Problem 72
Question
What is the original molarity of an aqueous solution of ammonia \(\left(\mathrm{NH}_{3}\right)\) whose \(\mathrm{pH}\) is 11.22 at \(25^{\circ} \mathrm{C}\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{NH}_{3}=1.8 \times 10^{-5}\right) ?\)
Step-by-Step Solution
Verified Answer
The original molarity of the ammonia solution is approximately 0.153 M.
1Step 1: Understand the Problem
You are given the pH and need to find the original molarity of an ammonia solution. You also have the base dissociation constant \( K_b \) for ammonia.
2Step 2: Convert pH to pOH
Use the relation \( ext{pH} + ext{pOH} = 14 \) to find the pOH. For a pH of 11.22, \( ext{pOH} = 14 - 11.22 = 2.78 \).
3Step 3: Calculate Hydroxide Ion Concentration
The concentration of hydroxide ions \([OH^-]\) is found using \( [OH^-] = 10^{- ext{pOH}} \). Therefore, \( [OH^-] = 10^{-2.78} \approx 1.66 \times 10^{-3} \text{ M} \).
4Step 4: Relate Hydroxide Ion Concentration to Ammonia Concentration
For the equilibrium \( NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \), the hydroxide ion concentration \([OH^-]\) is equal to the concentration of \( NH_4^+ \) at equilibrium.
5Step 5: Use the Base Dissociation Constant
The expression for \( K_b \) is \( K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \). Substitute the known values: \( 1.8 \times 10^{-5} = \frac{(1.66 \times 10^{-3})^2}{[NH_3]} \).
6Step 6: Solve for Ammonia Concentration
Rearrange the equation to find \([NH_3]\): \([NH_3] = \frac{(1.66 \times 10^{-3})^2}{1.8 \times 10^{-5}} \approx 0.153 \text{ M}\).
Key Concepts
Buffer SolutionEquilibrium ConstantAcid-Base EquilibriumAqueous Solution
Buffer Solution
A buffer solution is a special type of aqueous solution that helps maintain a stable pH, even when small amounts of acids or bases are added. This stability arises through the use of a weak acid and its conjugate base, or a weak base and its conjugate acid.
These solutions are particularly useful in chemical reactions and processes that require a constant pH level.
These solutions are particularly useful in chemical reactions and processes that require a constant pH level.
- Buffer solutions resist changes in pH because the weak acid/base present can neutralize added acids or bases.
- They often contain a mixture of a weak acid, such as acetic acid, and its conjugate base, acetate, or a weak base, like ammonia, and its conjugate acid, ammonium.
Equilibrium Constant
The equilibrium constant is a key concept in the study of chemical reactions and helps describe the balance of a reversible reaction at equilibrium.
For reactions in aqueous solutions, this constant helps determine the concentration of different species at equilibrium.
For reactions in aqueous solutions, this constant helps determine the concentration of different species at equilibrium.
- It is represented by different symbols, such as \( K_c \) for reactions in terms of concentration, and \( K_b \) specifically for base dissociation in an aqueous solution.
- The higher the value of the equilibrium constant, the more the products of the reaction are favored at equilibrium.
- For ammonia in this context, the base dissociation constant \( K_b \) provides insight into how readily ammonia forms \( NH_4^+ \) and \( OH^- \) in water.
Acid-Base Equilibrium
The concept of acid-base equilibrium is central to understanding the pH of solutions and how different species interact in aqueous environments.
It involves reactions where acids give up protons \( (H^+) \) and bases accept protons.
It involves reactions where acids give up protons \( (H^+) \) and bases accept protons.
- This equilibrium is dynamic, meaning the reactions constantly move forward and backward until a balance is achieved.
- Acid-base equilibrium calculations often involve using the acid or base dissociation constants (\( K_a \) or \( K_b \)) to predict the concentrations of all species in solution.
- In the context of our example, understanding the equilibrium helps us find the original concentration of ammonia by accounting for its dissociation into \( NH_4^+ \) and \( OH^- \).
Aqueous Solution
An aqueous solution is a solution where water acts as the solvent, dissolving the solute particles.
This is the most common medium for chemical reactions, especially those involving acids and bases.
This is the most common medium for chemical reactions, especially those involving acids and bases.
- Water is a polar solvent, meaning it can dissolve a wide range of substances due to its ability to form hydrogen bonds.
- In an aqueous solution, the dissociation of substances such as acids and bases is facilitated, which is essential for conducting electricity in solutions.
- For ammonia, water aids in its dissociation into \( NH_4^+ \) and \( OH^- \), which is a central point in solving for its concentration based on its pH level.
Other exercises in this chapter
Problem 69
Which of the following has a higher \(\mathrm{pH}\) : (a) \(1.0 \mathrm{M} \mathrm{NH}_{3}\), (b) \(0.20 \mathrm{M} \mathrm{NaOH}\left(K_{\mathrm{b}}\right.\) f
View solution Problem 71
The \(\mathrm{pH}\) of a \(0.30-\mathrm{M}\) solution of a weak base is 10.66 at \(25^{\circ} \mathrm{C}\). What is the \(K_{\mathrm{b}}\) of the base?
View solution Problem 73
Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(0.61-M\) aqueous solution of a weak base \(\mathrm{B}\) with a \(K_{\mathrm{b}}\) of \(1.5 \ti
View solution Problem 74
Determine the \(K_{\mathrm{b}}\) of a weak base if a \(0.19-M\) aqueous solution of the base at \(25^{\circ} \mathrm{C}\) has a \(\mathrm{pH}\) of 10.88 .
View solution