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.
Thank you!!!
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.
To find the pH of a mixture of two solutions, we need to consider the individual contributions from each component. In this case, we have a strong acid, HI, and a weak acid, NH4NO3.
First, let's find the contribution of the strong acid, HI.
1. Calculate the moles of HI:
Moles = Molarity x Volume
Moles = 0.0614 M x 0.0500 L = 0.00307 moles
2. Since HI is a strong acid, it completely dissociates in water, giving one H+ ion for every HI molecule. Therefore, the concentration of H+ ions contributed by HI is equal to the moles of HI.
Next, let's find the contribution of the weak acid, NH4NO3.
1. Calculate the moles of NH4NO3:
Moles = Molarity x Volume
Moles = 0.0845 M x 0.0750 L = 0.00634 moles
2. NH4NO3 is a salt, which means it disassociates into its constituent ions in water. In this case, NH4NO3 dissociates into NH4+ and NO3- ions. However, NH4+ is a weak acid and can partially dissociate, so we need to consider its equilibrium as well:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
The concentration of H+ ions contributed by NH4+ can be found using the equilibrium expression for the dissociation of NH4+:
[H+] = √(Ka x [NH4+])
The Ka (acid dissociation constant) for NH4+ is known to be 5.6 x 10^-10.
[NH4+] = Moles of NH4NO3 / Total Volume
= 0.00634 moles / (0.0500 L + 0.0750 L)
= 0.0422 M
Therefore, using the Ka value and the concentration of NH4+, we can calculate the concentration of H+ ions contributed by NH4+.
Now, we need to add the contributions of H+ ions from both HI and NH4+ to get the total concentration of H+ ions.
[H+]total = [H+]HI + [H+]NH4
Finally, we can calculate the pH using the formula: pH = -log10[H+].
Now, let's go through the calculations:
1. Contribution from HI:
[H+]HI = 0.00307 M
2. Contribution from NH4+:
[H+]NH4 = √(5.6 x 10^-10 x 0.0422 M)
3. Total [H+] concentration:
[H+]total = [H+]HI + [H+]NH4
4. pH:
pH = -log10[H+]total
Double-check these calculations and make sure you're using the correct values for concentration, volume, and equilibrium constant.