pH of salt solutions | Acids and bases | Chemistry | Khan Academy

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Examples of calculating pH of 0.25 M solution of sodium acetate, and calculating the pH of 0.050 M solution of ammonium chloride.
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Опубликовано:

24 сен 2024

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Комментарии : 36

@to181 6 лет назад

These videos are so wonderful. They aid me when I read my textbooks. I LOVE IT

@taylorcates1701 8 лет назад

Thank you so much!

@arnedesmet6356 7 лет назад

How do you know which part of the dissociated salt will react with water?

@SportifyTVofficial 7 лет назад

i think the conjugates of strong bases or acids will not react with water. In the first problem Na is conjugate of NaOH ,which is a strong base, so it won't react with water

@sarella1109 7 лет назад

Thank you so much! this is very helpful...

@jojobanks7809 4 года назад

How do you know which cations or anions are going to react appreciably with water ?

@rozana4470 4 года назад

yes l have the same question too

@NabojeetD 4 года назад

ions from strong acids and bases don't react. For example, HCl is a strong acid and so in a salt it won't react. Thats why in the NH4Cl example we ignored it. More specifically, if you have a reallly strong acid that means its very good at being an acid, and because its such a good acid the conjugate base of it is toooo weak to react. For strong bases its the same. The strong base NaOH just dissociates completely and doesn't react. Idk if my explanation is good or not, I'm just learning this stuff too. Main takeaway tho is if you see a cation or anion from a strong acid/base it won't do anything.

@JasonChan-e1g 3 года назад

@@NabojeetD yep. the anions from the 7 strong acids and group I & II cations do not change the pH of the solution.

@ShinySephiroth1 5 лет назад

I feel like this is great for labs and schoolwork, but an MCAT version of this would be better because the questions are quicker than this. Or maybe I'm completely off and I'm gonna bomb it.

@pedropierre9594 4 года назад

maybe ;/ because these problems can only be so difficult for a chemistry god trust me i know a few.

@ghadiermatariek8955 7 лет назад

awesome😃 thx

@khaledyasser8293 5 лет назад

Why is the concentration of CH3COO- the same as that of CH3COONa? Shouldn't that mean that CH3COONa dissociates completely in water? But if so why does CH3COOH not completely dissociate?

@bookishbassoonist2210 5 лет назад

khaled yasser it’s a weak acid, weak acids don’t disassociate in water. The same holds true for weak bases. The reason the Na is not included is because it’s a spectator ion, so it wouldn’t be included in the net ionic equation.

@doogydoogs 5 лет назад

Why did you not use Na for the table? Sorry I'm so confused.

@jeffreydaclan1035 2 года назад

same question, please answer

@christianjosephaquino3217 2 года назад

@@jeffreydaclan1035 Sodium acetate (CH3COOHNa) is a salt present in water (H2O). This is also known as a buffer solution. This just means it is a product of a weak acid (CH3COOH) and a strong base (NaOH). CH3COOH + NaOH = CH3COOHNa + H2O Imagine this: you add a weak acid and a strong base in water. Strong bases like NaOH dissociates 100% completely in water, we will ignore it because Na will be completely dissolved in water, and there will be no value of Na to compute. While weak acids like CH3COOH will only partially dissociate in water, this is why we solve for its molarity or pH. TLDR: If its ions come from a strong base/acid, ignore it. If its ions come from a weak base/acid, solve for it.

@parisamkh1 8 лет назад

thank youuuuuuuuuuuuuuuuuuuu. i watch all your videos. thanks

@taters1123 8 лет назад

Brilliant stuff

@Abkuyper 2 года назад

you always say Na+ will not react with water without the reason. I later learned it is a spectator ion.

@toolfrontman 4 года назад

Please correct me if I am wrong but I think the final calculations are incorrect in the terms of neglecting the X. Yes, in the previous videos X would represent the dissociated part of the weak acid/base, however in this case X represents the concertration of the weak acid that is formed. (Since acidic ion acts as a strong base). In my opinion the correct way is solving a whole quadratic equation