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The ionization constant of acetic acid is \(1.74 \times 10^{-5} .\) Calculate the degree of dissociation of acetic acid in its \(0.05 \mathrm{M}\) solution. Calculate the concentration of acetate ion in the solution and its \(\mathrm{pH}\).

Short Answer

Expert verified
Degree of dissociation: 0.0187; Concentration of acetate: 9.35 脳 10鈦烩伌 M; pH: 3.03.

Step by step solution

01

Write the Chemical Equation

The chemical equation for the dissociation of acetic acid (CH鈧僀OOH) in water is: \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
02

Define the Equilibrium Expression

The ionization constant (\(K_a\)) for acetic acid is given by the expression: \[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]Substitute the given ionization constant, \(K_a = 1.74 \times 10^{-5}\).
03

Set Up the Concentration Changes

Let the degree of dissociation be \(\alpha\). The initial concentration of acetic acid is \([\text{CH}_3\text{COOH}]_0 = 0.05 \text{ M}\). At equilibrium, we have:\[ [\text{CH}_3\text{COOH}] = 0.05(1-\alpha) \]\[ [\text{CH}_3\text{COO}^-] = [\text{H}^+] = 0.05\alpha \]
04

Substitute into the Equilibrium Expression

Insert these expressions into the \(K_a\) equation:\[ 1.74 \times 10^{-5} = \frac{(0.05\alpha)(0.05\alpha)}{0.05(1-\alpha)} \]This simplifies to:\[ 1.74 \times 10^{-5} = \frac{0.0025\alpha^2}{0.05 - 0.05\alpha} \]
05

Simplify and Solve for \(\alpha\)

Assume \(\alpha\) is small, so \(0.05 - 0.05\alpha \approx 0.05\). Therefore:\[ 1.74 \times 10^{-5} = \frac{0.0025\alpha^2}{0.05} \]Simplify to find \(\alpha\):\[ \alpha^2 = \frac{1.74 \times 10^{-5} \times 0.05}{0.0025} = 3.48 \times 10^{-4} \]\[ \alpha = \sqrt{3.48 \times 10^{-4}} \approx 0.0187 \]
06

Calculate Acetate Ion Concentration

The concentration of acetate ions,\([\text{CH}_3\text{COO}^-] = 0.05\alpha\) is:\[ [\text{CH}_3\text{COO}^-] \approx 0.05 \times 0.0187 = 9.35 \times 10^{-4} \text{ M} \]
07

Calculate the pH

Since \([\text{CH}_3\text{COO}^-] = [\text{H}^+]\), the concentration of \(\text{H}^+ \) ions is the same:\[ [\text{H}^+] = 9.35 \times 10^{-4} \text{ M} \]The pH is given by:\[ \text{pH} = -\log([\text{H}^+]) \]\[ \text{pH} = -\log(9.35 \times 10^{-4}) \approx 3.03 \]

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Key Concepts

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

Degree of Dissociation
When acetic acid (CH鈧僀OOH) dissolves in water, a small portion of its molecules dissociate into acetate ions (CH鈧僀OO鈦) and hydrogen ions (H鈦). The degree of dissociation, denoted by \( \alpha \), measures how much of the acid dissociates compared to the initial amount. For weak acids like acetic acid, \( \alpha \) is typically a small value, indicating that not all the acid molecules dissociate. To calculate the degree of dissociation, use:
  • Initial concentration: 0.05 M acetic acid
  • Equilibrium concentration of ions: determined by \( 0.05 \cdot \alpha \)
This helps in establishing how much of the acetic acid actually ionizes in solution.Using the approximation \( \alpha \approx 0.0187 \) (calculated through the dissociation constant), we find that approximately 1.87% of the acetic acid dissociates in the given solution.
Acetic Acid
Acetic acid, commonly known as vinegar's active ingredient, is a weak carboxylic acid with the chemical formula CH鈧僀OOH . Its weak acidic nature results from partial ionization in water, making it an excellent example for studying acid dissociation equilibria. In aqueous solutions, it significantly remains in the undissociated form, and only a small fraction releases hydrogen ions:
  • Weak acid characteristic: Does not completely ionize in water.
  • Importance in food and chemical industries due to its acidity and preservative qualities.
When dissolved, it establishes a dynamic equilibrium between the undissociated acetic acid and the produced ions. This equilibrium state is crucial for understanding its behavior in different solutions and concentrations.
Equilibrium Constant
The equilibrium constant, denoted as \( K_a \) for acids, quantifies the extent of ionization or dissociation of an acid in solution. For acetic acid, the given \( K_a \) value is \( 1.74 \times 10^{-5} \). This demonstrates its weakly acidic nature. The expression for acetic acid's ionization is given as:\[K_a = \frac{[CH鈧僀OO鈦籡[H^+]}{[CH鈧僀OOH]}\]
The equilibrium constant remains consistent under a specific set of conditions (temperature, pressure), serving as a critical parameter for calculating other properties, like the degree of dissociation.A low \( K_a \) value reflects partial ionization, aligning with the minimal \( \alpha \) value derived through calculations, emphasizing acetic acid's small dissociation capability.
Concentration Calculation
In chemistry, calculating concentrations of ions helps in understanding the behavior of solutions. For acetic acid, the concentration calculations involve determining the amount of acetate ions and hydrogen ions formed at equilibrium.Starting with the initial concentration of acetic acid at 0.05 M and degree of dissociation \( \alpha \approx 0.0187 \), the concentration of CH鈧僀OO鈦 is:\[[CH鈧僀OO^-] = 0.05 \times \alpha = 9.35 \times 10^{-4} \text{ M}\]This demonstrates practical use of the degree of dissociation, allowing for the evaluation of the strength and behavior of the solution.Moreover, since [CH鈧僀OO鈦籡 = [H鈦篯, the hydrogen ion concentration is also 9.35 \times 10^{-4} M. From here, we calculate the pH:\[pH = -\log(9.35 \times 10^{-4}) \approx 3.03\]These calculations offer insights into the solution's acidity and its dissociation dynamics.

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Most popular questions from this chapter

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (a) Write as expression for \(K\), for the above reaction. (b) How will the values of \(K_{r}\) and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst ?

Calculate the \(\mathrm{pH}\) of the resultant mixtures: a) \(10 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{Ca}[\mathrm{OH})_{2}+25 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) b) \(10 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) c) \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KOH}\)

Which of the followings are Lewis acids? \(\mathrm{H}_{2} \mathrm{O}, \mathrm{BF}_{3}, \mathrm{H}^{+}\), and \(\mathrm{NH}_{4}^{+}\)

Write the expression for the equilibrium constant, \(K_{c}\) for each of the following reactions: (i) \(\quad 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (ii) \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{CuO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) (iii) \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})\) (iv) \(\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s})\) (v) \(\mathrm{I}_{2}(\mathrm{~s})+5 \mathrm{~F}_{2} \rightleftharpoons 2 \mathrm{IF}_{5}\)

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

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