[muffins]

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 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

[Borek]

4 g of NaOH is not 10 moles.

[muffins]

Okay, sorry my bad, was a ridiculous miscalculation... But still, what IF it was 10 moles of NaOH?

[Borek]

If there is a huge excess, you can usually ignore everything else.

[muffins]

What do you mean by ignore everything else?
Which method is the correct one to calculate such a problem though?

[Borek]

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.

[muffins]

Thank you very much