[xshadow]

I have to prepare a  NH₃/NH₄⁺ buffer solution at pH = 10  with [NH₄⁺] = 0,05M

The starting solution are:
a) NH₃ solution, 2M
b) NH₄⁺ salt 1:1 , 0.1M

So I think I should use the H.H.equation:

pH= pka + log (  NH₃/NH₄⁺ )
10 = 9.25 + log [NH₃] - log (0.05)
[NH₃] = 10^(-0.55) = 0,28M


SO I have 250ml  of buffer solution  at pH= 10 with:
[NH₃]= 0,28 M
[NH₄⁺ ] =  0,05 M


Now I can find the correspondents moles in those 250ml (0,250l) :
mol NH₃ = 0, 28M * 0,250l = 0,07mol
mol NH₄⁺ = 0,05M * 0,250l =  0,0125 mol

Now I consider my  NH3 and NH4 starting solutions  in order to find the corresponding volume that cointains that number of moles I "need" :
V NH3 =  0,07mol / 2M = 0,035l
V NH4 = 0,0125 mol / 0,1 = 0,125l


SO I have to mix

0,035l of the starting NH3 solution with 0,125l of the starting NH4+ solution.

0,035l + 0,125l = 0,16l  (V NH3 + V NH4+ added)
0,250 - 0,16 = 0,09 ml 

SO I add ,at the end, 0,09ml of water in order to get 250ml
Is it correct?

THANKS

[Orcio_87]

You're right except for volume of water which should be 0,09 liter, not mililiter.

[xshadow]

You're right except for volume of water which should be 0,09 liter, not mililiter.

oh thanks

Now I  have another question...
If I want to prepare 250ml of that buffer solution but my starting NH₃ solution is 0,1M instead of 2M the volume I should add is:

V_NH3 = 0,07mol / 0,1M = 0,7l  = 700ml
SO If I have a  0,1M NH3 solution I can't prepare that 250ml buffer solution??? The NH3 volume I must add overcame the 250ml of buffer solution .


I think I can solve if I consider  another method:

Remembering my buffer solution data:  250ml buffer solution pH = 10 with concentration:

[NH₃] = 0,28M
[NH₄⁺] = 0,05M


Now if I consider that  the concentration in buffer solution pH = 10 (first post) should be::
[NH₃] /[NH₄⁺] = 0,28/0,05 =  (mol NH₃/0,250l) / ( mol NH₄⁺/ 0,250l)  = mol NH3/ mol NH4+


So:
mol NH₃=  [NH₃] (starting solution) *  V _NH3
mol NH₄⁺ =  [NH₄⁺] (starting solution) * V _NH4+

If  concentrations of the two "starting" solutions are:
[NH₃] = 0,1M
[NH₄⁺] (Cl-) = 0,2M


I'll get :
[NH₃] (starting solution) * V _NH3 / ( [ NH₄⁺] * V_NH4+) =  0,28 / 0,05

Also I know that :
V _NH3+ V__NH4+  = 0,250l

And I get (2 equation system  with VNH3 and V NH4+ as x and y):
from the first I get:
V _NH3 = 11,2 * V _NH4+

So:
11.2 *V_NH4+ +  V_NH4+  = 0,250l
12.2 * V_NH4+ =  0,250l
V_NH4+ =  0,250l  / 12.2 = 0,020l =  20ml

V _NH3 =   250ml - 20ml = 230 ml


IS it  right?
WIth the method of my first post I can't prepare a 250ml solution starting  from  a  0,1M NH3  solution   because I had to add  700ml of only NH₃ starting solution when I want a 250ml bubbfer solution

[Borek]

SO If I have a  0,1M NH3 solution I can't prepare that 250ml buffer solution??? The NH3 volume I must add overcame the 250ml of buffer solution

Nothing unusual. Just like you can't prepare 0.2M solution from 0.1M solution.

V _NH3 = 11,2 * V _NH4+

That's not what I get when I try to follow your math.

[xshadow]

SO If I have a  0,1M NH3 solution I can't prepare that 250ml buffer solution??? The NH3 volume I must add overcame the 250ml of buffer solution

Nothing unusual. Just like you can't prepare 0.2M solution from 0.1M solution.

V _NH3 = 11,2 * V _NH4+

That's not what I get when I try to follow your math.

mmhhh  I' llretry:

If  concentrations of the two "starting" solutions are:
[NH₃] = 0,1M
[NH₄⁺] (Cl-) = 0,2M


I'll get :
[NH₃] (starting solution) * V _NH3 / ( [ NH₄⁺] (starting solution) * V_NH4+) =  0,28 / 0,05    (i)

Also I know that :
V _NH3+ V__NH4+  = 0,250l  (ii)

And I get (2 equation system  with VNH3 and V NH4+ as x and y):
from (i) I get:
V _NH3 =   V _NH4+ * [ NH₄⁺]  / [NH₃] =    V _NH4+ * 0,2M  / 0,1M  = 2*  V _NH4+

SO I have
V_NH4+ +  V_NH3  = 0,250l
V_NH4+  + 2*  V _NH4+  = 0,250l

V_NH4+ =  0,250l /3  =  0,083l = 83ml NH₄⁺ (starting solution)
V _NH3 = 0,250l - 0,083l =  0,167l   = 167ml  NH3 starting solution


Now is it correct??
But if I  calculate the NH3 moles in this 250ml buffer solution I get:
0,167 l  * 0,1M = 0,0167
That  differ from the previous value of 0,07  that was the NH3 moles in 250ml of that buffer solution..so strange...
Thanks


 

[Borek]

OK, I think I know where you went wrong:

V _NH3+ V__NH4+  = 0,250l  (ii)

That's not a correct condition.

Your set of equations guarantees correct ratio of concentrations, but they won't be 0.28M and 0.05M (and you asked for the latter).