For the reaction of weak acid:

The Henderson-hasselbalch equation is:
![pH = pk_a+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_a%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20)
The concentration ratio at
is:
![pH = pk_1+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_1%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)
Substituting the values:
![3.2 = 4.1 + log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%203.2%20%3D%204.1%20%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)
![log \frac{[A^{-}]}{[HA]} = - 0.9](https://tex.z-dn.net/?f=%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20%3D%20-%200.9%20%20)
![\frac{[A^{-}]}{[HA]} = 0.126](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20%3D%200.126%20%20)
Now, for calculating the pk_2 value:
![pH = pk_2+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_2%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)


Thus,
.
When orthobromoanisole is treated with sodamide (NaNH₂) and ammonia (NH₃), benzyne is formed and the tripple bond is formed between C₂-C₃ carbon atoms of benzene ring. Then NH₂⁻ can attack at ortho position as well as meta position.
The reaction takes place in three steps. Step-1:The amide NH₂⁻ ion attacks the H
-atom that is ortho to C3
, generating a carbanion. Step-2:Loss of Br
- to form a benzyne intermediate.Step-3:Addition of NH₂⁻ ion. The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition of NH₂⁻.
Two isomeric products are formed: (1) <em>o</em>-methoxyaniline and (2) <em>m</em>-methoxyaniline. The structures of both the isomeric products are given in diagram.
The question is incomplete, the complete question is shown in the image attached to this answer.
Answer:
139.13
Explanation:
The average atomic mass of the element Likhitium is the sum of the relative abundance of all the isotopes of Likhitium.
We obtain the relative atomic mass of Likhitium as follows;
(44.7/100 * 138) + (52.3/100 * 139) + (0.5/100 * 140) + (2.5/100 * 141)
61.7 + 73.2 + 0.7 + 3.53 = 139.13
Hence the relative abundance of Likhitium is 139.13