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  #1  
7th May 2015, 12:24 PM
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Chemical Equilibrium JEE Questions

I have given Exam of 12th in PCM and now I am going to give Joint Entrance Exam (JEE) Main 2015 . I can solve all the Questions of Chemistry but I am weak in Chemical Equilibrium . Would you please provide me a book that publish only JEE Questions of Chemical Equilibrium so I will get confidence after solving them ?
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  #2  
4th September 2018, 03:54 PM
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Re: Chemical Equilibrium JEE Questions

Hi buddy here I am looking for Chemical Equilibrium Questions for JEE exam preparation so would you plz let me know from where I can do download it ??
  #3  
4th September 2018, 03:55 PM
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Join Date: Aug 2012
Re: Chemical Equilibrium JEE Questions

As you are asking for Chemical Equilibrium Questions for JEE e-am preparation so on your demand I am providing same here :

Question 1:

Calculate the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mi-ed, [Ka (CH3COOH)=10-5]
Solution:

CH3 COOH \rightleftharpoons CH3 COO_ + H+ (I)

NaOH → Na+ + OH-

H+ + OH_ \rightleftharpoonsH2O (II)

(I) + (II)

CH3COOH + OH- CH3COO- + H2O . (III)

0.05-- 0.05-- -

Keq of eq. (III) = Ka/Kw

conc. of H2O remain constant

109 = -/(0.05--)2

because value of eq. Const.is very high

here for 0.05

let 0.05-=a

109=0.05/a2

a = 7.07\times10-6

pOH= 6-log 7.07

pOH= 6 0.85

pH= 14-6+0.85 = 8.85

__________________________________________________ ______________________________________
Question 2:

Calculate the pH at the equivalence point of the titration between 0.1M CH3*COOH ( 25 ml) with 0.05 M NaOH. Ka (CH3COOH) = 1.8 \times 105.
Solution:

We have already seen that even though when CH3COOH is titrated with NaOH the reaction does not go to completion but instead reaches equilibrium. We can assume that the reaction is complete and then salt gets hydrolysed because, this assumption will help us to do the problem easily and it does not effect our answer.

[H^+]=\sqrt{\frac{K_wK_a}{C}}

First of all we would calculate the concentration of the salt, CH3COONa. For reaching equivalence point,

N1V1 = N2V2

0.1 25 = 0.05 V2

\Rightarrow V2 = 50 ml

Therefore [CH3COONa] = (0.1\times25)/75 =0.1/3

[H+] = \sqrt{\frac{10^-^14\times 1.8\times 10^-^5}{0.1/3}}

\Rightarrow[H+] = 2.32
Attached Files
File Type: doc Chemical Equilibrium JEE Questions.doc (56.0 KB, 87 views)

Last edited by pawan; 4th September 2018 at 03:58 PM.


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