BUFFER SOLUTION CALCULATIONS 1

[@aang7505]

Thanks so much! A lot of teachers/professors fail to go over these more complicated questions and the line of reasoning behind answering them. I really do appreciate the time you spent making this video!

[@zakkharsane3988]

I am teaching myself A levels chemistry at home, I don't go to any college, and your videos are the base of my learning, thank you.

[@sujata8957]

I don't know what I would do without this channel so much immense respect and gratitude for you sir!

[@MaChemGuy]

So nice of you to say :)

[@mestanhadi68]

Thanks very much!!! My own teacher didn’t explain anything about buffers, not even a single thing! I really appreciate that 🙏🏻

[@MaChemGuy]

mestan hadi Glad it helped 👍

[@niamhmilligan8924]

Just wanted to say thank you soo much for taking time out of your day to make these awesome videos! Theyve been super helpful over the last 2 years! Thank you!

[@MaChemGuy]

Niamh Milligan Hey you're very welcome. Thanks for getting in touch, it means a lot. Glad you've found them helpful. Good luck both in the exams and for the future. You'll have to let me know how you do. Cheers!

[@chinrxhh6707]

And I finally understand what I have to learn by watching your video.☺️☺️☺️.Short and full of informative video as well as simple explanation.I am glad that I had came across your channel yesterday and it saves my life.Can't say how much I appreciate for your video.

[@MaChemGuy]

Thanks for the feedback, really appreciate it. Glad you have found the videos and that you find them helpful. Check out my buffer solutions walkthroughs playlist so you can test your understanding on real exam questions 😀

[@chinrxhh6707]

@@MaChemGuy sure.😉😉😉.I wanna ask that do it is compulsory that the mol for weak acid and it's conjugate base must be approximately the same so that the buffer can perform well?

[@MaChemGuy]

@@chinrxhh6707 yes the closer they are in concentration, be better a buffer is at dealing with the addition of acid or alkali

[@chinrxhh6707]

@@MaChemGuy ok thanks for answering my question.☺️☺️☺️☺️.

[@damongeo840]

Really liked your lessons, thank you very much! A few things that I would like to notice is that cm^3 is used mostly for solids (metals, soil etc); for liquids ml is mostly used - which is the same size. Another thing is that the 45ml volume is not certain that it will be the final volume of the two mixed solutions but certainly the final volume is the final volume so this could not be stated by using the moles for a constant not known exactly volume. I am trying to learn about buffer solutions and your videos are very useful! Thank you again!

[@immaaug2416]

I wonder who pressed The dislike button up there😠 really!!!. Fuckn love how you teach👌

[@MaChemGuy]

Well your comment just cancelled them out. Cheers!

[@Haris1]

I didnt

[@geeekA91]

Im going on holiday....Kaacid over salt!!!!!!!!! Whooop loool i got that question right... Cant thank you enough sir! ☺️

[@MaChemGuy]

It's such a lovely place !

[@tajwar5859]

MaChemGuy for Oscar! you are a legend thank you :)

[@MaChemGuy]

Bless you!

[@aaliyadaulatzai]

Explained 10000x better than my chemistry teacher, many thanks

[@MaChemGuy]

Glad you approve :)

[@ProElite97]

Kaacidoversalt, fucking love that.

[@MaChemGuy]

ProElite97 It's such a wonderful place

[@leone41ll]

Lmao brilliant

[@jocelyn2953]

Thank youuuu!!!!! You probably just saved my chemistry grade

[@iman7510]

Thank you so much for your awesome videos. You have an amazing way of teaching. You don't only cover the basic content but you also go over exam style questions. Things I thought I couldn't do has been made easy with your videos so thank you SO much. :)

[@MaChemGuy]

You're VERY welcome. Thanks for your kind words :)

[@Rico7817]

I love you man, thanks for these videos they really do help us!

[@aqsa3248]

Thankyou! Such a great vid, will be able to answer buffer q's now! :)

[@sea5205]

Amazing video and great explanation! Thanks

[@mxiraajoliee7514]

THANK YOU SO MUCH. THIS IS GREAT!! KEEP UP THE GOOD WORK

[@malanhemal6574]

sir, is the mol remaining which was 0.0165 the mols at equilibrium for the acid?

[@Dischordian]

I like the Holiday to KaCidoverSalt bit - very good, can I say though that you are harbouring a misconception regarding [H+] and [A-] on your introductory 'slide'? They are in EQUAL concentration until EXTRA "A-" is ADDED. It's not possible for them to be in different concentrations just as a result of dissolving an acid regardless of how strong or weak it is because in order to form H+, there must have also been A- formed!!!

[@MaChemGuy]

+Dischordian I cover why they are different in my introduction to buffer solutions video. The assumption is that students watch this introductory video first and so know where the different concentrations of [H+] and [A-] come from as shown at the start of the calcs 1 video

[@pritchards100]

At 8:50 why do you divide by 0.045? I would have thought you divided the moles of ch3ch2cooh by 0.030 as that is what it's volume is. Thanks a lot for all your videos they're so helpful

[@pritchards100]

For buffer solutions do you use total volume of solution?

[@MaChemGuy]

Xoovady Baranacus total vol is 45cm3 Concs are measured according to this :)

[@jamesrichard1140]

Very nice, simplfied how to do a hard buffer calc

[@SupernovaHD12399]

It’s 6 years late but how did your exam go ? 🤣🤣🤣🤣

[@jamesrichard1140]

@@SupernovaHD12399 lol I got an A and did Biochem at uni

[@SupernovaHD12399]

@@jamesrichard1140 Ah nice one. Got my chemistry mock exam in a week and I cant figure out buffers lol

[@mrgg336]

@MaChemGuy Do you source your examples from the CGP textbook, and if so, would you recommend them for revision notes etc.

[@gregorypeters1178]

Is there a video for a question similar to 2015 F325 where it asked the PH of the new buffer on addition of Magnesium? I fail to understand why 0.25 moles of Mg added would mean the conc. of acid to decrease by 0.5. Thanks, loved your videos.

[@TheAbs23]

because if you worked out the equation it was for every Mg there was two molecules of acid. Because mg was 2+ and the acid was 1-, so o.25 x 2=0.50

[@gregorypeters1178]

+Khal Drogo I understand that in terms of moles but I struggle to understand it in terms on concentrstion

[@TheAbs23]

Im not quite sure what you mean there, could you elaborate please? (sorry)

[@gregorypeters1178]

+Khal Drogo You work out the moles and concentration of Magnesium which is 0.25 ergo 0.5 moles of acid is used. The part I don't get is why this means the concentration is reduced by 0.5. Perhaps hard to understand my question as I don't understand it myself..

[@TheAbs23]

i think i get what you're saying, i could be wrong here but you have 1 moldm-3 of the acid, and as you said 0.5 moles of the acid is USED. Therefore 0.5 moles of the acid has been turned into the conjugate base. Put it this way you've got 1 litre of water someone takes half and adds it to a litre of squash. So you're left with half a litre of water and 1.5 litres of squash

[@MP-hv1fq]

Do you not have to subtract A- from HA to find new moles of HA as there this is partial neutralisation?

[@MaChemGuy]

Mohammed Patel technically yes but it's ok not to as the dissociation is so very small

[@amoyholddown2298]

helps a lot

[@MH-fq4vy]

these videos actually make me like chemistry!

[@anetas777]

thank you sooo much for all your useful videos, you explain everything so clearly! forever grateful :))

[@MaChemGuy]

A Stoyanova You’re very welcome. Thanks for that, means a lot 🙏🏻

[@Digitalguru642]

Thanks for this video, it Really helped

[@MaChemGuy]

You're welcome. Glad to hear it's helped

[@yeshigurmey6628]

thank you sir,very helpful

[@promisengobese8007]

this help my exam are in week.thanks

[@duffybelle3502]

For Q3 are you not supposed to do 0.1mol/0.5dm^3 to get the concentration as you have just used the moles but the square brackets means concentration. Hope that made sense!!!

[@MaChemGuy]

DuffyBelle The M is an abbreviation for moldm-3

[@duffybelle3502]

Oh ok thanks!!!

[@SlaveofChrist_]

In my Chemistry class, it's taught to do the ICE Table before finding the acid for weak acids. But I see you didn't do this, is the ICE table required?

[@MaChemGuy]

Charles Coldman No because it's ok to assume the conc of the weak in the buffer equilibrium is the same as its original concentration due to the tiny extent of dissociation

[@SlaveofChrist_]

+MaChemGuy Thanks!

[@killerbobization]

Isnt the acid concentration on the last question suppose to be 0.2 not 0.1, since 0.1 is just the mol of the acid and not the concentration.

[@hnataraja]

Kris Grabys yeah that's what I thought? can someone explain please!!?

[@dramirashraaf]

Isn't 0.1M = 0.1 mol / dm3 ? So concentration of acid should be 0.1 right ? Or am i wrong 😅

[@MaChemGuy]

+Amir Ashraaf M is the shorthand for moldm-3

[@nraishanzr3333]

ever heard of the Henderson-hasselbalch equation? It works a lot faster than this trick you're doing. Still great video, thank you!

[@MaChemGuy]

Yeah I’ve made a video on it. I tend to find that the non maths students don’t like it. Glad you like the video 👍

[@johnnysaleh7267]

In the second question, there was no need to convert number of moles of acid and salt to concentrations since both exist in same solution.

[@MaChemGuy]

I know. Thanks

[@abhishekvarmabh6372]

Very helpful TY

[@oonagh-mariemicklethwaite7601]

Thankyou so so so much!!! So helpful

[@sineadryan4424]

Why is the optimum pH for the H2CO3/HCO3- buffer 6.36 when the pH of blood is 7.4?

[@guzmananne]

so why is the acid dissociation constant for the weak acid 1.6 x10 -4, where do you get that from?

[@MaChemGuy]

Anne Guzman I got it from a data book. In an exam it would be given in the question

[@skhan5336]

Hi you said the moles of sodium ethanoate is 0.0015. How did you come to this ?

[@Jex2Monsta]

shoulnt it be salt=h+/ka multiplied by acid. This is according to rearranging equations. And if u wanna find out n in pv=nrt, it is n=pv/rt. What the hell grh pls respond

[@nikkii3738]

it was really good but one req please put your camera at some height the ques were not visible properly.

[@ariextra9988]

Why can't we calculate [H+] using the square root Ka x [HA] method? If its in equilibrium, isn't it a weak acid?

[@MaChemGuy]

Salt conc too different to H+

[@paperplaneinnyc]

i calculated the amounts of salt and acid and put them into the h+ equation without subtracting salt from the acid, yet i still got the right pH in the end? any idea why? :/

[@kimberlyrodrigues7789]

For Q2, why is [CH3CH2COOH]=0.37 and not 0.6?

[@TheAbs23]

0.0165 x 1000 divided by 45cm = 0.3666666667

[@zakkharsane3988]

Thank you, thank you, thank you,

[@Schrecken]

why not use henderson hasslebach?

[@MaChemGuy]

Jim Hazen You can. I go through that in next video

[@damongeo840]

It is the same exactly thing.

[@patrickstar7693]

It gets always talked about the addition of a strong acid or strong base to a buffer... my question would be how to calculate the ph with the addition of a weak acid? Lets say the addition of acetic acid to a phosphate buffer

[@itsConnor28]

very helpful, thanks :)

[@gyrojomo]

Thanks.

[@aldanenaicker1983]

i love you man

[@bgtl1000]

what do you mean kaacidoversalt is not a place. #liar

[@sundusm2507]

rubbish science Teachers need to take a leaf out of your book on how to teach effectively

[@rashmitabaral8955]

Hm ab too hi padhale murkh

[@rashmitabaral8955]

U nonsense take a class i want to see ur ability

[@alikellaway3901]

Surely the [A-] is not a salt, it's just the negative ion from the dissociation of salts or original acid? The salt would surely be the combination of some kind of metal ion and the conjugate base of the dissociated acid? Confused me a bit. Hate to sound like a nerdy twat criticising.

[@XpertGamerXGYT]

I love you man

[@MaChemGuy]

Steady on!