Calculate the pH of a solution that is made by dissolving 19.343 grams of Be(NO32 (M.M. 133.020 g/mol) in enough water to make 175.0 mL of solution. pK, for Be(H2O),= 5.70. %3D
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.
![**Problem Statement:**
Calculate the pH of a solution that is made by dissolving 19.343 grams of Be(NO₃)₂ (M.M. 133.020 g/mol) in enough water to make 175.0 mL of solution. The pKa for Be(H₂O)₄²⁺ is 5.70.
**Solution Explanation:**
The goal of this problem is to calculate the pH of a solution made by dissolving beryllium nitrate in water. Let's follow the steps to determine the pH.
1. **Calculate moles of Be(NO₃)₂:**
\[
\text{Moles of Be(NO₃)₂} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{19.343 \text{ grams}}{133.020 \text{ g/mol}} = 0.145 \text{ moles}
\]
2. **Find molarity (M) of Be(NO₃)₂ solution:**
\[
\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} = \frac{0.145 \text{ moles}}{0.175 \text{ L}} = 0.829 \text{ M}
\]
3. **Determine the pKa of Be(H₂O)₄²⁺:**
Given pKₐ of Be(H₂O)₄²⁺ is 5.70.
\[
\text{pKa} = 5.70
\]
4. **Determine concentration of H\(^+\) ions:**
The relationship between Ka and pKa is given by,
\[
\text{Ka} = 10^{-\text{pKa}} = 10^{-5.70} = 2.00 \times 10^{-6}
\]
5. **Calculate the concentration of H\(^+\) ions using the equilibrium expression for Be(H₂O)₄²⁺:**
\[
\text{Ka} = \frac{[\text{H}^+][\text{Be(H₂O)_3OH}^+]}{[\text{Be(H₂O)_4}^2^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36e576be-50e4-40e4-8c7b-fee84775ba0f%2F60ae8106-d986-4de7-ad65-fdfe76a7f9e4%2Ffqxvawi_processed.jpeg&w=3840&q=75)
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