[Nedgy]

Basically, I got the same answer but in a different way to the video, was this just a fluke or was it correct?

First here's the video and I'm talking about the second part of the question @ 4:57 (there is some previous information before this given by the problem that you need to know if you want to solve the question).

https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/titrations/v/titration-of-a-weak-acid-with-a-strong-base?modal=1

100mL of 0.050M [NaOH] is added to 50mL of 0.200M [CH3COOH], what is the pH of the resultant solution? K_a = 1.8x10-5

So what I did was converted the molar concentration of NaOH and CH3COOH given into moles and then set up this equation:

NaOH(aq) + CH3COOH(aq) → H2O(l) + CH3COONa(aq)

Subtracted the moles of NaOH (5x10-3 mol) from CH3COOH (0.001 mol) and added that to the moles of CH3COONa (0 mol).

Result: left with 5x10-3 mol CH3COOH and gained 5x10-3 mol CH3COONa, wrote another equation that comments on the significant chemical species for helping determine H3O+ conc.

CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-.

Then I converted everything's moles into molar concentration. (using the combined solution of 100 mL + 50 mL). Both would be 0.033M.

At this point I used an ICE table to determine what the equilibrium concentrations would be.

Initial:

CH3COOH: 0.033M H3O+: 0M CH3COO-: 0.033M.

Change:

CH3COOH: -x H3O+: +x CH3COO-: +x

(my reasoning: reaction runs forwards as H3O+ initial = 0M).

Equlibrim:

CH3COOH: 0.033M -x H3O+: x CH3COO: 0.033M+x

Setting up K_a for acetic acid (and assuming x is negligible):

1.8x10-5 = x(0.033)/(0.033).

x = 1.8x10-5M, this is our H3O+ concentration at equilbrium. Plug that into pH = -log[H3O+]

pH = 4.74, same answer as the video but I didn't use Henderson-Hasselbalch equation, legit or fluke?

[AWK]

The way you calculated this task is completely equivalent to the H-H equation.

[Nedgy]

Perfectttt, thanks just the answer I was looking for, now if I just get both under my belt, I'll be more formidable  have a good day!

[Borek]

Compare https://www.chembuddy.com/?left=buffers&right=with-ICE-table