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.
![**Chemical Reaction Problem**
The given chemical reaction is:
\[ \text{Cl}^- + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NO}_3^- \]
**Problem Statement:**
If the reaction makes 0.003 g of AgCl, how many grams of Cl⁻ were in the water? (Show your work)
**Explanation:**
In this reaction, chloride ions (Cl⁻) react with silver nitrate (AgNO₃) to form silver chloride (AgCl) and nitrate ions (NO₃⁻). This particular question is asking for the mass of chloride ion (Cl⁻) that was needed to produce a given mass of silver chloride (AgCl).
To find out how many grams of chloride ions were in the water, we need to follow these steps:
1. Calculate the molar mass of AgCl.
2. Determine moles of AgCl produced.
3. Use stoichiometry to find moles of Cl⁻.
4. Convert moles of Cl⁻ to grams.
**Solution:**
1. **Calculate the molar mass of AgCl:**
\[ \text{Molar mass of AgCl} = \text{Molar mass of Ag} + \text{Molar mass of Cl} \]
\[ \text{= 107.87 g/mol (Ag) + 35.45 g/mol (Cl)} \]
\[ \text{= 143.32 g/mol} \]
2. **Determine moles of AgCl produced:**
\[ \text{Moles of AgCl} = \frac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}} \]
\[ \text{= } \frac{0.003 \text{ g}}{143.32 \text{ g/mol}} \]
\[ \text{= 2.09} \times 10^{-5} \text{ mol} \]
3. **Use stoichiometry to find moles of Cl⁻:**
According to the balanced equation, 1 mole of AgCl is produced from 1 mole of Cl⁻. Therefore, moles of Cl⁻ is the same as moles of AgCl:
\[ \text{Moles of Cl}^- = 2.09 \times 10^{-5} \text{ mol} \]
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