real

real

real

Real

real

Seems like he's not reading.

Physical pharmacy is bullshit 😂

He has 1 .3 mil subs tho

Facts bruh. Quality👌🏼😁😂

percent ionization is degree of ionization times 100 in order to get the percent. Whenever you want percent you multiple the number by 100 which makes the 10^-5 a 10^-3 because you cut two decimal spaces

Hey that sounds great! I just came across your comment, and I was curious how your plans are going. I hope they're going well!

How have your studies been going?

Hey I know it's been years, but I just came across your comment and was curious. How did IB chem go, and how have your studies been going since then?

welcome

That's awesome! What degree did you get?

lol

Because 78% of it is now base

thank you very much! @@ProfessorDaveExplains

Visit my mathematics tutorials if you need review on logarithms.

@@ProfessorDaveExplains I appreciate U!

I think the first dissociation would probably have the most impact on determining the pH, but you can go through the math for the others. Did you set up multiple ICE boxes like the professor alluded to in the video?

Hey I know it's been years, but I was just curious. How did the rest of this project go?

hmm, which part are you referring to? you could know the hydronium concentration because of the initial acid concentration if its a strong acid, or you could calculate hydronium concentration from 10^-pH if you have the pH. second part of the question i don't understand!

I understood what you meant now. Thanks Dave. The second question was how to calculate hydronium ph i got it.

Same question

You haven't used [A-] in the formula of Ka when you were calculating [HA], I guess that is the reason for the change in answer. You may calculate [HA] by subtracting 78 from 100 cause we don't know the value of [A-].

I did the same thing as you. This is an unusual use of the H-H equation as there's no indication this is a buffer, but must be in order to make sense. The 78% does not relate to the [H+] as you'd expect if the acid is 78% ionized and and as your/our calculation proves. Rather, it refers to the [A-]. This has nothing to do with what equation you're using (like the previous comment pointed out). To @Shreyansh Singh's point, the Ka equation is mislabeled in your comment, you're using the % ionization equation not the Ka equation. One way to think of this is to first dissociate the acid, THEN use equilibrium to establish the pH. This is a little unusual, since both the ionization AND equilibrium concentration of [H+] are the same process which is determined by Ka. Of course, you can just "plug and chug" and get on with your life but I'm a little uncomfortable with that process as you miss out on all the exciting chemistry concepts at work.

when phenol deprotonates you get one hydronium and one conjugate base!

That's bcz when equal amounts of NaOH and HCl are added, they neutralise each other completely, producing no H+ or OH- ions in the products, making solution acidic or basic, as: NaOH + HCl = Nacl + H2O Whereas when HCL is replaced by HOCL, the reaction is : NAOH + HOCL = Nacl + (OH) 2 Thus the (OH) 2 at the product side makes the solution basic.

acidity and basicity have only to do with the stability of the conjugate, and the CH3 anion is incredibly unstable, so hydrocarbons are not even remotely acidic. also, literally everything participates in dispersion interactions.

Thank you sir

if 78% is ionized then 22% is unionized

Professor Dave Explains ohhh haha why did I not think of that. Thank you Professor! Much love

@@ProfessorDaveExplains Sorry to comment on an old comment, but why is the ratio 0.78/0.22? Shouldn't it be 0.22/0.78 because [base]/[acid] not [acid]/[base]?

@@kiraevans9477 did you find any answer for that?

@@kiraevans9477 In the H-H equation it's [base]/[acid].

I think it's not correct bcuz the equation is conjugate base over acid so It will be 3.4-log(0.22÷0.78)

Well water can act as either an acid or a base, it's amphoteric. Phenol is significantly more acidic than water because of the resonance stabilization enjoyed by its conjugate base.

That's what this video explains.

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if 78% ionizes then 22% remains unionized

@@ProfessorDaveExplains ohh so the HA in Handerson equation is referring to the unionized. Dang I'm so dumb lol. Thank you

if 78% is ionized, 22% is not!

@@ProfessorDaveExplains how do we know which one is acid and which on is base?

the proton donor is the acid, the proton acceptor is the base. check out my tutorial on acids and bases!

I don't really understand the question. The blood uses carbonic acid and bicarbonate to maintain a strict pH level. These neutralize acids and bases that might enter solution, and then the equilibrium shifts to regenerate them, so they have a reasonable buffer capacity. This has worked for millions of years, so it must be a great system! The stomach has a different system and different pH.

so carbonic acid is produced by the blood on spot in order to resist the change in ph

it's not produced by the blood, it's just present in the bloodstream.

if acid ionizes you get conjugate base

1 - 0.78

well a buffer system is an equilibrium between a weak acid or base and its conjugate, so usually we make such a system by using a mixture of such a conjugate pair.

Anyone is free to answer if they'd like to, but please provide a clear explanation.

I died when I heard that

It's because he is Italiano!

Never mind. I realize you just multiply by 100 and it goes down two zeroes. Because... its decimal. My bad!

just the percentages that add up to 100

@@ProfessorDaveExplains Oh wow, thanks for the reply, from Professor Dave himself! Thanks a lot!

if 78% ionizes, then 22% remains as the acid!

Professor pls can you give me proof about what makes the solution a buffer and professor pls can you give me a better explanation on the solution to the comprehension because my friends has been bothering me on how I got my answer I and can't provide a solid explanation professor

And professor please what makes the acidic solution in the question a buffer sir

Professor please you said 22percent of acid remains so how is the base now 78percent also sir

buddy chill out with the comments. send me an email if you want. when an acid ionizes it becomes its conjugate base. 78% of it ionized, so 22% remains in the acid form. a buffer is a solution of a weak acid or base and its conjugate, as is explained in this clip.

great question! generally when the initial concentration and the Ka differ by a factor of at least 1000, the change will be negligible, since if it goes from 1.0 M to 0.9995 molar, there is no difference when we report to three or so significant figures. so when you see Ka values involving 10^-5, 10^-8, or even smaller, we can very safely make this assumption.

Thanks a lot professor, really appreciate the quick response!

if 78% is ionized, then 22% is not!

so that the .22 is the acid

yep

1.00 - 0.78

Sir where did you get the 1.00

1.00 represents 100%

nope