Chapter 14: Problem 36
Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 M C_{2} H_{5} N H_{2} / 0.25 M C_{2} H_{5} N H_{3} C l\) b. \(0.25 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\) c. \(0.50 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\)
Short Answer
Expert verified
The pH values for the buffered solutions are:
a. 11.05
b. 10.45
c. 10.75
Step by step solution
01
Identify the weak acid and its conjugate base
In each buffered solution, \(C_2H_5NH_2\) (ethylamine) acts as the weak base and \(C_{2}H_{5}NH_{3}C l\) (ethylammonium chloride) is its conjugate acid. The corresponding conjugate base of ethylammonium ion (\(C_{2}H_{5}NH_{3}^+\)) is ethylamine.
02
Find the pKa of the conjugate acid
We can find the pKa of ethylammonium ion by looking up the \(K_b\) value for ethylamine (\(C_2H_5NH_2\)) in a table or online resources and using the following relation:
\(K_w = K_a \times K_b\)
where \(K_w\) is the ion product constant of water (\(1.0 \times 10^{-14}\) at 25°C) and \(K_a\) is the acid dissociation constant of ethylammonium ion.
In this case, the \(K_b\) value of ethylamine (\(C_2H_5NH_2\)) is \(5.6 \times 10^{-4}\). To find the pKa of ethylammonium ion, we can calculate its \(K_a\) value and then convert it to pKa:
\(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-4}} \approx 1.79 \times 10^{-11}\)
pKa = -log10(\(K_a\)) = 10.75
03
Calculate the pH for each buffered solution
Using the Henderson-Hasselbalch equation, we can find the pH of each buffered solution.
a. pH = 10.75 + log10(\(\frac{0.50}{0.25}\)) = 10.75 + log10(2) = 10.75 + 0.301 = 11.05
b. pH = 10.75 + log10(\(\frac{0.25}{0.50}\)) = 10.75 + log10(0.5) = 10.75 - 0.301 = 10.45
c. pH = 10.75 + log10(\(\frac{0.50}{0.50}\)) = 10.75 + log10(1) = 10.75 + 0 = 10.75
04
Report the pH values for the buffered solutions
The pH of the three buffered solutions are as follows:
a. \(0.50 M C_{2} H_{5} N H_{2} / 0.25 M C_{2} H_{5} N H_{3} C l\) has a pH of 11.05.
b. \(0.25 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\) has a pH of 10.45.
c. \(0.50 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\) has a pH of 10.75.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is crucial for anyone studying chemistry, especially when it comes to buffered solutions. This equation serves as a simplified way to estimate the pH of a buffer solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid.
The equation is given as:
\[ \text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{\text{[Conjugate Base]}}{\text{[Acid]}} \right) \]
where \( \text{p}K_{\text{a}} \) is the negative logarithm of the acid dissociation constant (\( K_{\text{a}} \)) of the conjugate acid, and the concentrations are those of the conjugate base and the acid, respectively.
Using this formula, you can easily adjust the pH of a buffer by changing the ratio of the concentrations of the conjugate base to the acid. This understanding is incredibly helpful when preparing solutions for various laboratory applications or understanding biological systems where maintaining a near-constant pH is essential.
The equation is given as:
\[ \text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{\text{[Conjugate Base]}}{\text{[Acid]}} \right) \]
where \( \text{p}K_{\text{a}} \) is the negative logarithm of the acid dissociation constant (\( K_{\text{a}} \)) of the conjugate acid, and the concentrations are those of the conjugate base and the acid, respectively.
Using this formula, you can easily adjust the pH of a buffer by changing the ratio of the concentrations of the conjugate base to the acid. This understanding is incredibly helpful when preparing solutions for various laboratory applications or understanding biological systems where maintaining a near-constant pH is essential.
Conjugate Acid-Base Pair
A conjugate acid-base pair is a pair of substances that transform into each other by the loss or gain of a proton (\( H^+ \)). In a buffered solution, this pair consists of a weak acid and its conjugate base or a weak base and its conjugate acid. This pairing is fundamental in understanding buffer solutions because their ability to resist changes in pH when small amounts of acid or base are added is due to the presence of these two species.
For example, in the case of our exercise, ethylamine (\( C_2H_5NH_2 \)) is a weak base, and its conjugate acid is ethylammonium ion (\( C_2H_5NH_3^+ \)). When ethylamine accepts a proton, it becomes ethylammonium, and when ethylammonium loses a proton, it reverts back to ethylamine. These reversible reactions help maintain the stability of the pH in the buffer solution.
For example, in the case of our exercise, ethylamine (\( C_2H_5NH_2 \)) is a weak base, and its conjugate acid is ethylammonium ion (\( C_2H_5NH_3^+ \)). When ethylamine accepts a proton, it becomes ethylammonium, and when ethylammonium loses a proton, it reverts back to ethylamine. These reversible reactions help maintain the stability of the pH in the buffer solution.
Acid Dissociation Constant (Ka)
The acid dissociation constant, denoted as \( K_{\text{a}} \), is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation of a weak acid into its ions in an aqueous solution. A larger \( K_{\text{a}} \) value indicates a stronger acid because it demonstrates a greater tendency of the acid to lose its proton and form its conjugate base.
\[ K_{\text{a}} = \frac{\text{[Conjugate Base]} \times [H^+]}{\text{[Acid]}} \]
\( K_{\text{a}} \) is used to calculate the pKa, which is more commonly used in pH calculations, with the relationship:\[ \text{p}K_{\text{a}} = -\log( K_{\text{a}} ) \]
In the context of our exercise, the calculation of \( K_{\text{a}} \) for the conjugate acid is derived from the base dissociation constant \( K_{\text{b}} \) of ethylamine and the ion product constant of water, \( K_{\text{w}} \), using the formula \( K_{\text{w}} = K_{\text{a}} \times K_{\text{b}} \).
\[ K_{\text{a}} = \frac{\text{[Conjugate Base]} \times [H^+]}{\text{[Acid]}} \]
\( K_{\text{a}} \) is used to calculate the pKa, which is more commonly used in pH calculations, with the relationship:\[ \text{p}K_{\text{a}} = -\log( K_{\text{a}} ) \]
In the context of our exercise, the calculation of \( K_{\text{a}} \) for the conjugate acid is derived from the base dissociation constant \( K_{\text{b}} \) of ethylamine and the ion product constant of water, \( K_{\text{w}} \), using the formula \( K_{\text{w}} = K_{\text{a}} \times K_{\text{b}} \).
Base Dissociation Constant (Kb)
The base dissociation constant, or \( K_{\text{b}} \), mirrors the acid dissociation constant but for bases. It is the equilibrium constant for the dissociation of a weak base into its ions in an aqueous solution. A higher value of \( K_{\text{b}} \) corresponds to a stronger base.
For a reaction where the weak base (B) accepts a proton from water, forming its conjugate acid (BH+) and hydroxide ion (OH-), the base dissociation constant is represented as:
\[ K_{\text{b}} = \frac{\text{[BH^+]} \times [OH^-]}{\text{[B]}} \]
In our exercise, we are given the \( K_{\text{b}} \) for ethylamine, from which we calculate the \( K_{\text{a}} \) of the corresponding conjugate acid, using the relationship between \( K_{\text{w}} \) and \( K_{\text{a}} \), \( K_{\text{b}} \), as mentioned before.
For a reaction where the weak base (B) accepts a proton from water, forming its conjugate acid (BH+) and hydroxide ion (OH-), the base dissociation constant is represented as:
\[ K_{\text{b}} = \frac{\text{[BH^+]} \times [OH^-]}{\text{[B]}} \]
In our exercise, we are given the \( K_{\text{b}} \) for ethylamine, from which we calculate the \( K_{\text{a}} \) of the corresponding conjugate acid, using the relationship between \( K_{\text{w}} \) and \( K_{\text{a}} \), \( K_{\text{b}} \), as mentioned before.
pH Calculation
The pH of a solution is a measure of its acidity or alkalinity, defined as the negative logarithm of the hydrogen ion concentration \( [H^+] \).
\[ \text{pH} = -\log( [H^+] ) \]
Calculating the pH of a solution is often carried out in chemistry to understand the nature and behavior of substances in different environments. For buffered solutions, the pH is typically calculated using the Henderson-Hasselbalch equation, as it incorporates the acid and base properties through the \( pK_{\text{a}} \) and the ratio of concentrations of the conjugate base and acid.
In our given problem, we use the calculated pKa value and the ratios of the concentrations of the conjugate base (ethylamine) and the conjugate acid (ethylammonium ion) in the buffered solutions to determine their respective pH values. This is a prime example of utilizing the Henderson-Hasselbalch equation to find the pH without directly measuring the hydrogen ion concentration.
\[ \text{pH} = -\log( [H^+] ) \]
Calculating the pH of a solution is often carried out in chemistry to understand the nature and behavior of substances in different environments. For buffered solutions, the pH is typically calculated using the Henderson-Hasselbalch equation, as it incorporates the acid and base properties through the \( pK_{\text{a}} \) and the ratio of concentrations of the conjugate base and acid.
In our given problem, we use the calculated pKa value and the ratios of the concentrations of the conjugate base (ethylamine) and the conjugate acid (ethylammonium ion) in the buffered solutions to determine their respective pH values. This is a prime example of utilizing the Henderson-Hasselbalch equation to find the pH without directly measuring the hydrogen ion concentration.
