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**
To calculate the pH of a solution formed by dissolving 0.174 g of CsOH (Cesium Hydroxide) in enough water to make 935 mL of solution, follow these steps:
1. **Determine the number of moles of CsOH:**
- Molar Mass (M.M.) of CsOH = 149.912 g/mol
- Mass of CsOH = 0.174 g
Number of moles = \(\frac{\text{Mass}}{\text{Molar mass}}\)
\[
\text{Number of moles} = \frac{0.174 \text{ g}}{149.912 \text{ g/mol}} = 0.00116 \text{ mol}
\]
2. **Calculate the concentration of the CsOH solution:**
- Volume of solution = 935 mL = 0.935 L
Concentration (C) = \(\frac{\text{Number of moles}}{\text{Volume in liters}}\)
\[
C = \frac{0.00116 \text{ mol}}{0.935 \text{ L}} \approx 0.00124 \text{ M}
\]
3. **Determine the pOH of the solution:**
CsOH is a strong base and dissociates completely in water:
\[
\text{CsOH} \rightarrow \text{Cs}^+ + \text{OH}^-
\]
Therefore, the concentration of \(\text{OH}^-\) ions is the same as the concentration of CsOH:
\[
[\text{OH}^-] = 0.00124 \text{ M}
\]
pOH = \(-\log [\text{OH}^-]\)
\[
\text{pOH} = -\log (0.00124) \approx 2.91
\]
4. **Calculate the pH of the solution:**
Since pH + pOH = 14 (at 25°C),
\[
\text{pH} = 14 - \text{pOH}
\]
\[
\text{pH} = 14 - 2.91 \approx 11.09](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36e576be-50e4-40e4-8c7b-fee84775ba0f%2F2ad10b18-9ccf-45a3-b9b4-f40aedbe4f63%2Fdk0vd4l9_processed.jpeg&w=3840&q=75)
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