The base protonation constant K of azetidine (C²H²NH) is 1.5 × 10¯8. Calculate the pH of a 0.39 M solution of azetidine at 25 °C. Round your answer to 1 decimal place. pH = 0 X ?
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
![### Calculating the pH of an Azetidine Solution
#### Problem Statement
The base protonation constant \( K_b \) of azetidine (\( C_3H_6NH \)) is \( 1.5 \times 10^{-8} \). Calculate the pH of a 0.39 M solution of azetidine at 25 °C. Round your answer to 1 decimal place.
#### Given Data
- \( K_b \) for azetidine (\( C_3H_6NH \)) = \( 1.5 \times 10^{-8} \)
- Concentration of azetidine (\( C_3H_6NH \)) = 0.39 M
- Temperature = 25 °C
#### Calculation Steps
1. **Identify the initial concentration and \( K_b \).**
Since azetidine is a base, we use the base dissociation constant (\( K_b \)) to find the hydroxide ion (\( OH^- \)) concentration.
2. **Set up the expression for \( K_b \).**
\[
K_b = \frac{[OH^-][C_3H_6NH_2^+]}{[C_3H_6NH]}
\]
Given that \( [C_3H_6NH] \) = 0.39 M initially and assuming \( x \) is the concentration of \( OH^- \) formed:
\[
K_b = \frac{x^2}{0.39 - x} \approx \frac{x^2}{0.39}
\]
The approximation assumes \( x \) is much smaller than the initial concentration.
3. **Solve for \( x \).**
\[
1.5 \times 10^{-8} = \frac{x^2}{0.39}
\]
\[
x^2 = 1.5 \times 10^{-8} \times 0.39
\]
\[
x^2 = 5.85 \times 10^{-9}
\]
\[
x = \sqrt{5.85 \times 10^{-9}}
\]
\[
x \approx 7.65 \times 10^{-5}
\]
Here, \( x \) represents](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc60b291-b343-46d1-985f-51e0feae8cbd%2Fb939474f-9267-42e7-bd04-b6cb549c281b%2F1uo472k_processed.png&w=3840&q=75)
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