Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: xshadow on April 23, 2021, 08:47:02 AM
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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 :)
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You're right except for volume of water which should be 0,09 liter, not mililiter.
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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:
VNH3 = 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+] * VNH4+) = 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 *VNH4+ + VNH4+ = 0,250l
12.2 * VNH4+ = 0,250l
VNH4+ = 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
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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.
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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) * VNH4+) = 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
VNH4+ + VNH3 = 0,250l
VNH4+ + 2* V NH4+ = 0,250l
VNH4+ = 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
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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).