posted by Katie L on .
What would be the pH of a mixture of 50.0 mL of 0.0614 M HI and 75.0 Ml of 0.0845 M NH4NO3??
When i worked this question out, i got a basic pH of 12.42!! But I know that's not right b/c its a strong acid mixed with a weak acid (i think NH4NO3 is a salt of a reaction; i know it's midly acidic).
Can you please walk me through the steps so I can find out where my calculations went wrong?? Thank you so much!
Is that HI, as in hydroiodic acid? I'll assume so. That's a strong acid and ionizes 100%.
I haven't seen a problem like this because usually we see buffer solutions but I don't see this as a buffer nor do I see the salt and the strong acid reacting. Here is what I would do.
HI ==> H^+ + I^-
(HI) = 0.0614 x 50/125 = ??
(NH4NO3) = 0.0.0845 x 75/125 = ??
The NH4NO3 will hydrolyze to produce more acid.
NH4+ + H2O ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
If we call (NH3) = y, then (H3O^+) = y from the NH4NO3 + 0.0245 from the HI and the (NH4^+) = 0.0507 - y
Plug those into Ka to solve for y. I found y to be so small it can be neglected completely and pH is determined only by HI for which I found 1.61. Check my thinking. Check my arithmetic.
Yes, it's quite an odd problem. My teacher was attempting to show us something to do with polyprotic acids, but he also neglected to tell us that solving for y was a waste of time because its negligible. I was just confused because there's no conjugate base and I was not sure how it reacted.
I think that the NH4NO3 comes from the reaction of NH4OH and HNO3. But I'm not entirely sure.
Anyway, thank you!
That's what was confusing to me, too, no conjugate acid/base. The NH4NO3 is the salt of a weak base and a strong acid, so NH3 solution with NH4NO3 would be a buffer. And there was no polyprotic acid or base there either. By the way, you mentioned that NH4NO3 was the salt of NH4OH and HNO3. It has been proved now, and it has been years in the doing, that NH4OH does not exist as a compound. True, when NH3(g) is placed in water solution, it forms NH4^+ and OH^- but there is no NH4OH intermediate between NH3 and the ions. I taught for years and used NH4OH; I've had to adjust my thinking a little.