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Amongst the following hydroxides, the one which has the lowest value of \(\mathrm{K}_{\mathrm{sp}}\) at ordinary temperature (about \(\left.25^{\circ} \mathrm{C}\right)\) is: (a) \(\mathrm{Mg}(\mathrm{OH})_{2}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(\mathrm{Be}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
The hydroxide with the lowest (\(K_{sp}\)) is \(\mathrm{Be(OH)}_2\).

Step by step solution

01

Understanding Solubility Product

The solubility product (\(K_{sp}\)) is a constant that indicates the solubility of a compound under equilibrium conditions. A lower (\(K_{sp}\)) value indicates less solubility in water. Amongst hydroxides, \(K_{sp}\) values can be compared to determine which compound is least soluble.
02

Analyze Given Hydroxides

Look at the compounds to determine their (\(K_{sp}\)) values. Generally, alkaline earth metal hydroxides such as \(\mathrm{Mg(OH)}_2\), \(\mathrm{Ca(OH)}_2\), \(\mathrm{Ba(OH)}_2\), \(\mathrm{Be(OH)}_2\) follow trends in the periodic table where solubility increases down the group.
03

Assess Periodic Trends

Based on periodic trends, the solubility of Group 2 metal hydroxides increases from \(\mathrm{Be(OH)}_2\) to \(\mathrm{Ba(OH)}_2\). This implies that \(\mathrm{Be(OH)}_2\) should have the lowest (\(K_{sp}\)) among the hydroxides listed.
04

Identify Answer Based on Comparisons

Since \(\mathrm{Be(OH)}_2\) is placed before \(\mathrm{Mg(OH)}_2\), \(\mathrm{Ca(OH)}_2\), and \(\mathrm{Ba(OH)}_2\), its solubility product is expected to be less than these other compounds, making it the least soluble in this group, thus having the lowest (\(K_{sp}\)).

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Key Concepts

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

Solubility Trends
Solubility trends help us understand how likely a compound will dissolve in a solvent, usually water. When we talk about solubility product, or \( K_{sp} \), we are referring to the constant that represents this dissolved state at equilibrium. It's like a snapshot of how willing a compound is to split into its ions in water.
Solubility trends can show us which substances dissolve more easily or quickly compared to others. For instance, in a family of compounds or elements, such as group 2 hydroxides, we notice patterns. These patterns tell us that as you move down the group in the periodic table, from top to bottom, hydroxides generally become more soluble in water.
  • A smaller \( K_{sp} \) value indicates a substance is less soluble in water.
  • This is important when trying to figure out which compound will precipitate or dissolve.
  • The solubility product is intrinsic to each compound and is influenced by temperature.
Understanding solubility trends allows chemists to predict behaviors of compounds under various conditions, aiding in reactions, formulations or even in natural environmental processes.
Group 2 Hydroxides
The hydroxides of Group 2 elements are interesting because they show a clear trend in solubility aligned with their position on the periodic table. These elements are known as alkaline earth metals, and include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba).
The hydroxides of these elements, like \( \text{Mg(OH)}_2 \), \( \text{Ca(OH)}_2 \), or \( \text{Ba(OH)}_2 \), show increasing solubility with larger atomic numbers. This means beryllium hydroxide \( \text{Be(OH)}_2 \), falls at the less soluble end of the spectrum, while barium hydroxide \( \text{Ba(OH)}_2 \) is far more soluble.
  • Beryllium hydroxide is very sparingly soluble, making it have the lowest \( K_{sp} \) value.
  • Calcium and barium hydroxides are increasingly soluble, with \( \text{Ba(OH)}_2 \) being so soluble that it can actually make a noticeable concentration in solution.
  • The actual solubility is determined by the balance in their respective hydration energy and lattice energy.
This regular increase in solubility helps us understand and manipulate these compounds in various chemical applications, from industrial processes to environmental sciences.
Periodic Table Trends
The periodic table is a powerful tool for understanding chemical properties and behaviors of elements. For Group 2, the alkaline earth metals, we can see trends that tell us how elements will behave based on their position in the table.

One key trend involves the solubility of metal hydroxides, showing that as we move down the group, from beryllium to barium, the solubility of their hydroxides increases. This is because elements have increasing atomic size and decreasing lattice energies with larger atomic numbers, leading to easier breakdown of solid structures into ions in solution.
  • Atomic size influences how well a metal will pair with hydroxide ions, affecting solubility.
  • Electronegativity decreases down the group, causing weaker bindings with oxygen and easier solubility.
  • Higher atomic numbers often lead to better solubility, impacting the environmental and practical applications of these compounds.
Understanding these trends is crucial for predicting chemical behavior. It aids in everything from academic studies, such as performing experiments, to real-world applications like designing pharmaceuticals or understanding geological processes.

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Most popular questions from this chapter

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

In what manner will increase of pressure affect the following equation: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) Shift in the reverse direction (b) Shift in the forward direction (c) Increase in the yield of hydrogen (d) No effect

Which of the following reaction will be favoured at low pressure: (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\)

If equilibrium constants of reaction: \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) is \(\mathrm{K}_{1}\), and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\mathrm{K}_{2}\) then (a) \(\mathrm{K}_{1}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{1}=2 \mathrm{~K}_{2}\) (c) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}}_{1}\) (d) \(\mathrm{K}_{1}=\frac{1}{2} \mathrm{~K}_{2}\)

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\) The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by: (a) Low temperature and high pressure (b) High temperature and high pressure (c) Low temperature and low pressure (d) High temperature and low pressure

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