14B Acid-Base Titrations - Edexcel IAL Chemistry (Unit 4)

This video covers the content of Topic 14B Acid-Base Titrations in preparation for the Edexcel IAL Unit 4 Chemistry exam. Use the timestamps below to navigate the content in the video:
0:00:06 - Acid-Base Titrations
0:03:52 - Titration Curves
0:08:35 - Indicators
0:15:58 - Choosing the Correct Indicator
0:20:23 - Buffer Solutions
0:26:24 - How do Buffer Solutions work?
0:34:30 - Calculating the pH of a Buffer Solution
0:51:31 - Buffers in Blood and Food
0:54:48 - Buffer Solutions and pH Curves
0:59:51 - Past Paper Questions

[@joodosama9455]

Thank you for your hard work ❤️

[@60_069]

Thank you very much for uploading these ma'am. you're a life saver!!

[@3amakhathak767]

جزاك الله خير يا معلم

[@zainabashroff2999]

At 37:35 Isnt the total volume 45 + 50=95cm³ So the final pH would be 5.97

[@panadol2851]

I used 95cm^3 and still got ph 4.99

[@fakhirrehan8869]

smarty smarty

[@abcrm101]

I have finished all your videos on topics 11, 13, and 14 Thank you so much for putting so much effort into making these videos Having a few practice questions at the end of each topic is really nice since it allows me to recall the things that I have learned Hopefully, I would do better in Unit 4 this time ( I did not revise the content last time and jumped straight to the past papers and got only UMS 109/120, but I am aiming for an A* so I need a higher UMS in Unit 4 to pull the overall UMS of A2 up)

[@The_FutureDoc]

How did you do in A2? I am doing in June this year..

[@abcrm101]

@@The_FutureDoc hello :) hope it went well for you! I forgot about the result of my individual units, but I got A* in Chemistry at last

[@Thiakit]

Such a wonderful explanation. Thank you Miss Natalie !😊

[@AD-ch9jo]

great video,it all makes so much more sense now.Thanks! =)

[@varunliyer8594]

thank you miss Natalie your videos are vey good. i love it. please make more

[@SaraZorek-gs2nw]

Thank you so much for the amazing videos ❤❤❤ I really appreciate it

[@sarujan02thangavel93]

Thank you very much miss. Helped me a lot

[@itzmohd2186]

Amazing work💗

[@adnansiam8534]

love your class

[@juairyataskin3804]

thank you so much miss!

[@pudaminigomes619]

THANK YOU SO MUCH

[@Spoonbob_the_1]

This channel needs more subs and comments and shares please

[@saiyadun1172]

On 47:42,why cant I use the previous buffer formula? I get different answer for using -logKa+log[acid/salt}

[@afs204]

Hi miss i really appreciate all your videos. I just wanted to ask if we need to be able to know whether a molecule is an acid or a base and if its strong or weak just by looking at it?

[@Ai-pabellaaaa]

46:24 n(salt)=0.125+0.005 or it should be 0.05 as NaOH dissociated is 0.005

[@mehh_01]

Ye it should be 0.005

[@Sarah-yg3us]

In calculations for buffers how do we know if there's a large reservoir of acid?

[@rafiatbm7867]

For the calculation you did at 45:30 , how could you assume the fact that the acid told in the question is a monoprotic acid and not a diprotic acid ?? I mean if it was a diprotic acid then 0.0025 moles of acid would've reacted with 0.005 NaOH

[@MissNatalieChemistry]

At A Level, you would only ever be asked about a monoprotic acid. Diprotic acids are too complicated when it comes to the maths as sometimes it can involve quadratics to determine the pH value and this is above the scope of the course.

[@ayeshatauseef6892]

at 46:13 where youve just found the new moles for salt why have u used 0.05 instead of 0.005?

[@abcrm101]

28:50 I thought the ratio of these 2 substances would remain almost constant because both have a very high concentration (there is a very large reserve of them), instead of the concentration of one of them increase and one of them decrease

[@Wazguu]

That's what I learned too, because one increases and one decrease doesn't help keeping the ratio similar

[@dianamarissaleighton5613]

ratio remains *fairly* constant..?

[@abcrm101]

@@dianamarissaleighton5613 yeh probably Good luck with your IAL Chemistry :)

[@threeframes2856]

hi can u help me with topic 16 and 17 bcoz i cant find any videos on these topic

[@nusaibatasfiat5347]

hi! at 47:27 the question why didn't we divide the moles with volume again for concentration and put them directly in the [H+] equation, like we did in the previous questions?

[@ruchitalalla8578]

i have the same doubt

[@zainabxyz]

One q. So in the examples of buffer as in one u found the total vol and in the other one u found the excess , why did you do that?

[@yameroo3116]

❤ I wish u the heavens and further. Your Wildest Dreams. You are an angel and I thank u

[@the_crackin3610]

Thx

[@hanamostafa5268]

24:58 The Ka here is written depending on which equation is it the acid dissociation equation or salt dissociation cause it shouldn't be from the acid as we assumed that the dissociation of acid is negligible so its products should be negligible, Can you please help because I am a bit confused here about which equation we used to write Ka?

[@MissNatalieChemistry]

Ka is always written using the dissociation of the acid equation. Ka stands for the acid dissociation constant.

[@Abdul-bk5fj]

@@MissNatalieChemistry ok maam then how are we using the concentration of the ethanoate ion from the second equation? because if we were to use the concentration of the ethanoate ion from the first equation, it would be 0 or negligible due to negligible dissocation.

[@sara-wr5bk]

This video is so helpful! I just have a question - at 12:37, how does the expression in your calculation simplify to (H+) = 2.00 x 10^-4 through using (H+) = (In-) (in the second line of the calculation)? Or is that meant to be (HIn) = (In-) instead?

[@MissNatalieChemistry]

Very sorry you are absolutely correct! That should say [HIn] = [In-] so they will cancel each other out and [H+] can be calculated.

[@sara-wr5bk]

@@MissNatalieChemistry Ah okay, thank you :)

[@sajedaabdelnaser794]

omg i literally felt so stupid for not understanding how that came to be

[@Youtuber-by6hx]

I really appreciate your hard work! :) I have a question please after 42:35 can i use concentration of h+ squared instead of concentration of h+ times concentration of salt, which we got from naoh

[@MissNatalieChemistry]

No you cannot use [H+]^2 because that implies that the values are the same but they are not. For a buffer, you must keep the concentration of the acid and the salt separate.

[@fakhirrehan8869]

@@MissNatalieChemistry oye neechay wale comment ka jawab de

[@greengod3114]

@28:51 when salt decreases and acid increases the overall value is increased how is the ratio same here ?

[@aariqhasan2801]

Thery are in such high concentrations that any change is negligible. Hence the ratio remains relatively same.

[@zuhaibulhaq9572]

isnt it supposed to be 95 cm3?

[@sakeenasaleem231]

calculation of buffer solutions 34:30

[@fakhirrehan8869]

shukriya sakeena baji

[@kingofnothing7]

2:56

[@kingofnothing7]

Graphs starr 5:02

[@kingofnothing7]

11:14

[@kingofnothing7]

11:28

[@kingofnothing7]

19:13

[@kingofnothing7]

24:23

[@kasibfarhan5633]

Only an hour

[@MissNatalieChemistry]

I apologise! Use the timestamps and break it down to smaller parts rather than watching the whole thing.

[@modix699]

Well maybe you should be grateful for her awesome content instead of bitching about it

[@masumarahman2613]

@@MissNatalieChemistry You don't need to apologise for this though xD Infact, we should be super grateful that you actually took time to make detailed videos + these topics anyway come across as tough at the start! A HUGE help to self-studying kids so thankyou so muchh

[@fakhirrehan8869]

tu pagal hai

[@zebamaryam]

this is so useful! thank you! I just had a question, in 46:40 isn't the n(acid) = 0.1 - 0.05 and not 0.1 - 0.005?