Answer
425.1k+ views
Hint:
For making we take solid salt or mixture and solid $N{{a}_{2}}C{{O}_{3}}$ in a fixed ratio while preparing Sodium Carbonate extract of the mixture. As we use it in our practical work to make Sodium Carbonate extract of the mixture for wet test of acid radical. Because the heavy metal ions interfere in the test of acid radical.
Complete step by step answer:
Let us see how we prepare Sodium carbonate extract:
Take solid salt or mixture and $N{{a}_{2}}C{{O}_{3}}$ in the $3:1$ ratio in a hard glass boiling tube .Add to it about $20-30ml$distilled water and boil for $10-15C{{O}_{2}} i.e.$ $10-15$ minutes ,cool and filter the solution in a clean boiling tube .The filtrate is sodium carbonate extract (SCE).
When the mixtures are boiled with a strong solution of $N{{a}_{2}}C{{O}_{3}}$, double decomposition takes place.
It results in the formation of the carbonates of heavy metals radicals and sodium salts of acid radicals.
The sodium salts of the corresponding acid radical are soluble in water and thus passes into the solution. Sodium carbonate extract always contains unreacted sodium carbonate which has to be destroyed before using the extract to various wet tests. To do this the extract is neutralised by some suitable acids and is boiled to expect $C{{O}_{2}}$. The selection of the acid depends upon the radical to be identified. In making Sodium Carbonate extract we cannot use sodium bicarbonate as it is formed by double decomposition while remaining in the solution which we don’t want it to happen in that way.
Sodium Carbonate and mixture are taken in $3:1$ ratio while preparing Sodium carbonate extract.
Hence , option (A) is correct.
Note:
The sodium carbonate extract always contains unreacted sodium carbonate. Therefore the test for carbonate is not performed from sodium carbonate extract. And Use distilled water for the preparation of sodium carbonate extract. Sodium carbonate should be of analytical grade $i.e.$free from impurities.
For making we take solid salt or mixture and solid $N{{a}_{2}}C{{O}_{3}}$ in a fixed ratio while preparing Sodium Carbonate extract of the mixture. As we use it in our practical work to make Sodium Carbonate extract of the mixture for wet test of acid radical. Because the heavy metal ions interfere in the test of acid radical.
Complete step by step answer:
Let us see how we prepare Sodium carbonate extract:
Take solid salt or mixture and $N{{a}_{2}}C{{O}_{3}}$ in the $3:1$ ratio in a hard glass boiling tube .Add to it about $20-30ml$distilled water and boil for $10-15C{{O}_{2}} i.e.$ $10-15$ minutes ,cool and filter the solution in a clean boiling tube .The filtrate is sodium carbonate extract (SCE).
When the mixtures are boiled with a strong solution of $N{{a}_{2}}C{{O}_{3}}$, double decomposition takes place.
It results in the formation of the carbonates of heavy metals radicals and sodium salts of acid radicals.
The sodium salts of the corresponding acid radical are soluble in water and thus passes into the solution. Sodium carbonate extract always contains unreacted sodium carbonate which has to be destroyed before using the extract to various wet tests. To do this the extract is neutralised by some suitable acids and is boiled to expect $C{{O}_{2}}$. The selection of the acid depends upon the radical to be identified. In making Sodium Carbonate extract we cannot use sodium bicarbonate as it is formed by double decomposition while remaining in the solution which we don’t want it to happen in that way.
Sodium Carbonate and mixture are taken in $3:1$ ratio while preparing Sodium carbonate extract.
Hence , option (A) is correct.
Note:
The sodium carbonate extract always contains unreacted sodium carbonate. Therefore the test for carbonate is not performed from sodium carbonate extract. And Use distilled water for the preparation of sodium carbonate extract. Sodium carbonate should be of analytical grade $i.e.$free from impurities.
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