Why can you assume that the Fluoride ion is the concentration of the NaF compound? I understand why F- is the conjugate base for HF but why would NaF also be HF's conjugate base. Also, how do you know whether you are finding the Ka or Kb, because I thought if the product had a [OH-], we find Kb, but if it has a [H+] we're finding Ka. Thanks!
My textbook used the same initial concentration for the acid and the base so I was thinking the initial concentration of the acid was the same as the conjugate base (now in my head that doesn't make sense since it was a weak acid and there is no full dissociation) needless to say, thank you for the video. It helped me A LOT to see two different numbers for the two concentrations given.
it only works in the buffer zone, so if you reach the equivalence point it won't work. Some people can't tell which ones are buffers so some teachers prefer the long way which will give you the correct answer 100% of the time, doesn't matter if its in the buffer zone or past the half equivalence point
Sir can I ask this question I cannot understand this "A buffer solution contains acetic acid and sodium acetate. The concentration of each substance is .50M. Determine the pH of the solution"
No, the initial concentration of the fluoride ions is taken from the initial concentration of NaF. It is implied that this is an aqueous solution, and therefore Na and F will exist as ions, and since the problem said that there is 0.1M NaF, it can be deduced that the fluoride ion concentration will be 0.1M as well (1:1 mole ratio)
+M3HAN_ they dont cancel each other out bc you cant cancel out x's in division, they actually just are so small that they arent relevant to the problem anymore
Where did Na ( Sodium) gone? You didn't include it in the equation but you used the molarity. Is it because it's a solid. The ignoring of this element confused me. Shouldn't you have something like NaF + HF instead of HF + H2O ?
my teacher just did another very stupid and freaking hard way to understand... i find this alot easier.. but Ph is -log(H+) while u used -log (H3O+) how is that possible ??????
thanks, that was the most clear and helpful buffer tutorial I've seen on UA-cam.
You are an excellent teacher!!! And this is coming from a teacher:) I love how you simplify what you teach and make it understandable!
Nate, you're the only one who can teach me these concepts in a way I can understand. Also you're a straight up g. Keep keepin' it real.
HOW WAS THIS SO HARD TO UNDERSTAND IN CLASS
pH = pKa + log (NaF/HF). pH = -log Ka etc = 3.20 + log .5 = 3.20 - .301 = 2.9
thanks!
YAAASS I swear you just saved my good life. Blessings to you!!
Chem exam in the morning, this just saved me hours. Thank god for this video.
Why can you assume that the Fluoride ion is the concentration of the NaF compound?
I understand why F- is the conjugate base for HF but why would NaF also be HF's conjugate base.
Also, how do you know whether you are finding the Ka or Kb, because I thought if the product had a [OH-], we find Kb, but if it has a [H+] we're finding Ka.
Thanks!
Having the same question!
My textbook used the same initial concentration for the acid and the base so I was thinking the initial concentration of the acid was the same as the conjugate base (now in my head that doesn't make sense since it was a weak acid and there is no full dissociation) needless to say, thank you for the video. It helped me A LOT to see two different numbers for the two concentrations given.
Why are you using an ICE table? It would be a lot faster and easier to use the Henderson Hasselbalch equation.
+BeSharkbait MLG Did you watch the whole way through the video? he explains why at the en
d
Haha this is really helping me in College
So simple, thank you this has saved me
Also the assumption of x
Whats wrong with pH= pKa + log [base] / [acid] ?????
Nothing wrong at all... Its just another way to do it... lol
some profs dont allow the formula (like mine for example)
you can use it, but the henderson hasselbach equation is a good approximation to find pH of buffers easily. You can compare the results to verify.
it only works in the buffer zone, so if you reach the equivalence point it won't work. Some people can't tell which ones are buffers so some teachers prefer the long way which will give you the correct answer 100% of the time, doesn't matter if its in the buffer zone or past the half equivalence point
It,s a good website for understanding concepts. It,s a good way to teach.
Sir can I ask this question I cannot understand this
"A buffer solution contains acetic acid and sodium acetate. The concentration of each substance is .50M. Determine the pH of the solution"
How am i just finding this channel. Thank you
this is easier done using only one equation:
PH= PKA +log [A-]/[HA] which when putting the numbers gives 2.90
Do you know for sure that everytime you can delete the x variation both from the reactants and the products?
ICE Table is a savior!!! Thanks!!
Why did you assume that it goes forward?
Shouldn’t the pH have 3 decimal places though? Because the H3O concentration has 3 sig figs
couldnt u just plug into the henderson hasselback equation?
you are right but he is doing the buffer way...
So what happens if KI gets buffered by a mixture of COOHCH3 and COOCH3- ?
GOOD JOB!
But just can we cancel out the x which is in numerator??!
Guys I need help, please:
Acetic acid is added to water until the ph value reaches 4. What is the total concentration of the added acetic acid?
by using the Henderson Hasselbalch method, you can get answer easier and less complicated
hey! love your video! one question tho where does the H30 and the H2O come from? is it assumed?
Why dont you take into Le Chantliers principle as there was an original .1 mol of HF ? wont that affect the pH value ?
Wait what happened to the Na???
I always know when Nate is giving the tutorial because he always says "check it out". lol thanks Nate!!
I just learned what my class to an hour in 5 mins thank you
your hand is just the cutest :)
Why can we make assumption that it shift towards the right?
what do you call those undoing for the math near the end thats where it gets tricky...
The best chemistry ever
This might be a dumb question but how do you know that the initial concentration of NaF is equal to the initial concentration of the fluoride ions?
No, the initial concentration of the fluoride ions is taken from the initial concentration of NaF. It is implied that this is an aqueous solution, and therefore Na and F will exist as ions, and since the problem said that there is 0.1M NaF, it can be deduced that the fluoride ion concentration will be 0.1M as well (1:1 mole ratio)
Thank you so much!!! Super helpful!!!
why the "+x" and "-x" disappear magically?
+Brian Lock (神通) They cancel each other out. Its like saying what's 0 +1 -1 equal to?
+M3HAN_ they dont cancel each other out bc you cant cancel out x's in division, they actually just are so small that they arent relevant to the problem anymore
5% assumption
this might help someone else but when concentration/ka value>100 then you can cancel out the x's
Why do we start with 0,1 M of F- ?
+Top 10s [NaF] = [F-] because the NaF dissociates to equal concentrations of Na+ and F-
Give a lecture on buffer action.
Where did Na ( Sodium) gone? You didn't include it in the equation but you used the molarity. Is it because it's a solid. The ignoring of this element confused me. Shouldn't you have something like NaF + HF instead of HF + H2O ?
Sodium is a spectator ion
jose10garcia I hate spectator. They just disrupt the flow. Mojo disruptors!
MIND BLOWN lol i just remembered my professor saying that too, i was confused also. thanks for reminding me of it !lol
i think you did it wrong the first time cause you multipied by .2 before dividing by the .1.
Great work
In the next video he attempts to do an ICE table while writing on his pillow. What is that thing you're writing on?
I just simply used the Henderson Hasselback equation.
I sincerely appreciate
you Canadian man?
You're brilliant. thank you!
really helped me out! thank you very much!
Could have saved a lot of time by just using the Henderson Hasselbach equation, since you already have the acid and it's conjugate
Thank you!
Thank u so much, u made my day🙂🙂🙂
my teacher just did another very stupid and freaking hard way to understand...
i find this alot easier.. but Ph is -log(H+) while u used -log (H3O+)
how is that possible ??????
Mohamed Abdulatief H3O(hydronium) and H+ are the same thing, they indicate the same thing and are interchangeable.
Ohh.. i understood... Thanks for the reply
thank you
writing on carpet, what a badass
THANK U
Ur amazing!!!! Thanks
You da best!
Sorry I don't understand it's complicated 🤕
Thx Nate from a Nate
Wow that was very helpful. Thank you.
thank you!!!! :)
Thank you!