A solution is prepared at 25 °C that is initially 0.16M in ammonia (NH,), a weak base with K, =1.8 × 10 and 0.011 M in ammonium chloride (NH,Cl). Calculate the pH of the solution. Round your answer to 2 decimal places.
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 a Solution Containing Ammonia and Ammonium Chloride
**Problem Statement:**
A solution is prepared at \(25 \, ^\circ \mathrm{C}\) that is initially \(0.16\, M\) in ammonia \((\mathrm{NH}_3)\), a weak base with \(K_b = 1.8 \times 10^{-5}\), and \(0.011\, M\) in ammonium chloride \((\mathrm{NH}_4\mathrm{Cl})\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
To calculate the pH of the solution containing ammonia and ammonium chloride, follow these steps:
1. **Determine the concentration of \(\mathrm{OH}^-\)** using the given \(K_b\):
- Ammonia (\(\mathrm{NH}_3\)) is a weak base, which partially dissociates in water:
\[
\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-
\]
- Using the given \(K_b\) value:
\[
K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}
\]
Given \( [\mathrm{NH}_4^+] \) from \(\mathrm{NH}_4\mathrm{Cl} = 0.011\, M\) and initial \([\mathrm{NH}_3] = 0.16\, M\).
2. **Set up the equilibrium expression:**
\[
1.8 \times 10^{-5} = \frac{(0.011 + x)x}{0.16 - x}
\]
Because \(K_b\) is very small, \(x\), the amount dissociated, will be very small relative to the initial concentrations; thus, we can approximate \(0.011 + x \approx 0.011\) and \(0.16 - x \approx 0.16\).
3. **Simplify the equation:**
\[
1.8 \times 10^{-5}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F868a318f-0624-4ed5-b0cc-007ff8576a9b%2F5f93ccd7-8ddb-453d-91d7-414f55d05498%2Fzj04bqt_processed.png&w=3840&q=75)
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