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\(\mathrm{CH}_{3} \mathrm{COOH}\) is weaker acid than \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) It is due to: (a) More ionization (b) Less ionization (c) Covalent bond (d) Electrovalent bond

Short Answer

Expert verified
Less ionization (b).

Step by step solution

01

Understanding Ionization in Acids

Acids ionize in water to release hydrogen ions (H\(^+\)). Strong acids ionize completely, while weak acids only partially ionize.
02

Compare Ionization of CH3COOH and H2SO4

\(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid) is a weak acid that partially ionizes in solution, whereas \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) is a strong acid that ionizes completely in its first step and partially in the second.
03

Identifying the Correct Explanation

Since \(\mathrm{CH}_{3} \mathrm{COOH}\) is a weaker acid than \(\mathrm{H}_{2} \mathrm{SO}_{4}\), it must ionize less. Therefore, the correct answer is based on its partial ionization compared to \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

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Key Concepts

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

Weak Acids
Let's dive into the world of weak acids! A weak acid is an acid that doesn't fully ionize in solution.

This means they only release a small fraction of their hydrogen ions, leaving the rest of the acid molecules intact.
  • Example: Acetic acid (\( \mathrm{CH}_{3}\mathrm{COOH} \)) is a common weak acid. It partially ionizes to form \(\mathrm{CH}_{3}\mathrm{COO}^-\) and \( \mathrm{H}^+\) ions.
  • The degree of ionization is often less than 5% for most typical weak acids.
Weak acids have a higher pH than strong acids for the same concentration due to less \( \mathrm{H}^+ \) ion formation. This characteristic of weak acids makes them desirable for applications where less aggressive acidification is needed, such as in food preservation or in biological systems.
Strong Acids
Strong acids, in contrast to weak acids, ionize completely in solution. This means they release all their hydrogen ions, leaving no un-ionized acid molecules behind.

Complete ionization leads to strong acids having very low pH values when dissolved in water.
  • Example: Sulfuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4} \)) is a strong acid. Upon dissolving in water, it releases \( \mathrm{H}^+ \) ions fully in the first ionization step.
  • The high concentration of hydrogen ions makes strong acids highly reactive.
These properties make strong acids useful in various industrial processes, such as the manufacturing of fertilizers and explosives. Their ability to completely ionize makes them more aggressive than their weak counterparts.
Ionization in Acids
Ionization is a crucial process in understanding acid strength. When an acid dissolves in water, it breaks down to release hydrogen ions (\( \mathrm{H}^+ \).

The extent to which this happens differentiates weak from strong acids.
  • Complete ionization means every acidic hydrogen ion is released from the acid molecule, a hallmark of strong acids.
  • Partial ionization, characteristic of weak acids, means only some hydrogen ions are released, and many acid molecules remain whole.
The degree of ionization is measured through the acid dissociation constant (\( K_a \)). Larger \( K_a \) values indicate stronger acids since they signify more ionization. Understanding this ionization process helps us predict and explain the behavior of acids in various solutions and reactions.

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Most popular questions from this chapter

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

For the reaction, \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) the equilibrium concentration of \(\mathrm{H}_{2}, \mathrm{I}_{2}\) and \(\mathrm{HI}\) are \(8.0,3.0\) and \(28.0\) mole/litre, respectively, the equilibrium constant is: (a) \(28.34\) (b) \(32.66\) (c) \(34.78\) (d) \(38.88\)

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, 1.10 molesofNOClremained. Calculatetheequilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}):\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

For the reaction, \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the rate constants for the forward and backward reactions are found to be \(4.2 \times 10^{-2}\) and \(3.36 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. What is the equilibrium constant for the reaction: (a) \(11.5\) (b) \(12.5\) (c) \(8.0\) (d) \(6.0\)

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