Ionic Equilibrium L7 | Questions Of Buffer Solutions | JEE & NEET 2022 | Class 11 | Pahul Sir

[@CatalysisbyVedantu]

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[@zaradhafzar6484]

and also note my name sir....i'm a new student of catalysis.

[@divyansh_kala]

Homework Question Answer: α = 1 * 10⁻⁸ Solution: C6H5ONa = C6H5O⁻ + H3O⁺ 0.05 - 0.05α 0.05α 0.05α Ka= [C6H5O⁻] [H3O⁺] / [C6H5OH] = 1.0 X 10⁻¹⁰ 10⁻¹⁰ = ( 0.05α ) ( 0.01 + 0.05α ) / ( 0.05 - α ) 0.01 + 0.05α ≈ 0.01 & 0.05 - α ≈ 0.05 10⁻¹⁰ = ( 0.05α ) ( 0.01 ) / ( 0.05 ) 10⁻¹⁰ = ( α ) ( 0.01 ) α = 10⁻⁸ Thus the value of alpha as 1x10⁻⁸

[@saimadhavchakrala4430]

Yeah I got the same

[@davidmarczak4502]

Thank you from California!

[@ishithm7636]

Pahul sir....such an awesome teacher... But dont worry about live people...we are watching in night.👍👍

[@zaradhafzar6484]

yes...

[@sesharijyothimiryala7911]

Hats off sir

[@shalinisaxena3983]

Answer Degree of dissociation = 1×10^-8.

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@BhawnaSaharan-ly4sc]

1:00:36 1:00:38 1:00:40 1:00:40 1:00:40 1:00:42 1:00:42

[@kaviprasath2001]

Sir don't feel Becoz offline viewing is more than online class 😘😍🤩🥰

[@krishmacwan9078]

Respected Sir, the answer of H.W. question is 1x10^-8. Love from Gujarat!

[@sesharijyothimiryala7911]

Please tell the process

[@krishmacwan9078]

@@sesharijyothimiryala7911 SURE! K OF PHENOL= 1.0 X 10^-10 C6H50H + H2O = C6H50^- + H30^+ 0.05-X X X Ka=[C6H5O^-][H3O^+]/[C6H5OH] 1X10^-10=X^2/0.05-X (HERE, WE CAN NEGLECT X) FROM SOLVING THE ABOVE EQUATION, WE WILL GET 2.2x10^6 [C6H5O^-]=[H3O^+]=2.2x10^6 NOW, WE HAVE: C6H5ONa = C6H5O^- + Na^+ 0.01 0.01 C6H5ONa + H2O = C6H5O^- + H3O+ 0.05-0.05∝ 0.05∝ 0.05∝ AS ALPHA IS SMALL, WE CAN NEGLECT IT [C6H5OOH]=0.05 [C6H5O^-]=0.01 [H3O+]=0.05∝ Ka= [C6H5O^-][H3O^+]/[C6H5OH] = 0.01x0.05∝/0.05 1x10^-10=0.01x0.05∝/0.05 calculating this, we get value of alpha as 1x10^-8!

[@rhythemaggarwal2274]

Can u plzz clear my one doubt ??? Here it is ....

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@maviya3232]

Solution A and B wale question me sir ne ch3cooh ko poora dissociate krta hua kyu dikhaya

[@yogeshlale1552]

2:47

[@Dhruv-fu2os]

sir you teach really well

[@suhanimehta5766]

No doubt😇😊

[@Mahadevm123]

Absolutely true 👍

[@srirajgullapudi592]

Sir pls try to special question sessions

[@tusharjain7752]

Degree of dissociation is 1 x 10^-8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@abmohit3117]

Brilliant teaching sr👍👍

[@yugrajsingh7157]

Sir I had a doubt sir at first question B part we had (0.1M CH3COOH) sir how all of the acid dissociates because it is a weak base??

[@lavanya3443]

Amazing😄

[@ks._.art_]

Amazing teacher ..........🙏 Thank you sir

[@srm9104]

2022

[@ajithaakkineni7194]

sir answer for HW question is 1*10^(-8)

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@therichillout9837]

Sir dimerization formed

[@poonamkhairwa5443]

sir Alpha = 10^-4

[@rhythemaggarwal2274]

Sir plzz reply In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@HariKrishnan-vp5ed]

phenol is a very weak acid and that's why it gives high ph (low H+ concentration)..but still it loses H+, even though in a very low amount..so it is acidic and btw acidity is a relative term

[@sivoham2845]

1×10^-8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@sibisugan5026]

how pH of solution B is 4.74 in 47:24 ?? as the pH is pKa + log [salt]/[acid] how sir??

[@prathamshrivastava6641]

Bro that is for the buffer solution haven't you had watched previous videos

[@kgagan8639]

Answer to hw question Alpha=1×10^-8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@aviralmishraofficial1626]

By method I told yesterday answer is coming unbelievable, 1.97x10^-2

[@laxmanshinde916]

30.55

[@lavanya3443]

ans-10^-8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@maviya3232]

Solution A and B wale question me sir ne ch3cooh ko poora dissociate krta hua kyu dikhaya?

[@abhiramtippabhotla]

D

[@ishithm7636]

Sir alpha is >

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@vanshtyagiit491]

10^-8

[@kevoncoke1664]

D sir

[@StudyTime-ib9xt]

Sir answer is 13.7

[@samhithbeeravelli9412]

damn that BGM!!!

[@manisha8259]

What's the name??

[@diyasabose2029]

Ans is- 10^-8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@therichillout9837]

244

[@nehamamfanclubvedanthumath9470]

10 to the power of -8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer

[@raihanshaik]

Bhayanak chapter hai

[@StudyTime-ib9xt]

Sir answer is 13.78

[@adithiyechury1605]

Degree of dissociation alpha = 1 * 10^ -8

[@rhythemaggarwal2274]

In the last ques ....the calculation for pH goes like 10 + (log 1 - log 5) Which eventually gives pH to be 9.4 But how's that possible Bcz the buffer is acidic buffer