'The ionization constant of dimethyl amine is \(5.4 \times 10^{-4} .\) Calculate its degree of ionization in its \(0.02 \mathrm{M}\) solution. What percentage of dimethyl amine is ionized if the solution is also \(0.1 \mathrm{M}\) in \(\mathrm{NaOH}\).

Short Answer

Expert verified
Degree of ionization without NaOH is 3.29 x 10^-3 and the percentage is 16.45%. With the addition of 0.1 M NaOH, the ionization of dimethyl amine is negligible and the percentage is effectively 0%.

Step by step solution

01

Understanding Ionization Constant

The ionization constant, represented as Ka, is a measure of the strength of a weak acid or a weak base. In the case of dimethyl amine, which is a weak base, the base ionization constant is given as Kb instead of Ka, and it represents the equilibrium constant for its ionization in water to form hydroxide ions (OH-) and the corresponding amine cation.
02

Write the Ionization Equation

Write the balanced chemical equation for the ionization of dimethyl amine (DMA) in water: DMA + H2O -> DMAH+ + OH-. Let the degree of ionization be represented by 'x', which is the amount of DMA that ionizes per mole. At equilibrium, the concentration of DMAH+ and OH- will each be 'x' and the concentration of DMA will be 0.02 - x.
03

Set Up the Ionization Expression

Express the ionization constant in terms of the degree of ionization. The expression for Kb is: Kb = [DMAH+][OH-]/[DMA]. Substituting the terms with 'x' gives us Kb = x^2 / (0.02 - x).
04

Use the Approximation

Since dimethyl amine is a weak base, it ionizes very little. Therefore, we can assume x << 0.02, allowing us to approximate the denominator in the expression for Kb as 0.02. The expression now simplifies to Kb = x^2 / 0.02.
05

Solve for Degree of Ionization

Now, plug in the value of Kb and solve for 'x'. Kb = 5.4 x 10^-4, so 5.4 x 10^-4 = x^2 / 0.02. Solving for x gives x = sqrt(5.4 x 10^-4 * 0.02), which is the degree of ionization.
06

Calculate the Degree of Ionization

Perform the calculation to find the value of 'x'. x = sqrt(5.4 x 10^-4 * 0.02) = sqrt(1.08 x 10^-5) = 3.29 x 10^-3.
07

Calculate Ionization as a Percentage

Convert the degree of ionization to a percentage by multiplying the ratio of the ionized amine to the initial concentration by 100%. Percentage Ionization = (x / 0.02) x 100% = (3.29 x 10^-3 / 0.02) x 100%.
08

Influence of Added NaOH

When NaOH is added, it supplies additional OH- ions that suppress the ionization of the weak base due to the common ion effect. The additional OH- from NaOH (0.1 M) far exceeds the amount produced by the weak base's ionization, thus the degree of ionization of DMA will be significantly smaller and can be considered negligible in comparison to that provided by NaOH.
09

Calculate Ionization with NaOH

With the common ion effect, the ionization of DMA is so small that nearly none of the DMA is ionized. Therefore, the percentage ionized in the presence of 0.1 M NaOH is effectively 0%.

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Key Concepts

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

Weak Base Ionization
Ionization of weak bases, such as dimethyl amine, is a critical concept in understanding how such substances behave in solution. Weak bases do not completely disassociate in water, which results in an equilibrium between the base, its ionized form, and the hydroxide ions. The ionization can be represented by a balanced chemical equation:

DMA + H2O ⇌ DMAH+ + OH-.

In this equation, DMA represents dimethyl amine, and DMAH+ is the conjugate acid of DMA. During this ionization process, only a fraction of the base converts into hydroxide ions (OH-) and the conjugate acid. This fraction is quantified by the degree of ionization, 'x', meaning that only 'x' moles of OH- are produced per mole of the weak base in solution. To understand the extent of ionization, the base ionization constant (Kb) is used, which offers a measure of the equilibrium between the reactants and products.
Common Ion Effect in Ionization
The common ion effect plays a pivotal role in controlling the ionization equilibrium of weak acids and bases. It occurs when a solution containing a weak base (or acid) also contains another compound that provides an ion identical to the one produced by the ionization of the weak base. For instance, when NaOH is added to a dimethyl amine solution, it releases hydroxide ions (OH-) into the solution, which are common to the ionization product of dimethyl amine.

This influx of common ions shifts the equilibrium, according to Le Chatelier's principle, towards the left, reducing the ionization of the weak base. Consequently, the presence of a strong base like NaOH significantly suppresses the ionization of dimethyl amine, as the system seeks to reduce the concentration of the common ion, OH-, by reducing the extent of DMA ionization.
Percentage Ionization Calculation
Calculating the percentage ionization provides insight into how much of a weak base has ionized in a solution. To determine this, the degree of ionization must first be established by using the ionization constant and the initial concentration of the base. In mathematical terms:

Percentage Ionization = \(\frac{x}{[DMA]_{initial}} \times 100%\)

In this formula, 'x' stands for the degree of ionization (the concentration of ionized base at equilibrium), and [DMA]initial is the initial concentration of the weak base before ionization. After calculating 'x' (as detailed in the step-by-step solution), it can be converted into a percentage to reflect the proportion of the base that has been ionized. However, when other sources provide a common ion, such as NaOH in the exercise, the percentage ionization can be significantly affected or even rendered negligible due to the suppression of ionization by the common ion effect.

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Most popular questions from this chapter

The \(p K_{a}\) of acetic acid and \(p K_{b}\) of ammonium hydroxide are \(4.70\) and \(4.75\) respectively. Calculate the hydrolysis constant of ammonium acetate at \(298 \mathrm{~K}\) and also the degree of hydrolysis and \(\mathrm{pH}\) of its (a) \(0.01 M\) and (b) \(0.04 \mathrm{M}\) solutions.

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Calculate the \(\mathrm{pH}\) of a solution of given mixtures; (a) \(\left(4 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}+6 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COONa}\right)\) in \(100 \mathrm{~mL}\) of mixture; \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}\) (b) \(5 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MOH}+250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MCl}\); \(K_{\mathrm{a}}\) for \(M \mathrm{OH}=1.8 \times 10^{-5}\) (c) \(\left(0.25\right.\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}+0.35\) mole of \(\mathrm{CH}_{3} \mathrm{COONa}\) ) in \(500 \mathrm{ml}\) mixture; \(K_{\text {a }}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=3.6 \times 10^{-4}\)

The \(K_{\mathrm{w}}\) for \(2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}\) changes from \(10^{-14}\) at \(25^{\circ} \mathrm{C}\) to \(9.62 \times 10^{-14}\) at \(60^{\circ} \mathrm{C} .\) What is \(\mathrm{pH}\) of water at \(60^{\circ} \mathrm{C} ?\) What happens to its neutrality?

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