Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sharbeldam on March 26, 2021, 10:37:48 AM
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If we add 0.008 moles of strong base like NaOH to 1 liter of ammonia with concentration of 15.3M , what would the pH be?
Now in all my years of studying chem alone, i have never encountered such problem, usually we can ignore the OH- that comes from the weak base but not in this case since its very concentrated so what is the way?
NH3+H2O <----> NH4++OH-
i can do an ice table for the weak base alone, i'd get the concentration of OH- alone and then the moles, and then can i just add the moles of OH- from the strong base to that and calculate new concentration and then pOH and then pH?
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You need the equilibrium constant for the reaction NH3 + H2O =》NH4 + + OH-
If you add Hydroxide the equlibrium will pushed to the left side, ammonia gas will pushed out.
I think it works like an alkaline buffer. Probably can use HH equation. In the logarithmen you have decrease of NH4+ And increase of NH3
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i have the kb, can you tell me the values for the HH formula? since i dont have NH4+?
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You know the dissociation of the ammonia in water. You have 15.3 mol/ l total.
pH = pka + log(cNH4+/cNH3). The change will be (cNH4+ - cOH-)/ (cNH3 + cOH-)
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so i calculate the concentration of NH4+ and NH3 in equilibrium first alone using the ice table of the weak base, and then i do what you said in HH equation?
also 1 more question, why isnt my method correct? to just add the moles of OH- together from weak base and addition of OH-?
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NH3 + H2O ---> NH4+ + OH-
K = [NH4+] x [OH-] / [NH3] = 1,77 x 10^-5
Concentration of NH3 will not change, it will still be around 15,3 M.
Concentration of NH4+ is concentration of ammonia x it's dissociation degree.
Concentration of OH- is concentration of ammonia x it's dissociation degree plus concentration of NaOH (it is fully dissociated).
You need only to calculate quadratic equation (ax2 + bx + c = 0).
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Thank you first of all,
Is this also right in the picture? i can just find the initial concentration and hence moles of [OH]- and just simple add the moles of the hydroxide solid that i added and then calculate pH.
also you said NH3 doesnt change, but if we added OH- doesnt the reaction shift to the left and NH3 should increase a little?
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Is this also right in the picture? i can just find the initial concentration and hence moles of [OH]- and just simple add the moles of the hydroxide solid that i added and then calculate pH.
No, because as you said:
if we added OH- doesnt the reaction shift to the left
(your result will differ from true)
also you said NH3 doesnt change, but if we added OH- doesnt the reaction shift to the left and NH3 should increase a little?
Yes, but it is very small change (less than 1 %) :)
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like this just to be sure yes?
ahhh no, it should shift to the left, im frickin confused
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Now it is correct.
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I really understand but its still a bit confusing because before the addition of OH-, what i have in the flask? NH3, NH4+ and OH-, NH3 Is 15.3 all the way, NH4+ and OH- are both x (that we can calculate).
after the addition that x should become smaller because the reaction goes to the left, so NH3 increase a tiny bit but we ignore that, but the concentration of NH4+ should decrease (x-y) and the OH- should also decrease so it would be like x+0.008-y, do you get my point?
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Maybe it isn't "0,008 mol of NaOH added to 15,3 M NH3" but rather a "0,008 M solution of NaOH and 15,3 M solution of NH3".
Maybe you should look at this from that point.
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it is 0.008 moles solid NaOH. but it's okay we can deal with moles, dont see the issue, especially that it's 1 liter solution of ammonia
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Maybe it is 1 liter and 0,008 moles, maybe 0,5 liter and 0,002 moles NaOH....
It is still question about pH, not about dissolving NaOH in ammonia solution.
and the OH- should also decrease so it would be like x+0.008-y, do you get my point?
I understand, it is wrong.
Maybe (?) it can be calculated like that but it will be more difficult.
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the OH- should also decrease so it would be like x+0.008-y
0.008 is definitely larger than y, so the OH- increases.