Game:

The Legend of Zelda: Ocarina of Time (1998)

Duration: 10:46

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This is a video I made for my chemistry class. It gets pretty detailed talking about buffers, so proceed with caution. Mildly funny included! Also, as you can tell from the first two seconds of the video, graphic design is my passion.

Weak acids in this video are red, weak bases are green, strong acids are lighter red, strong bases are teal, and very weak acids/bases (e.g. Cl⁻, H₂O) are in dark grey.

All the background music is created by me using LMMS. The clarinet in one of the songs is from a soundfont of The Legend of Zelda: Ocarina of Time.

Extra clarification:
At 3:56, the reason H₂O isn't accounted for in the equilibrium fractions is because only aqueous solutions and gases are accounted for in equilibrium constants. H₂O is a pure liquid, and thus isn't accounted for because its concentration can be assumed to be constant.
Whenever H⁺ or H₃O⁺ are mentioned, they both just mean the same thing. Sometimes H₂O picks up H⁺ to form H₃O⁺, and sometimes it doesn't. The K constant for H⁺ + H₂O ⇌ H₃O⁺ is exactly 1, so you can pretty much substitute H₃O⁺ for H⁺.
When I say the Na⁺ or Cl⁻ ions do essentially nothing once dissociated, the reason is because they wouldn't favorably react with H₂O, H⁺, OH⁻, HA, or A⁻. Let's use Na⁺ as an example.
- In Na⁺ + H₂O ⇌ NaOH + H⁺, the Ka for Na⁺ is Kw/Kb for NaOH, and since NaOH is a strong base with a Kb of 10²⁰, the Ka for Na⁺ is much less than the Kw, which is 1*10⁻¹⁴, meaning Na⁺ is a very weak acid.
- In Na⁺ + C₂H₃O₂⁻ ⇌ NaC₂H₃O₂, since it's just the reverse reaction for NaC₂H₃O₂ ⇌ Na⁺ + C₂H₃O₂⁻, and since NaC₂H₃O₂ is highly soluble, Na⁺ reacting back with C₂H₃O₂⁻ is highly unfavorable.
- In Na⁺ + OH⁻ ⇌ NaOH, since it's just the reverse reaction for NaOH ⇌ Na⁺ + OH⁻, the K constant is 1/Kb, which is 1*10⁻²⁰, meaning Na⁺ doesn't want to pick up OH⁻ either.
- However, the weak acid HC₂H₃O₂ WILL favorably react with OH⁻, because in HC₂H₃O₂ + OH⁻ ⇌ C₂H₃O₂⁻ + H₂O, this reaction is just the reverse reaction of C₂H₃O₂⁻ + H₂O ⇌ HC₂H₃O₂ + OH⁻, which means the K constant for this reaction is 1/Kb for C₂H₃O₂⁻, which is 1/(5.6*10⁻¹⁰), which is 1.8*10¹⁸, so HC₂H₃O₂ will favorably react with a free supply of OH⁻, despite not wanting to react directly with H₂O due to it being a weak acid.

Timestamps:
0:00 Intro/Bloodstream buffer
0:51 Definition
1:12 Experiment: pH change
2:25 Methods to making a buffer
2:47 Experiment: Acetic acid + sodium acetate
3:45 Equilibrium constants
4:52 HA and A⁻ must be weak
5:20 Experiment: Titrating acetic acid + NaOH
8:00 Titration graphs + Henderson-Hasselbalch equation
8:58 Strong + strong titration
9:45 Summary/Outro

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