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Read the following statement and explanation and answer as per the options given below : Assertion : Sulphate is estimated as \(\mathrm{BaSO}_{4}\) and not as \(\mathrm{MgSO}_{4}\). Reason : Ionic radius of \(\mathrm{Mg}^{2+}\) is smaller than that of \(\mathrm{Ba}^{2+}\). (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct.

Short Answer

Expert verified
(b) Both assertion and reason are correct, but reason is not the correct explanation of the assertion.

Step by step solution

01

Analyzing the Assertion

The assertion states that sulfate is estimated as \( \text{BaSO}_4 \) and not as \( \text{MgSO}_4 \). This is true because \( \text{BaSO}_4 \) is an insoluble compound, commonly used in laboratory techniques to precipitate sulfate ions for quantitative analysis. \( \text{MgSO}_4 \), on the other hand, is soluble in water and cannot be used for this purpose.
02

Analyzing the Reason

The reason given is that the ionic radius of \( \text{Mg}^{2+} \) is smaller than that of \( \text{Ba}^{2+} \). This is correct; \( \text{Mg}^{2+} \) indeed has a smaller ionic radius compared to \( \text{Ba}^{2+} \), due to its position in the periodic table.
03

Evaluating the Explanation

While both the assertion and reason are correct statements, the size difference in ionic radii is unrelated to why sulfate is estimated using \( \text{BaSO}_4 \) rather than \( \text{MgSO}_4 \). The real reason has to do with their solubility characteristics. Therefore, the reason is not the correct explanation for the assertion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulphate Estimation
In chemistry, accurately determining the amount of sulphate present in a sample is crucial for various analyses. To achieve this, chemists often rely on a method called sulphate estimation. This involves precipitating sulphate ions from a solution in the form of an insoluble salt. Among many possible compounds, barium sulphate (\(\text{BaSO}_4\)) is commonly used for this purpose. This is due to its low solubility in water, which ensures a substantial amount of sulphate can be removed from the solution in solid form.
  • Barium sulphate doesn't dissolve in water easily.
  • The sulphate ion (\(\text{SO}_4^{2-}\)) binds effectively with the barium ion (\(\text{Ba}^{2+}\)).
Once precipitated, the barium sulphate can be filtered out and weighed to determine the sulphate content in the original sample. This process is not only efficient but also reliable for analytical chemistry purposes.
Ionic Radius
The ionic radius is a term used to describe the size of an ion. Ionic size is crucial because it affects how ions interact with one another and with molecules around them. In the periodic table, ionic radii typically decrease across a period from left to right, and increase down a group.
The ionic radius plays an essential role in determining the chemical and physical properties of compounds. For instance, in the exercise above, while the assertion was related to sulphate estimation, the reason mentioned the ionic radius of \(\text{Mg}^{2+}\) and \(\text{Ba}^{2+}\).
  • \(\text{Mg}^{2+}\) has a smaller ionic radius compared to \(\text{Ba}^{2+}\) due to its position higher up in the periodic table.
  • This smaller size can affect the solubility of its compounds, but not necessarily their use in sulphate estimation.
Solubility of Compounds
Solubility refers to the ability of a substance to dissolve in a solvent, like water. It significantly impacts how compounds are used in chemical reactions and analyses. Understanding solubility is vital for predicting and controlling the behavior of chemical species in different environments.
For the task of estimating sulphates, the solubility characteristics of the compounds are key. Compounds like barium sulphate (\(\text{BaSO}_4\)) are preferred for sulphate precipitation because they are quite insoluble in water.
  • Very low solubility means barium sulphate stays as a solid rather than dissolving.
  • This allows accurate measurement of the sulphate ions in a solution because the precipitation process is complete and consistent.
In contrast, magnesium sulphate (\(\text{MgSO}_4\)) is highly soluble in water. This solubility makes it less useful for direct sulphate estimation in a lab setting since it does not precipitate out of a solution easily.

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Most popular questions from this chapter

A white solid is either \(\mathrm{Na}_{2} \mathrm{O}\) or \(\mathrm{Na}_{2} \mathrm{O}_{2}\). A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid. (i) Identify the substance and explain with balanced equation. (ii) Explain what would happen to the red litmus if the white solid were the other compound.

\([\mathrm{X}]+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow[\mathrm{Y}]\) a colourless gas with irritating smell, \([\mathrm{Y}]+\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\) green solution. \([\mathrm{X}]\) and \([\mathrm{Y}]\) are: (a) \(\mathrm{SO}_{3}^{2-}, \mathrm{SO}_{2}\) (b) \(\mathrm{Cl}^{-}, \mathrm{HCl}\) (c) \(\mathrm{S}^{2-}, \mathrm{H}_{2} \mathrm{~S}\) (d) \(\mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}\)

With reference to aqua regia, choose the correct option(s) (a) Reaction of gold with aqua regia produces \(\mathrm{NO}_{2}\) in the absence of air (b) Reaction of gold with aqua regia produces an anion having Au in \(+3\) oxidation state (c) Aqua regia is prepared by mixing conc. HCl and conc. \(\mathrm{HNO}_{3}\) in \(3: 1\) \((v / v)\) ratio (d) The yellow colour of aqua regia is due to the presence of \(\mathrm{NOCl}\) and \(\mathrm{Cl}_{2}\)

The correct option(s) to distinguish nitrate salts of \(\mathrm{Mn}^{2+}\) and \(\mathrm{Cu}^{2+}\) taken separately is (are) (a) \(\mathrm{Mn}^{2+}\) shows the characteristic green colour in the flame test (b) Only \(\mathrm{Cu}^{2+}\) shows the formation of precipitate by passing \(\mathrm{H}_{2} \mathrm{~S}\) in acidic medium (c) Only \(\mathrm{Mn}^{2+}\) shows the formation of precipitate by passing \(\mathrm{H}_{2} \mathrm{~S}\) in faintly basic medium (d) \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) has higher reduction potential than \(\mathrm{Mn}^{2+} / \mathrm{Mn}\) (measured under similar conditions)

A light bluish green crystalline compound responds to the following tests : (i) Its aqueous solution gives a brown precipitate or colour with alkaline \(\mathrm{K}_{2}\left[\mathrm{HgI}_{4}\right]\) solution. (ii) Its aqueous solution gives a blue colour with \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) solution. (iii) Its solution in hydrochloric acid gives a white precipitate with \(\mathrm{BaCl}_{2}\) solution. Identify the ions present and suggest the formula of the compound.

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