into account?

[xshadow]

Calculate stoichiometry of complexation, forget about ammonia released from complex, and calculate concentration of ammonia not involved in complex (0.1 - 2 x 0.01 - is it so difficult?). Complex is not involved in dissociation equilibrium of ammonia. Ammonia released in the process of complex dissociation is order of10^-8 M. Is it worthy to take it into account?

So if the starting C_NH3 is  0.1M ...i have:

[NH₃']= 0.1 -2*0.01 = 0.08 M   

Where:
[NH₃']=  [NH₃] + [NH4⁺]   (the not complexed ligand)

But now in order to calculate pH I have to use this balance mass:
[NH₃']=  [NH₃] + [NH4⁺]

So in this case is equal to calculate the pH of a water solution of NH₃ 0,08M   
In this case I don't consider the Ag(NH3)₂ contribute to [NH₃] because 10^-8 is forgettable compared to 0,08M  in the calculation of pH....

is correct??
Thanks

[xshadow]

Calculate stoichiometry of complexation, forget about ammonia released from complex, and calculate concentration of ammonia not involved in complex (0.1 - 2 x 0.01 - is it so difficult?). Complex is not involved in dissociation equilibrium of ammonia. Ammonia released in the process of complex dissociation is order of10^-8 M. Is it worthy to take it into account?

So if the starting C_NH3 is  0.1M ...i have:

[NH₃']= 0.1 -2*0.01 = 0.08 M   

Where:
[NH₃']=  [NH₃] + [NH4⁺]   (the not complexed ligand)

But now in order to calculate pH I have to use this balance mass:
[NH₃']=  [NH₃] + [NH4⁺]

So in this case pH of a solution NH3/AgNO3 is equal to calculate the pH of a water solution of NH₃ 0,08M   
In this case I don't consider the Ag(NH3)₂ contribute to [NH₃] because 10^-8 is forgettable compared to 0,08M  in the calculation of pH....

is correct??
Thanks

[AWK]

Yes, calculate pH 0.08 M NH₃. For this concentration you may use the most approximated formula.

[xshadow]

Thanks.. I get a pH value of 11.101

Only one  things:
This way to proceed to calculate pH in a solution where an and acid/base is also a ligand for a metal is a general way??

I.e to calculate pH when I have also a complexometric equilibrium can I always calculate pH of this solution considering as analytical concentration for the acid base reaction:  C_L = [L']

Where L is the ligand.
[L']= [L]+[HL⁺]  (if the conjugate acid is monoprotic)

And  [L'] can be calculated from the TOTAL initial concentration of ligand put in solution C_{L_initial} using the stechiometry of the complexation reaction  ( but in this case I usually suppose that the complexation reaction is quantitative....why?? because it has a huge  K/β constant reaction??? )

That's all...
Thankss )

[AWK]

There a two different methods of calculating pH in the case of such complexes.
1. You have pure complex in water, eg [Ag(NH₃)₃]⁺ -  then only ammonia from complex dissociation may  change pH of solution.
2. The complex is in solution in the presence of an excess of ligand (ammonia) - then, in majority of cases, you can safely neglect dissociation of complex, and treat both equilibria separately. This is your case. Only for a very, very low excess of ammonia you should take into account both equilibria simultaneously. After approximated calculations (much easier) you can always check if more complicated calculations are needed.

And a very special case is when you dissolve Ag₂O in ammonia (usually not found in textbooks).

[xshadow]

Thank you AWK

And in order do calculate [Ag⁺]   is correct to suppose that:

0.1= C_ligand=C_NH3=[NH₃] + [NH₄+] + [AgNH₃]+ 2[Ag(NH3]₂] 

Became:
0.1=C_NH3=[NH₃]  + [AgNH₃]+ 2[Ag(NH3]₂] 

I.e I can neglect [NH₄+] compared to [NH₃] because  the pH of the solution AgNO₃ / NH₃  is about 11.1

At this pH value i have that [NH₃] >> [NH₄+] ...so [NH3] + [NH4+]≈ [NH3]

[AWK]

0.1= Cligand=CNH3=[NH3]

?


Mixture of NH₃ and NH₄⁺ may work a buffer, but the ratio of concentrations should be from 1:10 to 10:1. This is not your case. Of course for very precise calculations all these concentrations are taken into account, but change of pH usually is negligible (much lower that can be measured). Moreover, such calculations are quite complicated with calculator.

For this approximation you can always check Ag(NH₃)⁺ from β₁ and [NH₄⁺]≈[OH^-] and decide that this approximation satisfied you or not. If not, you can enter an iterative process for which calculator is sufficient.

[xshadow]

0.1= Cligand=CNH3=[NH3]

?


Mixture of NH₃ and NH₄⁺ may work a buffer, but the ratio of concentrations should be from 1:10 to 10:1. This is not your case. Of course for very precise calculations all these concentrations are taken into account, but change of pH usually is negligible (much lower that can be measured). Moreover, such calculations are quite complicated with calculator.

For this approximation you can always check Ag(NH₃)⁺ from β₁ and [NH₄⁺]≈[OH^-] and decide that this approximation satisfied you or not. If not, you can enter an iterative process for which calculator is sufficient.

I mean in order to calculate [Ag⁺]  I need to write the metal balance and the ligand one:

0.1= C_ligand=C_NH3=[NH₃] + [NH₄+] + [AgNH₃]+ 2[Ag(NH₃]₂]   (ligand balance)

The metal one is:
C_Ag+ = [Ag⁺] + [AgNH₃+] + [ Ag(NH3)₂ +]

Now I can suppose that:

[Ag(NH3)₂] >> [AgNH₃]]   (because K₂ >> k₁
[NH₃] >> [NH₄] ....because the pH is about 11...and at this value NH₃ >> NH₄+ (the diagram concentration vs pH says this)

So I get:

C_NH3 = 0,1 = [NH₃] +2[Ag(NH3)₂]
C_Ag+= 0,01 = [Ag⁺] +  [Ag(NH3)₂] = [Ag(NH3)₂]

Because C_NH3 / C_AgNO3 ≈ 15-20 ( or better equal or >)....so I can neglect [Ag⁺] compared to the equilibrium concnetration of  [Ag(NH3)₂]

So  I have 2 equation in  two variable...I can get the exact value of  [NH₃] and then to know [Ag⁺] I could use this formula:
α_Ag+* C_Ag+ =  C_Ag+* 1/1+β₁ [NH₃] + β₂[NH₃]²

I have a

[AWK]

I don't understand how to calculate pH and [Ag+] of a solution of:

NH3 0,10M
AgNO3  10-2M
(pkb NH3 =4.7 , logK1=3.32, logK2 =3.89 )

K1 ----->     Ag++ NH3   >>>   [AgNH3]+
K2 ----->     Ag+ + AgNH3+  >>>  Ag(NH3)2 +

I should use the  two "mass" balance of the ligand NH3 and [Ag+] :

CNH3= [NH3] + [NH4+]  +  [AgNH3]+ + 2[ Ag(NH3)2 +]

CAg+ = [Ag+] + [AgNH3+] + [ Ag(NH3)2 +]

In the excess of ammonia concentration of Ag⁺ is ~10^-8 and concentration of [Ag(NH₃)]⁺ ~10^-6. Both concetrations can be neglected in the presence of 10^-2 [Ag(NH₃)₂]⁺
Concentration of NH₃ in this solution is about 70 times greater then NH₄⁺ which means that neglecting NH₄⁺ gives you about 1.5 % error in concentration of hydroxide ion. This gives you error less then 0.01 in value of pH.
All my information you may check during calculations. Use as much approximation as I showed you (and K₁·K₂)

[xshadow]

Thanks