
Which of the following will not be soluble in sodium hydrogen carbonate ?
a.) 2, 4, 6 -trinitrophenol
b.) Benzoic acid
c.) o- Nitrophenol
d.) Benzene sulphonic acid
Answer
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Hint: The sodium hydrogen carbonate is formed from ${H_2}C{O_3}$ where one hydrogen atom is replaced by sodium. The acids that are stronger than ${H_2}C{O_3}$ will react with sodium hydrogen carbonate to give sodium salts which are water soluble along with evolution of carbon dioxide. The molecules from above which are stronger than ${H_2}C{O_3}$ will give soluble products.
Complete answer:
We know that the acids that are stronger than ${H_2}C{O_3}$ react with sodium bicarbonate to give carbon dioxide. Such acids are soluble in sodium hydrogen carbonate.
Now, let us see one by one the options given to us.
The first option given is 2, 4, 6 -trinitrophenol. This is a very strong acid due to the presence of three nitro groups. It is even stronger than ${H_2}C{O_3}$. So, it reacts with sodium hydrogen carbonate to give carbon dioxide and is thus, soluble in sodium hydrogen carbonate. So, this is the correct answer.
The second option given is Benzoic acid. The Benzoic acid is also stronger than ${H_2}C{O_3}$. It reacts with sodium hydrogen carbonate to form water soluble sodium salt and evolve carbon dioxide. So, even option b.) is the correct answer.
The third molecule given is o- Nitrophenol. The o- Nitrophenol is less acidic than ${H_2}C{O_3}$. So, it does not form a water soluble product. This option is incorrect.
The fourth molecule given is Benzene sulphonic acid. The benzene sulphonic acid is also stronger than ${H_2}C{O_3}$. It reacts with sodium hydrogen carbonate to form water soluble sodium salt and evolve carbon dioxide. So, even the option ds.) is the correct answer.
Thus, the correct answer is the option a.), b.) and d.).
Note:
The o- Nitrophenol is less acidic than ${H_2}C{O_3}$. This is because in o- Nitrophenol, the oxygen atom of nitro group forms hydrogen bonding with hydrogen of phenol group. As a result, the hydrogen atom phenol can not be donated and thus, the molecule is less acidic in nature.
Complete answer:
We know that the acids that are stronger than ${H_2}C{O_3}$ react with sodium bicarbonate to give carbon dioxide. Such acids are soluble in sodium hydrogen carbonate.
Now, let us see one by one the options given to us.
The first option given is 2, 4, 6 -trinitrophenol. This is a very strong acid due to the presence of three nitro groups. It is even stronger than ${H_2}C{O_3}$. So, it reacts with sodium hydrogen carbonate to give carbon dioxide and is thus, soluble in sodium hydrogen carbonate. So, this is the correct answer.
The second option given is Benzoic acid. The Benzoic acid is also stronger than ${H_2}C{O_3}$. It reacts with sodium hydrogen carbonate to form water soluble sodium salt and evolve carbon dioxide. So, even option b.) is the correct answer.
The third molecule given is o- Nitrophenol. The o- Nitrophenol is less acidic than ${H_2}C{O_3}$. So, it does not form a water soluble product. This option is incorrect.
The fourth molecule given is Benzene sulphonic acid. The benzene sulphonic acid is also stronger than ${H_2}C{O_3}$. It reacts with sodium hydrogen carbonate to form water soluble sodium salt and evolve carbon dioxide. So, even the option ds.) is the correct answer.
Thus, the correct answer is the option a.), b.) and d.).
Note:
The o- Nitrophenol is less acidic than ${H_2}C{O_3}$. This is because in o- Nitrophenol, the oxygen atom of nitro group forms hydrogen bonding with hydrogen of phenol group. As a result, the hydrogen atom phenol can not be donated and thus, the molecule is less acidic in nature.
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