Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: muffins on March 30, 2014, 02:52:19 PM
-
First of all, I'm really sorry if my question is stupid... I just need some sort of clarification about this..
I'm given this reaction: NH3 + H2O ::equil:: NH4+ + OH-
Data: the solution is at pH=8.95 ; [NH3]=0.5M ; [NH4+] = 1M ;
4g of NaOH is added to the solution (to make a 1L solution). What will the new pH be?
4g of NaOH=10mols ; [NaOH]=10M
If we use HH, we get pH to be 10.25 (pKa=9.25)
My question is... can this be calculated differently? Considering that 10 moles of NaOH reacted with 1mol ammonium ion, we get 9 excess moles of NaOH = [OH-]=9M. From here onwards, calculating the pOH = -log[OH-]=-0.95 and so the final pH would be 14-(-0.95)~15.
Can anyone please explain to me why the second method is incorrect?
Thanks so much
-
4 g of NaOH is not 10 moles.
-
Okay, sorry my bad, was a ridiculous miscalculation... But still, what IF it was 10 moles of NaOH?
-
If there is a huge excess, you can usually ignore everything else.
-
What do you mean by ignore everything else?
Which method is the correct one to calculate such a problem though?
-
In the solution in which there is a tenfold excess of NaOH you can safely assume NH3 has no effect on pH.
Or, you can use ICE table - assume initial OH- concentration to be that of NaOH, and see how far NH3 reaction with water goes.
-
Thank you very much