
Sulphuric acid is not used for preparation of original solution in analysis of basic radicals because?
(A) It is a reducing agent
(B) It forms insoluble sulphates with certain radicals
(C) It forms a soluble complex
(D) It is oxidizing in nature
Answer
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Hint : The chemical formula of Sulphuric acid is $ {H_2}S{O_4} $ . It is a mineral acid of sulphur, oxygen and hydrogen. It is a colorless, odorless compound which is soluble in water. It is also known as oil of vitriol.
Complete step by step answer
Sulphuric acid is a covalent compound with a molecular weight of $ 98g $ . When it is dissolved in water, it releases heat. Being a strong acid it dissociates completely in water into $ {H^ + } $ and $ HSO_4^ - $ ion.
The analysis of basic radicals is an example of qualitative analysis. It is a three-step process. The first step is the formation of an original solution. The second step is the separation of basic radicals into different groups, and the last step is the analysis of the precipitate for each group.
The original solution is a transparent, coloured solution of salt which is prepared by dissolving the salt or mixture in a suitable solvent. It could be cold water, hot water, Conc. $ HCl $ or dilute $ HCl $ .
First a small amount of substance is treated with cold or hot water, if it gets dissolved then fine otherwise it is dissolved in $ HCl $ .
An original solution cannot be prepared by sulphuric acid because of two reasons. One is its oxidizing nature and second is it forms insoluble sulphates on reaction with some basic radicals like $ P{b^{2 + }},C{a^{2 + }},S{r^{2 + }} $ . Hence, it is not used in preparing original solutions.
Therefore, option (B) and (D) are correct.
Note
Basic radicals are those species that are formed upon the loss of an hydroxide ion form the compound i.e. $ O{H^ - } $ . They are positively charged radicals. Every inorganic compound is composed of two parts that are called radicals. The positive charged one is called basic and the negatively charged one is called acidic.
Complete step by step answer
Sulphuric acid is a covalent compound with a molecular weight of $ 98g $ . When it is dissolved in water, it releases heat. Being a strong acid it dissociates completely in water into $ {H^ + } $ and $ HSO_4^ - $ ion.
The analysis of basic radicals is an example of qualitative analysis. It is a three-step process. The first step is the formation of an original solution. The second step is the separation of basic radicals into different groups, and the last step is the analysis of the precipitate for each group.
The original solution is a transparent, coloured solution of salt which is prepared by dissolving the salt or mixture in a suitable solvent. It could be cold water, hot water, Conc. $ HCl $ or dilute $ HCl $ .
First a small amount of substance is treated with cold or hot water, if it gets dissolved then fine otherwise it is dissolved in $ HCl $ .
An original solution cannot be prepared by sulphuric acid because of two reasons. One is its oxidizing nature and second is it forms insoluble sulphates on reaction with some basic radicals like $ P{b^{2 + }},C{a^{2 + }},S{r^{2 + }} $ . Hence, it is not used in preparing original solutions.
Therefore, option (B) and (D) are correct.
Note
Basic radicals are those species that are formed upon the loss of an hydroxide ion form the compound i.e. $ O{H^ - } $ . They are positively charged radicals. Every inorganic compound is composed of two parts that are called radicals. The positive charged one is called basic and the negatively charged one is called acidic.
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