Problem 16
Question
When \(\mathrm{K}_{2} \mathrm{O}\) is added to water, the solution is basic because it contains a significant concentration of (a) \(\mathrm{OH}^{-}\) (b) \(\mathrm{K}^{+}\) (c) \(\mathrm{O}^{3-}\) (d) \(\mathrm{O}_{2}^{2-}\)
Step-by-Step Solution
Verified Answer
The solution is basic because it contains \\(\text{OH}^-\\).
1Step 1: Understand the Reaction
When potassium oxide (\(\text{K}_2\text{O}\)) is added to water, it reacts to form a solution. We need to write the chemical equation for this reaction to determine which ions are produced.
2Step 2: Write the Chemical Equation
The chemical reaction when \(\text{K}_2\text{O}\) is added to water is:\[ \text{K}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{K}^+ + 2\text{OH}^- \]This shows that potassium ions (\(\text{K}^+\)) and hydroxide ions (\(\text{OH}^-\)) are formed in the solution.
3Step 3: Identify the Ion Responsible for Basicity
Now, observe the ions formed. A solution is basic if there is a significant concentration of hydroxide ions \(\text{OH}^-\). Here, \(\text{OH}^-\) ions are produced in the reaction, making the solution basic.
Key Concepts
Potassium Oxide Reaction with WaterChemical Reactions with WaterHydroxide Ions and Basicity
Potassium Oxide Reaction with Water
When potassium oxide, known chemically as \(\text{K}_2\text{O}\), is introduced to water, an interesting chemical transformation occurs. This oxide of potassium dissolves and reacts with water to produce potassium ions \(\text{K}^+\) and hydroxide ions \(\text{OH}^-\). The balanced chemical equation representing this reaction is written as: \[ \text{K}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{K}^+ + 2\text{OH}^- \] This equation tells us about the products of the reaction and emphasizes the generation of hydroxide ions, which is crucial for understanding the resulting properties of the solution. Potassium oxide is known as a basic oxide, meaning it generally reacts with water to form a basic solution. This is largely due to the presence of those generated \(\text{OH}^-\) ions.
Chemical Reactions with Water
Chemical reactions involving water are fundamental and form the basis of many processes both in nature and in industrial applications. When a compound reacts with water, it can either donate or accept protons, or it can dissociate into ions. The reaction includes several important types:
- Acidic oxides, which produce H\(^+\) ions in water, making an acidic solution.
- Basic oxides, like \(\text{K}_2\text{O}\), that release \(\text{OH}^-\) ions in water, making the solution basic.
- Amphoteric oxides that can behave as either acids or bases depending on the reacting substances.
Hydroxide Ions and Basicity
The concept of basicity is intertwined closely with the presence of hydroxide ions in a solution. A basic solution is characterized by its ability to accept protons, which is manifested through the generation of \(\text{OH}^-\). In water, hydroxide ions increase the pH of the solution, thereby making it basic or alkaline. The link between \(\text{OH}^-\) ions and basicity is evident in reactions like those where substances like potassium oxide dissolve in water.
- These hydroxide ions are crucial for many chemical processes, such as neutralization reactions.
- In industrial and laboratory applications, they are responsible for alkalinity, assisting in cleaning agents and chemical synthesis.
Other exercises in this chapter
Problem 13
Which of the following does not precipitate in the Solvay process, for the manufacture of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (a) \(\mathrm{CO}_{2}\) (b) \(\m
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Excess of \(\mathrm{Na}^{+}\)ions in human system causes (a) anaemia (b) high blood pressure (c) diabetes (d) low blood pressure
View solution Problem 17
Crystalline sodium chloride is a bad conductor of electricity because (a) it contains free molecules (b) the ions present in it are not free to move (c) it does
View solution Problem 19
When sodium is treated with sufficient oxygen/air, the product obtained is (a) \(\mathrm{NaO}_{2}\) (b) \(\mathrm{NaO}\) (c) \(\mathrm{Na}_{2} \mathrm{O}\) (d)
View solution