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If \(\mathrm{pKa}=-\log \mathrm{K}=4\) and \(\mathrm{K}=\mathrm{Cx}^{2}\) then van't Hoff factor weak monobasic acid when \(\mathrm{C}=0.01 \mathrm{M}\) is (a) \(1.02\) (b) \(1.01\) (c) \(1.20\) (d) \(1.10\)

Short Answer

Expert verified
The correct van't Hoff factor is (d) 1.10.

Step by step solution

01

Understand the Relationship

We know the relationship between pKa and Ka is given as \( \mathrm{pKa} = -\log \mathrm{K} \). Thus, \( \mathrm{K} = K_a \) for an acid that dissociates as \( A^- \rightleftarrows H^+ + A^- \). Given \( \mathrm{pKa} = 4 \), then \( K_a = 10^{-4} \).
02

Relate Concentration to Dissociation

The dissociation of the weak acid is described by \( \mathrm{K} = \mathrm{C}x^2 \), where \( x \) is the degree of dissociation. Given \( \mathrm{K_a} = 10^{-4} \) and \( \mathrm{C} = 0.01 \), substitute these into the equation to find \( x \): \( 10^{-4} = 0.01x^2 \).
03

Solve for x (Degree of Dissociation)

Rearrange the equation \( 10^{-4} = 0.01x^2 \) to solve for \( x^2 = \frac{10^{-4}}{0.01} \). Simplify to get \( x^2 = 10^{-2} \). Thus, \( x = \sqrt{10^{-2}} = 0.1 \).
04

Calculate the van't Hoff Factor i

The van't Hoff factor \( i \) for a weak acid can be calculated as \( i = 1 + x \), since ions are produced in the form of \( H^+ \) and \( A^- \). Substitute \( x = 0.1 \) into \( i = 1+0.1 \), giving \( i = 1.1 \).
05

Conclusion and Correct Answer

The calculated van't Hoff factor is \( i = 1.1 \). Therefore, the correct answer is option (d) \( 1.10 \).

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Key Concepts

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

Understanding Weak Monobasic Acids
A weak monobasic acid is an acid that can donate only one proton (\( H^+ \)), meaning it has only one ionizable hydrogen atom. These types of acids do not completely dissociate in solution. When placed in water, some of the acid molecules stay intact, while others dissociate into ions.This incomplete dissociation distinguishes weak acids from strong ones, making it important to calculate their dissociation behavior accurately. Weak acids, like acetic acid (\( ext{CH}_3 ext{COOH} \)), typically have equilibria that lie toward the undissociated form. In practice, this means their solutions have fewer hydrogen ions compared to stronger acids, which directly affects properties like pH and conductance.
Understanding pKa Calculation
The pKa value of an acid is a numerical representation of its acid strength. It is calculated as the negative logarithm of the acid dissociation constant (\( K_a \)). Mathematically, it is represented by the equation: \[\mathrm{pKa} = -\log K_a\]A lower pKa value indicates a stronger acid because it relates to a higher concentration of dissociated ions at equilibrium. In our exercise, the given pKa is 4, translating to a \( K_a \) of \( 10^{-4} \).The concept of pKa is crucial in comparing the strength of different acids and predicting their behavior in chemical reactions.For instance, with a given pKa, one can assess how effective an acid will be in transferring its proton to water or any other base in solution.
Exploring the Degree of Dissociation
The degree of dissociation represents the fraction of a weak acid that dissociates into ions in a solution. It is often denoted by the symbol \( x \). In the context of this exercise, it is calculated using the relation \( K = Cx^2 \), where \( C \) is the concentration of the acid.Once you know the degree of dissociation, you can understand how much of the acid contributes to forming the ion pairs in the solution. The stronger the acid, the higher \( x \) tends to be, because more molecules are dissociating.In detailed calculations, you'd use \( K_a = Cx^2 \) to find \( x \) and thereby determine how much of the acid is present as ions, directly influencing properties like pH and osmotic pressure.
Role of Concentration of Acid
The concentration of the acid in solution, denoted as \( C \), plays a pivotal role in determining the extent of its dissociation. in a dilute solution, even a weak acid may show a relatively high degree of dissociation because the equilibrium is shifted towards ion formation. In the exercise, the given concentration, \( C = 0.01 \text{ M} \), is used in conjunction with the dissociation equation, allowing the calculation of the degree of dissociation, \( x \).The concentration of an acid can alter the pH of the solution drastically. the knowledge of both the concentration and the dissociation constant \( K_a \) allows for the calculation of the practical behavior of the solution.Ultimately, understanding how concentration affects dissociation and, therefore, traits like pH and conductivity is essential for predicting the behavior of acids in practical scenarios.

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Most popular questions from this chapter

When \(\mathrm{NaNO}_{3}(\mathrm{~d}=2.0 \mathrm{~g} / \mathrm{cc})\) is heated in a closed vessel of \(100 \mathrm{ml}\), oxygen is liberated and \(\mathrm{NaNO}_{2}\) \((\mathrm{d}=1.5 \mathrm{~g} / \mathrm{cc})\) is left behind as per the reaction \(2 \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NaNO}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) .\) At equilibrium the volumes of NaNO, left and NaNO \(_{3}\) left and NaNO, produced are very small and can be neglected. Which of the following is a correct statement about this equilibrium? (a) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) favours reverse reaction. (b) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{2}\) favours forward reaction. (c) Increasing temperature favours reverse reaction. (d) None of these.

The equilibrium constant \(\mathrm{K}_{\mathrm{P}_{1}}\) and \(\mathrm{K}_{\mathrm{p}}\) for the reactions, \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g})\) and \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\) respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(\mathrm{X}\) and \(\mathrm{Z}\) be equal then the ratio of total pressures at these equilibria is: (a) \(1: 36\) (b) \(1: 9\) (c) \(1: 3\) (d) \(1: 1\)

At \(25^{\circ} \mathrm{C}\), the standard emf of a cell having reaction involving two electron exchange is found to be \(0.295 \mathrm{~V} .\) The equilibrium constant of the reaction is approximately (a) \(9.50 \times 10^{9}\) (b) \(1 \times 10^{10}\) (c) 10 (d) \(9.51 \times 10^{\top}\)

In which of the following reactions, equilibrium is independent of pressure? (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons{\rightleftharpoons} \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)

In which of the following reactions, the concentration of reactant is equal to concentration of product at equilibrium \((\mathrm{K}=\) equilibrium constant \() ?\) (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.01\) (b) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=1\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=10\) (d) \(L \rightleftharpoons \quad J ;=0.025\)

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