Textbook Problem 6D13

[SophiaJenny3I]

Hi everyone! Hope everyone is doing well! Could someone please walk me through their logic for this problem? I don't know why I can't figure out some of the pH values given the information. Thank you!

[Juliet Cushing_2H]

I was confused on this one too! Turns out you have to use the table of Ka values from tables 6C.1 and 6C.2 (in the chapter rather than in the section for these practice problems). Hope this helps out!

[Yasmina Zaarour 1G]

^ Yes you need to use those tables. Also, remember that for weak acids and bases you need to create ICE tables, but for strong acids and bases you can assume that the reaction goes to completion (so, the concentration of H3O+ for the HCl solution is equal to the original concentration of HCl. For some of the other ones, the table might give values for only the conjugate acid/base of the given compound, so you should convert to the desired value using Ka*Kb=Kw.

[Zaid Bustami 1B]

Hey Sophia!

Since this problem doesn't really provide acid or base equilibrium constants right off the bat, I'd start off with identifying which ones are acids or bases and which ones are strong or weak.

a) HCl is a strong acid
b) CH3NH3Cl is a salt that will dissociate into CH3NH3+ and Cl-. The chloride won't contribute to pH because it's a conjugate base of a strong acid (HCl), but the CH3NH3+ will because it's a conjugate acid of a weak base (CH3NH2 - remember nitrogenous bases are weak). Thus CH3NH3+ is a weak acid.
c) CH3COOH is a weak acid
d) C6H5NH2 is a weak base (it has that same nitrogen group that we saw in b).

Based on what we know so far, ranking the solutions in terms of increasing pH means we need to rank them from the strongest acid to the strongest base.
a ? b ? c < d
We can determine whether a, b, or c will have a lower pH because they all involve acids. Even if HCl is strong it's present in a pretty low concentration, so it's not enough yet just knowing it's strong to say it has the lowest pH. Thus, we need to actually calculate the pH's of those solutions (note we also need to look up Ka values for the weak species)

a) pH = -log(1.0x10^-5) = 5
b) Ka = Kw/Kb = 1x10^-14/3.6x10^-4 = 2.8x10^-11
CH3NH3+ is a monoprotic acid (it only gives up one proton), so whatever concentration CH3NH3+ decreases by is equal and opposite to the concentration that its conjugate (CH3NH2) will increase by. You can apply the small K approximation here.
Ka = [CH3NH2][H+]/[CH3NH3+]
2.8x10^-11 = (x)(x)/(0.2-x) ≈ x^2/0.2
x = 2.4x10^-6 M = [H+]
pH = -log(2.4x10^-6) = 5.63
c) Ka = 1.8 x 10-5
Doing exactly what we did for b), we get pH = 2.72

Now we can rank all the species.
c < a < b < c

[SophiaJenny3I]

Thank you so much everyone for your detailed explanations! This helps a lot :)

[Alexis Sanft 1E]

Does anyone know how to determine if something is a weak acid or base, or are we just supposed to have a list memorized, and if so how do we obtain that list?