Indicators for Acids Titration

[Dhamnekar Win,odd]

 

How to choose indiacator here for these above mentioned Acids titation?

 Is the below-given  chemical compound structure of 4- nitrophenol same as given in the above question?

 

Edited April 27, 2022 by Dhamnekar Win,odd

[exchemist]

11 minutes ago, Dhamnekar Win,odd said:

 

How to choose indiacator here for these above mentioned Acids titation?

 Is the below-given  chemical compound structure of 4- nitrophenol same as given in the above question?

 

What do you think about the relative tendency to release H+, for these 3 molecules? 

[chenbeier]

45 minutes ago, Dhamnekar Win,odd said:

 

How to choose indiacator here for these above mentioned Acids titation?

 Is the below-given  chemical compound structure of 4- nitrophenol same as given in the above question?

 

Yes it is

[exchemist]

6 hours ago, Dhamnekar Win,odd said:

 

How to choose indiacator here for these above mentioned Acids titation?

 Is the below-given  chemical compound structure of 4- nitrophenol same as given in the above question?

 

The structure you have drawn for 4-nitrophenol is not quite right. You can't show 5 full bonds from nitrogen, as the n=2 valence shell has only 4 orbitals in it (s + 3x p.) 

It is often shown as as a resonance hybrid of a pair of structures with a double bond to one O atom and a single dative bond to the other. This gives the 2 O atoms a partial net -ve charge and the N atom a +ve charge, which is what gives the nitro group its electron-withdrawing properties on the benzene ring.  

Sometimes it is drawn as a mix of the 2 structures, with a single solid line to each O atom plus a dotted line to each, giving a bond order of 1 1/2 to each, for example in this Wiki article: https://en.wikipedia.org/wiki/Nitro_compound. The polarity of the nitro group, resulting from the dative bond, is crucial to understanding its influence. 

    

[Dhamnekar Win,odd]

 The following question is related to original question to some extent.  Hence I am asking here.☺️

What is the explanation for the following acid-base behavior of Methyl Orange, an Indicator?

 

 I am working on the original question. Any Chemistry help will be accepted.

   

[chenbeier]

The drawing on the right molecule is wrong. We have here mesomerie behaviour, the aromatic System is broken  in the left Ring.

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Methylorange_Indicator.svg/480px-Methylorange_Indicator.svg.png

Edited April 28, 2022 by chenbeier

[Dhamnekar Win,odd]

1)In the titration of 0.010 M HNO₃  (Nitric acid), the reaction equation is NaOH + HNO₃  → NaNO₃ + H₂O. At the equivalence point, [NaNO₃] = 0.005 M. At the equivalence point, equal volume of  hydroxide solution have been  added to the acid solution and all  of the hydronium ions resulting from HNO₃ have  been  converted to water. The solution at this point is simply an  aqueous solution of sodium nitrate.  Na⁺  + OH^-  + H₃O⁺  + (NO₃)^-   ⇌ 2H₂O + Na⁺  + (NO₃)^-.

Since neither sodium ions nor nitrate ions reacts with water, the solution is neither acidic nor basic; that is [H₃O⁺] = [OH^-] = 10^-7 M. pH at the equivalence point is therefore 7.0  Indicator such as Phenol red (C₁₉H₁₄O₅S) must be used which can respond to a pH of 7, the exact pH at the equivalence point. 

2) In the titration of 0.010 M of ClCH₂COOH (Chloroacetic acid), the reaction equation is ClCH₂CO₂^-  + H₂O  →  ClCH₂CO₂H  + OH^- 

x = [OH^-] = [ClCH₂CO₂H] ;  0.0050 - x = [ ClCH₂COO^-];

K_b = [math] \frac {x^2}{0.0050 - x}= \frac {K_w}{K_a}= \frac {10^{-14}}{1.4 \times 10^{-3}}= 7.1 \times 10 ^{-12} [/math]

x= 1.9 × 10^-7 M = [OH^-]; pOH = -log(1.9 × 10^-7 ) =6.72 ; pH =7.28.

Indicator such as Phenol red (C₁₉H₁₄O₅S) must be used which can respond to a pH of 7.28, the exact pH at the equivalence point.

 

3)  In the titration of 0.010 M of O₂NC₆H₄OH (o-nitrophenol), the reaction equation is O₂NC₆H₄O^- + H₂O → O₂NC₆H₄OH + OH^-.

x= [OH^-] = [ O₂NC₆H₄OH] and 0.0050 - x = [O₂NC₆H₄O^-];

K_b = [math] \frac{x^2}{0.0050 - x}= \frac{K_w}{K_a}= \frac{10^{-14}}{6.8 \times 10^{-8}} = 1.4 \times 10^{-7} [/math]

x = 2.6 × 10^-5  M ; pOH = -log (2.6 × 10^-5 ) = 4.58; pH = 9.42

Indiactor such as Phenolphthalein must be used  

Edited April 29, 2022 by Dhamnekar Win,odd