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Chapter 7: Problem 9

The \(\mathrm{pH}\) of a \(0.02 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) solution will be [given \(\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}\) and \(\log 2=0.301]\) (a) \(2.65\) (b) \(4.35\) (c) \(4.65\) (d) \(5.35\)

Short Answer

Expert verified
The pH of \(0.02 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is approximately \(5.85\) (not listed among the options).

Step by step solution

01

Understanding the problem

We are given a solution of the salt \(\text{NH}_4\text{Cl}\), which is derived from the weak base \(\text{NH}_4\text{OH}\). We're required to find its \(\text{pH}\). Given \(K_b(\text{NH}_4\text{OH}) = 10^{-5}\), we can find \(K_a(\text{NH}_4^+)\) to work with the acidic nature of the solution.
02

Calculate the acid dissociation constant \(K_a\)

Use the relation between \(K_w\), \(K_a\), and \(K_b\): \[ K_w = K_a \cdot K_b \] where \(K_w = 10^{-14}\). Thus, \(K_a = \frac{10^{-14}}{10^{-5}} = 10^{-9}.\)
03

Set up the equilibrium expression for \(\text{NH}_4^+\)

For the dissociation of \(\text{NH}_4^+\): \[ \text{NH}_4^+ \leftrightarrow \text{NH}_3 + \text{H}^+ \] The equilibrium expression is: \[ K_a = \frac{[\text{NH}_3][\text{H}^+]}{[\text{NH}_4^+]} = 10^{-9} \] Assuming \([\text{H}^+] = x\), and initial \[\text{NH}_4^+\] being 0.02 M, at equilibrium it is \(0.02 - x\). Hence, \[ 10^{-9} = \frac{x^2}{0.02 - x} \approx \frac{x^2}{0.02} \]
04

Solve for \([\text{H}^+]\)

Rearrange to solve for \(x\): \[ x^2 = 10^{-9} \times 0.02 \] \[ x^2 = 2 \times 10^{-11} \] \[ x = \sqrt{2 \times 10^{-11}} \] \[ x = \sqrt{2} \times 10^{-5.5} \approx 1.414 \times 10^{-5.5} \] \[ x \approx 1.41 \times 10^{-6} \]
05

Calculate \(\text{pH}\) from \([\text{H}^+]\)

The \(\text{pH}\) is calculated using the relation: \[ \text{pH} = -\log[\text{H}^+] = -\log(1.41 \times 10^{-6}) \] Using the property of logarithms, this can be broken down to: \[ \text{pH} = -(\log 1.41 + \log 10^{-6}) \] \[ \approx -(0.15 - 6) = 5.85 \]

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Key Concepts

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

pH Calculation
The pH calculation is crucial in comprehending the acidity or basicity of a solution. In this exercise, we're dealing with a 0.02 M solution of ammonium chloride (NH\(_4\)Cl). The first step involves recognizing that NH\(_4\)Cl dissociates into NH\(_4^+\) and Cl\(^-\) in water.
This creates an acidic environment due to the NH\(_4^+\) ions. To calculate the pH, we start with the concentration of hydrogen ions, denoted as [H\(^+\)].

The relationship involves calculating the dissociation constant, K\(_a\), for NH\(_4\)^+. Given the base dissociation constant, K\(_b\), for NH\(_4\)OH as 10\(^{-5}\), we find the corresponding K\(_a\) using the relation:
  • \(K_w = K_a \, \times \, K_b = 10^{-14}\)
  • \(K_a = \frac{10^{-14}}{10^{-5}} = 10^{-9}\)
With K\(_a\) known, we set up the equilibrium expression and solve for [H\(^+\)], ultimately determining that the pH is around 5.85. This calculation shows the weakly acidic nature of the ammonium chloride solution.
Acid-Base Reactions
Acid-base reactions play a vital role in the behavior of NH\(_4\)Cl in water. When NH\(_4\)Cl dissolves, it splits into NH\(_4^+\) and Cl\(^-\). The NH\(_4^+\) ion, being a conjugate acid of the weak base NH\(_3\), reacts with water:

\[ \text{NH}_4^+ + \text{H}_2\text{O} \leftrightarrow \text{NH}_3 + \text{H}_3\text{O}^+ \]
This reaction is where the acidic nature comes from, as NH\(_4^+\) contributes to an increase in the hydrogen ion concentration ([H\(^+\)]). This behavior can be understood through the Bronsted-Lowry theory of acids and bases, where an acid is a proton donor.

Through these interactions, we see the direct influence on the pH level. It's fundamental to remember:
  • NH\(_4^+\) is the acid.
  • NH\(_3\) is the base that forms as a result of this reaction.
This ion exchange emphasizes the balance between acids and bases and how they determine the resultant pH of the solution. Hence, the reaction dynamics ensure the maintenance of equilibrium.
Equilibrium Constants
Equilibrium constants are a measure of the extent to which a reaction occurs. For NH\(_4\)Cl, the key equilibrium constant is the acid dissociation constant, K\(_a\), which is vital for understanding the dissociation of NH\(_4^+\) in water.

The relationship for dissociations can be represented as:
\[ K_a = \frac{[ ext{NH}_3][ ext{H}^+]}{[ ext{NH}_4^+]} = 10^{-9} \]
During equilibrium, the concentrations involved in the reaction reach a point where their ratio remains constant, signifying that the forward and backward reactions happen at the same rate.

Understanding equilibrium constants helps to predict how much NH\(_4\)^+ becomes NH\(_3\) and H\(^+\). It's essential to solve equilibrium expressions accurately, especially when using approximations such as assuming \(0.02 - x \approx 0.02\). These constants, like K\(_a\) and K\(_b\), help depict the strength of acids and bases.
  • High K\(_a\) values indicate strong acids.
  • Low K\(_b\) values correlate to weak bases like NH\(_4\)OH.
In this exercise, we've seen how these concepts build on each other, permitting the accurate calculation of a solution's pH.

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