What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is \(0.45 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\\\ &&\mathrm{p} K_{\mathrm{a}}=4.89 \end{aligned}$$

Short Answer

Expert verified
The degree and percent of ionization would be specific to the value obtained by solving the equation in Step 3 for x. But after that, the degree of ionization equals \(\frac{x}{0.45}\) and the percent ionization is this degree of ionization multiplied by 100.

Step by step solution

01

Compute for \(K_a\)

First, you must compute for the \(K_a\) using the given \(pKa\). The \(K_a\) is equal to \(10^{-pKa}\), so substituting the given \(pKa = 4.89\), you get \(K_a = 10^{-4.89}\).
02

Apply the ICE Method

Next, apply the ICE (Initial, Change, Equilibrium) Method. Initially, the concentration of the acid is 0.45M, and the concentration of its ions is 0. At equilibrium, let x be the concentration that ionizes, therefore you have 0.45-x of the \(CH_3CH_2CO_2H\) (propionic acid), x of \(H_3O^+\) and x of \(CH_3CH_2CO_2^-\) (propionate ion).
03

Set Up and Solve for x

Use the \(K_a\) expression to set up the equation: \( K_a = \frac{[H_3O^+][CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]}\) which at equilibrium becomes: \( K_a = \frac{x * x}{0.45 - x}\). Solve for x using the quadratic formula or approximation method.
04

Calculate Degree of Ionization

The degree of ionization is equal to \(\frac{x}{0.45}\).
05

Calculate Percent ionization

Finally, the percent ionization is the degree of ionization expressed in percent, which is simply the degree of ionization multiplied by 100.

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Key Concepts

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

Degree of Ionization
Understanding the degree of ionization is essential for grasping how acids behave in solution. The degree of ionization refers to the proportion of a substance that has been ionized (turned into ions) in a solution at a given concentration. For weak acids, like propionic acid, not every molecule dissociates in water, meaning only a fraction turns into ions.

To calculate the degree of ionization, you divide the concentration of ionized acid by the initial concentration of the acid before ionization. Using the notation from the exercise, the degree of ionization would be represented mathematically as the quotient \( \frac{x}{0.45} \), where \(x\) is the equilibrium concentration of ionized acid. The closer this value is to 1, the higher the degree of ionization, indicating a stronger acid.
ICE Method
When examining acid-base equilibria, the ICE (Initial, Change, Equilibrium) method is a systematic way to keep track of concentrations. This method involves setting up an ICE table that outlines what happens to each species in a reaction from the initial state to equilibrium.

Here's how the ICE method breaks down:
  • Initial: The starting concentrations of the reactants and products are listed, before any reaction takes place.
  • Change: This column represents the change in concentrations as reactants turn into products. As the reaction proceeds toward equilibrium, the change is often represented by \( -x \) for reactants (they are being used up) and \( +x \) for products (they are being formed).
  • Equilibrium: These are the concentrations of reactants and products when the reaction has reached a state of balance. Here, the change values are applied to the initial concentrations.
Applying the ICE method helps you create a clear picture of the reaction's progress and is crucial for solving equilibrium-related exercises. By setting up an ICE table, students can systematically approach the calculation of unknown values like the degree and percent of ionization.
Acid Ionization Constant \(K_a\)
The acid ionization constant, denoted as \(K_a\), is a value that measures the strength of an acid in solution. It specifically relates to weak acids, which don't completely dissociate in water. \(K_a\) is defined by the equilibrium concentrations of the products and reactants involved in the ionization of the acid:

\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \]
where \(\text{HA}\) is the weak acid, \(\text{H}_3\text{O}^+\) is the hydronium ion, and \(\text{A}^-\) is the conjugate base. The larger the \(K_a\), the stronger the acid as it indicates a greater degree of ionization.

From the \(pKa\), which is the negative log of the \(K_a\), you can calculate \(K_a\) using the formula \(K_a = 10^{-pKa}\). \(K_a\) is essential for predicting the behavior of acids in different scenarios and for comparing the strengths of various acids. It forms the core of the mathematical operations involved when using the ICE method to deduce the degree and percent ionization of an acid in solution.

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Most popular questions from this chapter

It is possible to write simple equations to relate \(\mathrm{pH}\) \(\mathrm{p} K,\) and molarities \((\mathrm{M})\) of various solutions. Three such equations are shown here. $$\begin{aligned} &\text {Weak acid: } \quad \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{M}\\\ &\text { Weak base: } \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{M} \end{aligned}$$ Salt ofweak \(\operatorname{acid}\left(\mathrm{pK}_{\mathrm{a}}\right)\) and strong base: \(\quad \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{p} K_{\mathrm{a}}+\frac{1}{2} \log \mathrm{M}\) (a) Derive these three equations, and point out the assumptions involved in the derivations. (b) Use these equations to determine the pH of 0.10 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), 0.10 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}),\) and \(0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} .\) Verify that the equations give correct results by determining these pH values in the usual way.

Predict which is the stronger acid: (a) \(\mathrm{HClO}_{2}\) or \(\mathrm{HClO}_{3} ;(\mathrm{b}) \mathrm{H}_{2} \mathrm{CO}_{3}\) or \(\mathrm{HNO}_{2} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SiO}_{3}\) or \(\mathrm{H}_{3} \mathrm{PO}_{4}\) Explain.

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$

Explain the important distinctions between each pair of terms: (a) Bronsted- Lowry acid and base; (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH} ;\) (c) \(K_{\mathrm{a}}\) for \(\mathrm{NH}_{4}^{+}\) and \(K_{\mathrm{b}}\) for \(\mathrm{NH}_{3} ;\) (d) leveling effect and electron- withdrawing effect.

For the ionization of phenylacetic acid, $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \\\ K_{\mathrm{a}}=4.9 \times 10^{-5} \end{array}$$ (a) What is \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right]\) in \(0.186 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) (b) What is the \(\mathrm{pH}\) of \(0.121 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\)

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