Professor Dave, you’re a lifesaver. If I get accepted for transfer at my dream university, graduate, then have the life I’ve always wanted I want you to ordain my wedding.
If you were my real professor i'd have 105 in chem you explain so well give great examples and you make sure we understood what you showed us by making us do an exercise ourselves. What great way of teaching!
@Usern Last You are right! However this is not equal for children or I should rather say not for EVERY child. Guidance is important until a child matures and so the teacher can be a strong factor in the students success.
In my 13+ years of experience, I can tell you that some students need more support than the others. World is not fair! But we can make it work by publishing quality videos for the needed.
Molarity times Volume (MV)before titration= MV after titration 75/1000 x 0.5 = 165/1000 x M M= 0.2272727... This also worked and my teacher told us to use this method
Wait, did I just finally understand the topic in under 2 minutes and 40 seconds? Yes. Thank you so much prof Dave! You made it so much easier to understand!
I was stressing really bad over my homework because the other resources my teacher gave weren't helping, but your video made it so incredibly clear. Thank you!!
Your videos are great...I recommend my students to view them quite often. However, in this case it may have been better to use a monoprotic acid in the example...
You are my go-to channel for chemistry help! Keep making content they're really informative and helpful! In the future videos, can you please make some on organic chemistry? Such as naming and drawing hydrocarbons in functional groups, or anything related to organic chem. Thank you and keep doing what you do best!
1:40 Mr. Dave, perhaps you missed the point on using the stoicheometrical ratios. Besides you did not even balance the equation. Have a look at this: 2NaOH + H2SO4 ---> 2H2O + Na2aSO4 That implies you had to multiply 0.0348mol with 2, then divide with its volume. Because we need 2 miles of the base to neutralize just 1 mole of the acid, suggesting the resilience of the acid. So, 0.0348mol×2= 0.0696mol Then 0.0696mol/0.025L = 2.784M
Hi Prof. Dave, I have a question: Should we not consider the balanced equation of Sulfuric acid and Sodium Hydroxide when we do the calculation? The ratio of acid/base should be 1:2, then the mole of base should be 0.0348 x 2 = 0.0696...please advice thanks!
@@ProfessorDaveExplains Thanks! I am doing Cambridge Chemistry A/AS level review presently... from the other references I gathered, it's a general rule to consider the balanced equation. But I guess, it depends on the curriculum not sure.
@@ProfessorDaveExplains Hi Dave, I love your videos because they're very easy to follow. I would just like to correct this one. It is true that sulfuric acid is strongly acidic for the first proton, but that is relative to water acting as a weak base. In this case, you have a strong base, NaOH, reacting with H2SO4. It is strong enough to ionize the second proton, so the ratio is 1 NaOH : 2 H2SO4. Another example would be titration of acetic acid, a weak acid, to a strong base. You can still remove the proton from the acetic acid even if it is weakly acidic.
@@ProfessorDaveExplains At a glance it seems that the 3.0 M sulfuric acid also as 3.0 N sulfuric acid. but Sulfuric acid is easily deprotonated twice by NaOH, so it is 3.0 M but 6.0 N. Therefore the NaOH would actually be 1.39x2 M. pKas are -2.88 and 1.99. while the second proton is not strongly acidic, its still easily deprotonated by the strong base NaOH. Otherwise would be like saying Acetic acid pKa = 4.75 is not deprotonated by NaOH because its the first is a weak acid... Treating 1M H2SO4 as 1N H2SO4 can have catastrophic consequences :s I also second IC Duays comment.
@@ProfessorDaveExplains The second proton (the one of Hydrogensulfate) will still react with the hydroxide anions in the solution. Thus, the stochimetric factor of 2 has to be taken into account.
50 ml of a solution which is 0.050 M in the acid HA (pka = 3.80) and 0.10 m in HB (pka = 8.20) is a titrated with 0.2 M NaOH. calculate the pH at the first equivalence point, and second equivalence point. Please explain such question including this!
Great video but would you please consider updating the graphics to show the proper glassware (change the beaker to an Erlenmeyer Flask) since you need to swirl to mix.
How is that first math example a 1:1 mole ratio? Wouldn't balancing the equation between the acid and base result in a 1:2 ratio between sulfuric acid and sodium hydroxide? H2SO4 + 2 NaOH --> 2H2O + Na2SO4 Thus making the answer half of what you have listed...
+Matthew Geistfeld it's true i maybe should have used a monoprotic acid, but the Ka for the second ionization for sulfuric acid is tremendously smaller than the first, so it's not at all the same as considering it to be twice the volume of acid.
I'm not an expert by any means, so don't listen to closely to me, but I THINK it depends from substance to substance. The different solutes and solutions have different ratios. (Not 100% sure though, don't @ me if I'm wrong pls)
You don’t worry about coefficients in molarity. As long as you do Molarity = moles/ liters. If you solve molarity by m1v1= m2v2 then you do worry about coefficients. Don’t use the second method. You’ll be wrong.
even though its a diprotic acid, the second proton is nowhere near as acidic as the first, so we are just looking at neutralizing the results of the first deprotonation.
I think so too because the stoichiometrical mole ratio in the example for NaOH and H2SO4 is 2 to 1 when the equation is balanced. Therefore the solution should have been 2.784M and not 1.39M.
I thought the stoichiometric ratio of H2S04 to NaOH is 1:2. why is the moles of acid equal to the moles of base? I'm confused about the conversion part can you pls clarify. thanks!
Can we use NaOH in flask and HCL in burette and use phenolphthalein indicator.? Is it fine to have colour change from pink to colourless????? Pls answer me😢😢😢😢😢
because that's the point where the analyte is precisely neutralized, reaching the equivalence point where the change in pH is such that the indicator changes color.
@@ProfessorDaveExplains Thanks. I need to complete this experiment as part of my Chemistry course. However, I am a little unsure about selecting an indicator for my chosen base/acid. Can I message you for advice?
Agh, the prof wants us to answer concentration of .1M NaOH (which is being titrated INTO KHP soln), how can we do this? It doesn't make sense to figure out M of NaOH instead of M of KHP.
I get it but why is it that sometimes you change the volume of the acid to liters and sometimes you leave it as mLs? I googled a few of my practice questions and the way you do it here gives me a different answers than what I find online. Can M also be equal to mol/mL or only mol/L
I want to cry with joy at how easily you explained this, thank you for alleviating a huge frustration!!!
Professor Dave, you’re a lifesaver. If I get accepted for transfer at my dream university, graduate, then have the life I’ve always wanted I want you to ordain my wedding.
Did you?
@@TheNaikas its been a year how he finna do all that in this amount of time
@@omarqacle girl u neva know
Update us
u marry yet?
If you were my real professor i'd have 105 in chem you explain so well give great examples and you make sure we understood what you showed us by making us do an exercise ourselves. What great way of teaching!
@Usern Last You are right! However this is not equal for children or I should rather say not for EVERY child. Guidance is important until a child matures and so the teacher can be a strong factor in the students success.
In my 13+ years of experience, I can tell you that some students need more support than the others. World is not fair! But we can make it work by publishing quality videos for the needed.
Hey I know it's been years, but I just came across your comment and was curious. How have your studies been going?
@@chemistrywithcoffee1381
You know, that's pretty fair.
@@chemistrywithcoffee1381 That's equity, not equality
Brilliant teaching. Really appreciate the thought and effort made into making these videos.
Molarity times Volume (MV)before titration= MV after titration
75/1000 x 0.5 = 165/1000 x M
M= 0.2272727...
This also worked and my teacher told us to use this method
But only for monoprotic substances
i think this only works with niche acid-base titrations?
along with dilution problems - c1v1=c2v2
Coincidence. Only true when the stoichiometric ratio is 1:1
In fact, his video mistakes this when H2SO4 reacts with NaOH
@@cheddar6020 It is just an example dude, however I am curious, why wouldn't it react? If I remember correctly, they both are soluble
Assuming 1:1 molar ratio. If we balance the equation it's 1:2 giving us 2.784M
Wait, did I just finally understand the topic in under 2 minutes and 40 seconds? Yes. Thank you so much prof Dave! You made it so much easier to understand!
thanks pookie i missed like a week of school and didn’t even understand this lab
i want the 5 hour version of the jingle
Imagine a yr12 chemistry class in Australia watching someone from the United States.
you forgot to do molar ratio for the first problem. you should've multiplied the moles of H2So4 by 2 due to the 1:2 ratio. Thank you
I was also thinking the same thing
agreed
i don’t think so? doesn’t sulfuric acid only have one strong ion? when it ionizes the solution will be mostly hso4- and h+
Thank you for NOT over complicating your explanations. 🙏🏽
8 years and these videos are still saving my life.
On the example, the mol of base should be two times the acid so the mol of base is 0.0696
Thank you so much, your videos have been super helpful as I have been studying for my big chemistry exam!
my pleasure! you can do it!
SELL GOLD AND SILVER BUY PLATINUM PRICE IS LOW NOW PROMOTE PLATINUM PRICE IS RISING LIKE GOLD....
I missed the part where you explained the titration?
I was stressing really bad over my homework because the other resources my teacher gave weren't helping, but your video made it so incredibly clear. Thank you!!
Sir your theme song and your explanation both are the best.
Your videos are great...I recommend my students to view them quite often. However, in this case it may have been better to use a monoprotic acid in the example...
Yes, this is my concern too.
breef, 1/a.n(titrator)=1/b.n(titratee), such that the formula of both the titrator and the titratee is a.titrator and b.titratee
Acid-base titration? More like "This is amazing information!" Thanks for sharing another very high quality video.
This is exactly what I was looking for, tysm!
For real i couldnt get this concept this clearly anywhere else.🎉🎉
hard carry after 8 years, thank you so much!
Thenk you professor you saved my life i was having a seminar and i got 10/10❤️
You are my go-to channel for chemistry help! Keep making content they're really informative and helpful! In the future videos, can you please make some on organic chemistry? Such as naming and drawing hydrocarbons in functional groups, or anything related to organic chem. Thank you and keep doing what you do best!
+TheSecretBuster my friend, i have an entire organic chemistry playlist! 38 clips. check it out.
Professor Dave Explains Oh haha just saw them, sorry for the confusion.
You might want to fix your mole ratio at time = 1:38. Its a 1:2 the base would be twice the 0.0348 and hence double 1.39 as indicated.
1:40 Mr. Dave, perhaps you missed the point on using the stoicheometrical ratios. Besides you did not even balance the equation. Have a look at this:
2NaOH + H2SO4 ---> 2H2O + Na2aSO4
That implies you had to multiply 0.0348mol with 2, then divide with its volume. Because we need 2 miles of the base to neutralize just 1 mole of the acid, suggesting the resilience of the acid.
So, 0.0348mol×2= 0.0696mol
Then 0.0696mol/0.025L
= 2.784M
THANKS FOR HELPING ME MAKE BIODIESEL
Sir your video was very helpful
I understood everything
I wish that u were my chemistry teacher
Hi Prof. Dave, I have a question: Should we not consider the balanced equation of Sulfuric acid and Sodium Hydroxide when we do the calculation? The ratio of acid/base should be 1:2, then the mole of base should be 0.0348 x 2 = 0.0696...please advice thanks!
it's only strongly acidic for the first proton
@@ProfessorDaveExplains Thanks! I am doing Cambridge Chemistry A/AS level review presently... from the other references I gathered, it's a general rule to consider the balanced equation. But I guess, it depends on the curriculum not sure.
@@ProfessorDaveExplains
Hi Dave, I love your videos because they're very easy to follow. I would just like to correct this one.
It is true that sulfuric acid is strongly acidic for the first proton, but that is relative to water acting as a weak base. In this case, you have a strong base, NaOH, reacting with H2SO4. It is strong enough to ionize the second proton, so the ratio is 1 NaOH : 2 H2SO4.
Another example would be titration of acetic acid, a weak acid, to a strong base. You can still remove the proton from the acetic acid even if it is weakly acidic.
@@ProfessorDaveExplains At a glance it seems that the 3.0 M sulfuric acid also as 3.0 N sulfuric acid. but Sulfuric acid is easily deprotonated twice by NaOH, so it is 3.0 M but 6.0 N. Therefore the NaOH would actually be 1.39x2 M.
pKas are -2.88 and 1.99.
while the second proton is not strongly acidic, its still easily deprotonated by the strong base NaOH. Otherwise would be like saying Acetic acid pKa = 4.75 is not deprotonated by NaOH because its the first is a weak acid...
Treating 1M H2SO4 as 1N H2SO4 can have catastrophic consequences :s
I also second IC Duays comment.
@@ProfessorDaveExplains
The second proton (the one of Hydrogensulfate) will still react with the hydroxide anions in the solution. Thus, the stochimetric factor of 2 has to be taken into account.
It’s so much clearer now that I have gotten your explanation, thank you!
that was quite
helpful,keep on making such videos
Very helpful. Thank you for the clear explanation Professor Dave!
W you just dropped the exact answer for what i'm gonna need to do for my lab
50 ml of a solution which is 0.050 M in the acid HA (pka = 3.80) and 0.10 m in HB (pka = 8.20) is a titrated with 0.2 M NaOH. calculate the pH at the first equivalence point, and second equivalence point.
Please explain such question including this!
Thanq sir u make the concept very easy to understand with in short time
This man is saving my degree 😭
Great video but would you please consider updating the graphics to show the proper glassware (change the beaker to an Erlenmeyer Flask) since you need to swirl to mix.
you can do it with a beaker and a magnetic stir bar, actually that's the most common way since it allows for constant stirring
@@ProfessorDaveExplains but even then the flask is preferred.
You are wrong. Conc of naoh is 2.784 since it's a 1:2 ratio
How is that first math example a 1:1 mole ratio? Wouldn't balancing the equation between the acid and base result in a 1:2 ratio between sulfuric acid and sodium hydroxide? H2SO4 + 2 NaOH --> 2H2O + Na2SO4
Thus making the answer half of what you have listed...
+Matthew Geistfeld it's true i maybe should have used a monoprotic acid, but the Ka for the second ionization for sulfuric acid is tremendously smaller than the first, so it's not at all the same as considering it to be twice the volume of acid.
Thanks!
Professor Dave Explains I understand this but googling other examples of this same problem they do a 1:2 ratio. Why is that?
FINALLY IVE FOUND CHEMISTRY JESUS!
Why i don't understand the will neutralizer 1:58
How did you get the 11.6mL??
You're my literal lord and savior thank you so much
watching before my chemistry exam
Thank you, it is really useful
1:35
why is the mol ratio for stoichemetry 1 to 1? Isn't it supposed to be 1:2
I'm not an expert by any means, so don't listen to closely to me, but I THINK it depends from substance to substance. The different solutes and solutions have different ratios. (Not 100% sure though, don't @ me if I'm wrong pls)
You don’t worry about coefficients in molarity. As long as you do Molarity = moles/ liters. If you solve molarity by m1v1= m2v2 then you do worry about coefficients. Don’t use the second method. You’ll be wrong.
I liked the all the video but especially the backgrounds!!!
Aside from indicators, how could we determine the endpoint of acid-base titrations?
Thank you papa susej🙌🏻
Why 11.6ml ???
Is it really that easy
shouldn't it be 0.0696 mol base?
even though its a diprotic acid, the second proton is nowhere near as acidic as the first, so we are just looking at neutralizing the results of the first deprotonation.
@@ProfessorDaveExplains It's 98% as acidic and a balanced reaction will consider both protons.
Thanks Professor Dave!
For NaoH concentration the volume should be (.0116+.025) L not .025L.
concentration before the titration.
One more thing, maybe include the balanced chemical equation too as part of the solution to the proposed problem. :-)
I think so too because the stoichiometrical mole ratio in the example for NaOH and H2SO4 is 2 to 1 when the equation is balanced. Therefore the solution should have been 2.784M and not 1.39M.
TRUE, the ans is wrong
Thank you professor Daves
I thought the stoichiometric ratio of H2S04 to NaOH is 1:2. why is the moles of acid equal to the moles of base? I'm confused about the conversion part can you pls clarify. thanks!
I think he just made a mistake, I found ratio is 1:2 also.
Thank u prof u save my day
Sir You are the best teacher
I want to knowWhat are three attributes of a successful titrimetric analysis?
Dam you did more for me in 3 min than my prof did in a 4 page document
Thank you for helping me understand this💕
I did M1V1 = M2V2, solved for M2, got the same answer.
Thanks for the formula
EPIC VID PROFESSOR
Also, next semester I'm supposed to take Physical Chemistry. If possible, videos on that would be greatly appreciated.
its 2am and im on campus a day before the exam but im patting myself on the back for getting the comprehension problem right haha
you're a lifesaver
i thought the ratio of naoh to h2so4 is 2:1.. from the equation..
Wow.. So my university just stretched this simple concept out to a 45 minute explanation AND made it unnecessarily confusing.
I like that method, but I just used good ole c1v1=c2v2
Thanks dave
professor dave i’m in love with you
Thank you
But can u talk about neutralization titration
Pls
Whats the different between acid-base titration and acidimetry?
Sir you are like ranbir kapoor.
Bro what where did animal come from? 😂
On a titration curve, is base always on the x axis?
Professor Dave,
Do you have videos for biochemistry?
+Nicole Henry those will be coming soon! probably over the summer.
omgg yes please!!! i finish organic this semester and I am scared for biochem
Iam thinking that 1 mole of H2SO4 reacts with 2 mole of NAOH
should it not be 4 significant figures? the 165.0 had 4
Can we use NaOH in flask and HCL in burette and use phenolphthalein indicator.? Is it fine to have colour change from pink to colourless?????
Pls answer me😢😢😢😢😢
why did it take 2 other professors each a full day and a lab to explain this to me and have me not understand?
Shouldn’t you times 2 because there are 2 moles of NaOH for every H3SO4? why isn’t this the case?
Really good
hi teacher , why 11.6??? please
How many tags do you put on your videos? The quality of them is so often exceptional that something seems amiss.
pls make on Arrhenius,Bronstein and lewis concept literally need it
check my video on acids and bases
Thank you
Thank you chemistry jesus
Brother, the stoichiometric ratio is not 1:1 for NaOH and H2SO4...
Sir I would like to ask as to why the graph of volume of base vs pH is a curve not a straight line??
logarithmic relationship!
When you try to find the point of inflection of the graph......
yes!!!! calculus>chemistry
is the 3.0m H2SO4 the indicator?
wait.....how acid and base can be together they just react right!
Could you please explain the ratio part? Why does X moles of the acid equal the same number of moles of the base?
because that's the point where the analyte is precisely neutralized, reaching the equivalence point where the change in pH is such that the indicator changes color.
@@ProfessorDaveExplains Thanks. I need to complete this experiment as part of my Chemistry course. However, I am a little unsure about selecting an indicator for my chosen base/acid. Can I message you for advice?
sure you can email!
THANKS
Thank you!
I still can't fathom how we know when we reach an equivalence point with an indicator, all I get is we just make it change color.. ach.
Agh, the prof wants us to answer concentration of .1M NaOH (which is being titrated INTO KHP soln), how can we do this? It doesn't make sense to figure out M of NaOH instead of M of KHP.
I get it but why is it that sometimes you change the volume of the acid to liters and sometimes you leave it as mLs? I googled a few of my practice questions and the way you do it here gives me a different answers than what I find online. Can M also be equal to mol/mL or only mol/L
M is mol/L, it's just that for certain calculations volumes in mL are ok
Gigachad teacher
What topic are the calculations from that confused me slightly