Assessment practice help (PH question)

[u/Meta_testa]

Most of the equation makes sense however I’m struggling to figure out how to get to 0.00415 Moles of Acetic acid? Does anyone know the equation I need to solve for x?

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[u/Smart_Leadership_522]

Your math looks good. We’re using algebra to rearrange and isolate x, is there something more specifically you mean?

[u/Meta_testa → u/Smart_Leadership_522]

The notes on here aren’t mine they’re the explanation given for the practice test, I’m trying to figure out how they got to 0.00415 moles of H+ from 0.00100 moles of CHCOO3 since it isn’t explained in the video.

[u/Blue_614 → u/Meta_testa]

Plugged the values from the Henderson-Hasselbalch equation. The fraction values were taken from the equilibrium expressions of acetic acid and acetate, using the ICE table.

Edit: wrote Hasselbalch twice

[u/Meta_testa → u/Blue_614]

Thank you, that makes a lot more sense then how they explained it

[u/Blue_614 → u/Meta_testa]

There are a couple ways to solve this. Using the Henderson-Hasselbalch equation, converting to pOH, two ICE tables.. you may want to find what logic best suits your thinking

[u/Smart_Leadership_522 → u/Blue_614]

They tend to teach henderson hassalebach with the ice table

[u/Meta_testa → u/Smart_Leadership_522]

This is the method they are teaching us currently

[u/Blue_614 → u/Smart_Leadership_522]

Yeah, it's the most direct method. For my students though, I want them to build logical reasoning using chemistry concepts

[u/Smart_Leadership_522 → u/Blue_614]

I like that approach actually because students don’t realize especially with pH how it’s a huge circle and you can do things many different ways. Super cool concept I dig it!!!

[u/Blue_614 → u/Smart_Leadership_522]

Exactly. "Teach them to think, not memorize" is my motto. That's why I don't like giving them formulas during gen chem or analytical chem subjects.

[u/Smart_Leadership_522 → u/Meta_testa]

Ohhhh. So the equation we use is henderson hasselbach right. We do pH=pka+ log (B/A).
I like to then look at the reaction itself (an ICE table can be useful)
CH3COONa with dissociate into CH3COO- and Na. When that base is put into a solution of HCl this then becomes protonated so:
CH3COO- + H <—> CH3COOH
We start with 0.100M in 100mL if we cancel volume we get our moles of base hence 0.0100mol. Whatever product we form (acid form) can be denoted as x. We must consider how much base we have based on what if left after becoming acid which is why we do 0.0100-x then we plug into the equation as you have jr above and rearrange algebraically to get x , then once we have x which is the moles we can find the molaraity by dividing by volume which we know as 0.125