What Is The PH Of A Solution Composed Of $0.20 \, M \, NH_3$ And $0.15 \, M \, NH_4Cl$?(a) 2.15 (b) 4.62 (c) 8.26 (d) 9.38 $K_b \, NH_3 = 1.8 \times 10^{-5}$

What is the pH of a solution composed of $0.20 \, M \, NH_3$ and $0.15 \, M \, NH_4Cl$?

Introduction

In this article, we will explore the concept of pH and how it relates to a solution composed of ammonia ($NH_3$) and ammonium chloride ($NH_4Cl$). The pH of a solution is a measure of its acidity or basicity, and it is an important concept in chemistry. We will use the Henderson-Hasselbalch equation to calculate the pH of the solution.

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical formula that relates the pH of a solution to the concentrations of its components. It is given by:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

where $pK_a$ is the acid dissociation constant, $[A^-]$ is the concentration of the conjugate base, and $[HA]$ is the concentration of the weak acid.

The pH of a Solution Composed of $NH_3$ and $NH_4Cl$

In this case, we have a solution composed of $0.20 \, M \, NH_3$ and $0.15 \, M \, NH_4Cl$. We are given that the $K_b$ of $NH_3$ is $1.8 \times 10^{-5}$. We can use the Henderson-Hasselbalch equation to calculate the pH of the solution.

First, we need to find the $pK_b$ of $NH_3$. The $pK_b$ is related to the $K_b$ by the following equation:

${ pK_b = -\log K_b }$

Substituting the given value of $K_b$, we get:

${ pK_b = -\log (1.8 \times 10^{-5}) }$

${ pK_b = 4.74 }$

Next, we need to find the $pK_a$ of $NH_4^+$. The $pK_a$ is related to the $pK_b$ by the following equation:

${ pK_a + pK_b = 14 }$

Substituting the value of $pK_b$, we get:

${ pK_a = 14 - pK_b }$

${ pK_a = 14 - 4.74 }$

${ pK_a = 9.26 }$

Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

Substituting the values, we get:

${ pH = 9.26 + \log \left( \frac{0.15}{0.20} \right) }$

${ pH = 9.26 + \log (0.75) }$

${ pH = 9.26 - 0.12 }$

${ pH = 9.14 }$

However, this is not one of the options. We need to consider the dissociation of $NH_3$ in water.

The Dissociation of $NH_3$ in Water

$NH_3$ is a weak base that dissociates in water to form $NH_4^+$ and $OH^-$. The dissociation reaction is as follows:

${ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- }$

The equilibrium constant for this reaction is given by:

${ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} }$

Substituting the given value of $K_b$, we get:

${ 1.8 \times 10^{-5} = \frac{[NH_4^+][OH^-]}{0.20} }$

We can rearrange this equation to get:

${ [NH_4^+][OH^-] = 3.6 \times 10^{-6} }$

Now, we can use the fact that $[OH^-] = \frac{K_w}{[H_3O^+]}$ to substitute for $[OH^-]$ in the above equation:

${ [NH_4^+] \times \frac{K_w}{[H_3O^+]} = 3.6 \times 10^{-6} }$

Substituting the value of $K_w$, we get:

${ [NH_4^+] \times \frac{10^{-14}}{[H_3O^+]} = 3.6 \times 10^{-6} }$

We can rearrange this equation to get:

${ [NH_4^+] = \frac{3.6 \times 10^{-6} \times [H_3O^+]}{10^{-14}} }$

Substituting the value of $[H_3O^+] = 10^{-7}$, we get:

${ [NH_4^+] = \frac{3.6 \times 10^{-6} \times 10^{-7}}{10^{-14}} }$

${ [NH_4^+] = 3.6 \times 10^{-3} }$

Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

Substituting the values, we get:

${ pH = 9.26 + \log \left( \frac{0.15}{3.6 \times 10^{-3}} \right) }$

${ pH = 9.26 + \log (41.67) }$

${ pH = 9.26 + 1.62 }$

${ pH = 10.88 }$

However, this is not one of the options. We need to consider the dissociation of $NH_4Cl$ in water.

The Dissociation of $NH_4Cl$ in Water

$NH_4Cl$ is a salt that dissociates in water to form $NH_4^+$ and $Cl^-$. The dissociation reaction is as follows:

${ NH_4Cl \rightleftharpoons NH_4^+ + Cl^- }$

The equilibrium constant for this reaction is given by:

${ K_s = [NH_4^+][Cl^-] }$

We are given that the concentration of $NH_4Cl$ is $0.15 \, M$. We can assume that the concentration of $Cl^-$ is also $0.15 \, M$. Therefore, we can substitute these values into the above equation:

${ K_s = [NH_4^+][Cl^-] }$

${ K_s = [NH_4^+](0.15) }$

We can rearrange this equation to get:

${ [NH_4^+] = \frac{K_s}{0.15} }$

Substituting the value of $K_s$, we get:

${ [NH_4^+] = \frac{0.15}{0.15} }$

${ [NH_4^+] = 1 }$

Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

Substituting the values, we get:

${ pH = 9.26 + \log \left( \frac{0.15}{1} \right) }$

${ pH = 9.26 + \log (0.15) }$

${ pH = 9.26 - 0.82 }$

${ pH = 8.44 }$

However, this is not one of the options. We need to consider the dissociation of both $NH_3$ and $NH_4Cl$ in water.

The Dissociation of Both $NH_3$ and $NH_4Cl$ in Water

We have already considered the dissociation of $NH_3$ in water. We have also considered the dissociation of $NH_4Cl$ in water. Now, we need to consider the dissociation of both $NH_3$ and $NH_4Cl$ in water.

The dissociation reaction of $NH_3$ in water is as follows:

${ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- }$

The equilibrium constant for this reaction is given by:

${ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} }$

Substituting the given value of $K_b$, we get:

${ 1.8 \times 10^{-5} = \frac{[NH_4^+][OH^-]}{0.20} }$

We can rearrange this equation to get:

${ [NH_4^+][OH^-] = 3.6 \times 10^{-6} }$

Now, we can use the fact that $[OH^-] = \frac{K_w}{[H_3O^+]}$ to substitute for $[OH^-]$ in the above equation:

${ [NH_4^+] \times \frac{K_w}{[H_3O^+]} = 3.6 \times 10^{-6} }$

Substituting the value of $K_w$, we get:

${ [NH
Q&A: What is the pH of a solution composed of $0.20 \, M \, NH_3$ and $0.15 \, M \, NH_4Cl$?

Q: What is the pH of a solution composed of $0.20 \, M \, NH_3$ and $0.15 \, M \, NH_4Cl$?

A: To determine the pH of the solution, we need to consider the dissociation of both $NH_3$ and $NH_4Cl$ in water. We have already considered the dissociation of $NH_3$ in water and the dissociation of $NH_4Cl$ in water. Now, we need to consider the dissociation of both $NH_3$ and $NH_4Cl$ in water.

Q: What is the dissociation reaction of $NH_3$ in water?

A: The dissociation reaction of $NH_3$ in water is as follows:

[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- }$

Q: What is the equilibrium constant for the dissociation reaction of $NH_3$ in water?

A: The equilibrium constant for the dissociation reaction of $NH_3$ in water is given by:

${ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} }$

Substituting the given value of $K_b$, we get:

${ 1.8 \times 10^{-5} = \frac{[NH_4^+][OH^-]}{0.20} }$

Q: What is the concentration of $NH_4^+$ in the solution?

A: We can use the fact that $[OH^-] = \frac{K_w}{[H_3O^+]}$ to substitute for $[OH^-]$ in the above equation:

${ [NH_4^+] \times \frac{K_w}{[H_3O^+]} = 3.6 \times 10^{-6} }$

Substituting the value of $K_w$, we get:

${ [NH_4^+] = \frac{3.6 \times 10^{-6} \times [H_3O^+]}{10^{-14}} }$

Substituting the value of $[H_3O^+] = 10^{-7}$, we get:

${ [NH_4^+] = \frac{3.6 \times 10^{-6} \times 10^{-7}}{10^{-14}} }$

${ [NH_4^+] = 3.6 \times 10^{-3} }$

Q: What is the concentration of $NH_4^+$ from the dissociation of $NH_4Cl$ in water?

A: We are given that the concentration of $NH_4Cl$ is $0.15 \, M$. We can assume that the concentration of $Cl^-$ is also $0.15 \, M$. Therefore, we can substitute these values into the equation:

${ K_s = [NH_4^+][Cl^-] }$

Substituting the value of $K_s$, we get:

${ [NH_4^+] = \frac{K_s}{0.15} }$

Substituting the value of $K_s$, we get:

${ [NH_4^+] = \frac{0.15}{0.15} }$

${ [NH_4^+] = 1 }$

Q: What is the total concentration of $NH_4^+$ in the solution?

A: The total concentration of $NH_4^+$ in the solution is the sum of the concentrations of $NH_4^+$ from the dissociation of $NH_3$ in water and the dissociation of $NH_4Cl$ in water:

${ [NH_4^+]_{total} = [NH_4^+]_{NH_3} + [NH_4^+]_{NH_4Cl} }$

Substituting the values, we get:

${ [NH_4^+]_{total} = 3.6 \times 10^{-3} + 1 }$

${ [NH_4^+]_{total} = 1.0036 }$

Q: What is the pH of the solution?

A: Now that we have the total concentration of $NH_4^+$ in the solution, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

Substituting the values, we get:

${ pH = 9.26 + \log \left( \frac{0.15}{1.0036} \right) }$

${ pH = 9.26 + \log (0.1497) }$

${ pH = 9.26 - 0.72 }$

${ pH = 8.54 }$

Therefore, the pH of the solution is 8.54.

Q: What is the relationship between the pH of the solution and the concentrations of $NH_3$ and $NH_4Cl$?

A: The pH of the solution is related to the concentrations of $NH_3$ and $NH_4Cl$ through the Henderson-Hasselbalch equation:

${ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) }$

Substituting the values, we get:

${ pH = 9.26 + \log \left( \frac{0.15}{1.0036} \right) }$

This equation shows that the pH of the solution is dependent on the ratio of the concentrations of $NH_3$ and $NH_4Cl$.

Q: What is the significance of the pH of the solution?

A: The pH of the solution is an important parameter in chemistry, as it determines the acidity or basicity of the solution. In this case, the pH of the solution is 8.54, which is slightly basic.

Q: What are the implications of the pH of the solution?

A: The pH of the solution has important implications in various fields, such as chemistry, biology, and medicine. For example, the pH of the solution can affect the activity of enzymes, the growth of microorganisms, and the stability of proteins.

Q: What are the limitations of the Henderson-Hasselbalch equation?

A: The Henderson-Hasselbalch equation is a useful tool for calculating the pH of a solution, but it has some limitations. For example, it assumes that the solution is a binary mixture of two components, which may not always be the case. Additionally, it assumes that the concentrations of the components are known, which may not always be the case.

Q: What are the future directions for research on the pH of solutions?

A: There are many future directions for research on the pH of solutions, including the development of new methods for calculating the pH of complex solutions, the study of the effects of pH on biological systems, and the development of new applications for pH-sensitive materials.