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Chapter 14: Problem 165

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\). a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) b. Calculate the \(\mathrm{pH}\) of this solution.

Short Answer

Expert verified
a. The major species in the solution are: i. C鈧侶鈧匩H鈧�, iii. OH鈦�, iv. H鈧侽, and v. C鈧侶鈧匩H鈧冣伜. b. The pH of the solution is approximately 12.23.

Step by step solution

01

Write the reaction with water

First, we write the reaction of C鈧侶鈧匩H鈧� with water, which is a weak base, and the reaction will look like this: \[C_{2}H_{5}NH_{2} + H_{2}O \rightleftharpoons C_{2}H_{5}NH_{3}^{+} + OH^{-}\]
02

Set up the ICE table

Now, set up an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations of reactants and products. Let x be the change in concentration. | | C鈧侶鈧匩H鈧� | H鈧侽 | C鈧侶鈧匩H鈧冣伜 | OH鈦� | |------- |-------- |--------------|----------|------| | Initial | 0.67 M | - | 0 M | 0 M | | Change | -x | - | +x | +x | | Equilib | 0.67 - x | - | x | x |
03

Write the expression for Kb

Write the expression for the base ionization constant (Kb) and plug in the values from the equilibrium row. \[K_{b} = \frac{[C_{2}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{2}H_{5}NH_{2}]}\] \[5.6\times 10^{-4} =\frac{x\cdot x}{(0.67-x)}\]
04

Make an assumption and solve for x

Since C鈧侶鈧匩H鈧� is a weak base, we can assume that the change in concentration (x) will be insignificant compared to 0.67 M. So, we can simplify the expression: \[5.6\times 10^{-4} =\frac{x^2}{0.67}\] Solve for x: \[x^2 = (5.6\times 10^{-4})\cdot 0.67\] \[x = \sqrt{(5.6\times 10^{-4})\cdot 0.67}\] \[x = 0.01709\]
05

Identify the major species

x represents the concentration of both C鈧侶鈧匩H鈧冣伜 and OH鈦� at equilibrium. The concentration of C鈧侶鈧匩H鈧� at equilibrium is 0.67 - x. Now compare these values and the given species to identify the major ones: i. C鈧侶鈧匩H鈧�: 0.67 - 0.01709 = 0.65291 M ii. H鈦�: not in the reaction. iii. OH鈦�: 0.01709 M iv. H鈧侽: present in large excess as the solvent. v. C鈧侶鈧匩H鈧冣伜: 0.01709 M So, the major species are: C鈧侶鈧匩H鈧�, OH鈦�, H鈧侽 and C鈧侶鈧匩H鈧冣伜.
06

Calculate the pOH

We have the concentration of OH鈦� ions which is equal to 0.01709 M. Now, we can determine the pOH using the relationship: \[pOH = -\log{[OH^{-}]}\] \[pOH = -\log{(0.01709)}\] \[pOH = 1.767\]
07

Calculate the pH

Lastly, we will use the relationship between pH, pOH, and pKw(=14) to determine the pH of the solution: \[pH = pKw - pOH\] \[pH = 14 - 1.767\] \[pH = 12.233\] So, the pH of this solution is approximately 12.23.

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Key Concepts

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

Weak Bases
Weak bases are chemical species that partially accept protons, unlike strong bases which completely dissociate in water. Ethylamine oindent
  • In the context of the exercise, ethylamine (\(\text{C}_{2} \text{H}_{5} \text{NH}_{2}\)) is a weak base.
  • It doesn't fully ionize in water; this characteristic allows us to use equilibrium concepts to study its behavior in solution.
When added to water, ethylamine establishes an equilibrium by forming \(\text{C}_{2} \text{H}_{5} \text{NH}_{3}^{+}\) and hydroxide ions (\(\text{OH}^{-}\)).

This reaction doesn't go to completion, and that's why the equilibrium concept is so important to understanding weak bases. This allows us to analyze the balance between reactants and products that form in such systems.
pH Calculation
pH is a measure of the acidity or basicity of a solution.
  • For weak bases like ethylamine, the pH can be calculated by first determining the pOH, which is related to the concentration of hydroxide ions.
  • From the exercise, the concentration of \(\text{OH}^{-}\) is calculated using the ICE table and the equilibrium expression for the base.
After finding the concentration, you plug it into the logarithmic function: \[pOH = -\log{[\text{OH}^{-}]}\]With the pOH value determined, the pH can be calculated using the formula:\[pH = 14 - pOH\]This helps us convert the measurement from a base perspective (pOH) to the more commonly used pH scale. Remember, the lower the pH value, the more acidic the solution, while a higher pH indicates a more basic solution.

Understanding these calculations helps us understand the properties of the solution and how weak bases behave in aqueous environments.
Equilibrium Constant
The equilibrium constant, specifically here the base ionization constant (\(K_b\)), is a critical factor in understanding the strength of a weak base.
  • For ethylamine in the exercise, \(K_b\) is given as \(5.6 \times 10^{-4}\).
  • This constant determines how readily the base forms products in an equilibrium state.
The expression for \(K_b\) is set up based on the equilibrium equation, involving the concentrations of \(\text{C}_{2} \text{H}_{5} \text{NH}_{2}\), \(\text{C}_{2} \text{H}_{5} \text{NH}_{3}^{+}\), and \(OH^{-}\).

Using the equilibrium constant allows us to solve for unknown concentrations, such as \(x\), which is crucial in finding the major species in a solution and calculating pH. The assumption made (that \(x\) is small compared to the initial concentration) simplifies our calculations by reducing the complexity of the equation. Understanding the role of the equilibrium constant establishes a strong foundation for further exploring acid-base reactions and their implications in various contexts.

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Most popular questions from this chapter

An acid HX is \(25 \%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M}\), calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\).

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{HNO}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. \(\mathrm{HF}\) g. h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Ch芒telier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.)

The \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) hydrochloric acid is not \(8.00\). The correct \(\mathrm{pH}\) can be calculated by considering the relationship between the molarities of the three principal ions in the solution \(\left(\mathrm{H}^{+}, \mathrm{Cl}^{-}\right.\), and \(\left.\mathrm{OH}^{-}\right) .\) These molarities can be calculated from algebraic equations that can be derived from the considerations given below. a. The solution is electrically neutral. b. The hydrochloric acid can be assumed to be \(100 \%\) ionized. c. The product of the molarities of the hydronium ions and the hydroxide ions must equal \(K_{\mathrm{w}}\). Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-8}-\mathrm{M} \mathrm{HCl}\) solution.

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)
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