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### Topic: preparing 250ml Nh3/NH4+ buffer solution pH= 10  (Read 441 times)

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##### preparing 250ml Nh3/NH4+ buffer solution pH= 10
« on: April 23, 2021, 08:47:02 AM »
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_Dojek

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##### Re: preparing 250ml Nh3/NH4+ buffer solution pH= 10
« Reply #1 on: April 23, 2021, 09:52:53 AM »
You're right except for volume of water which should be 0,09 liter, not mililiter.

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##### Re: preparing 250ml Nh3/NH4+ buffer solution pH= 10
« Reply #2 on: April 24, 2021, 07:59:49 AM »
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

« Last Edit: April 24, 2021, 08:52:22 AM by xshadow »

#### Borek

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##### Re: preparing 250ml Nh3/NH4+ buffer solution pH= 10
« Reply #3 on: April 24, 2021, 09:27:16 AM »
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.

Quote
V NH3 = 11,2 * V NH4+

That's not what I get when I try to follow your math.
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##### Re: preparing 250ml Nh3/NH4+ buffer solution pH= 10
« Reply #4 on: April 24, 2021, 11:15:17 AM »
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.

Quote
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

« Last Edit: April 24, 2021, 11:56:36 AM by xshadow »

#### Borek

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##### Re: preparing 250ml Nh3/NH4+ buffer solution pH= 10
« Reply #5 on: April 24, 2021, 12:55:15 PM »
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).
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