(SHOW CALCULATIONS) You are titrating 25.00 mL of HCI with Ca(OH)2. If 11.20 mL of 0.450 M Ca(OH)₂ is needed to neutralize the acid, what is the molarity of the HCI solution? 2 HCl(aq) + Ca(OH)₂(aq) → 2H₂O(l) +CaCl₂(aq)
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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.
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![### Titration Problem for Determining Molarity
**Problem Statement:**
You are titrating 25.00 mL of HCl with Ca(OH)₂. If 11.20 mL of 0.450 M Ca(OH)₂ is needed to neutralize the acid, what is the molarity of the HCl solution?
**Chemical Equation:**
\[ 2 \text{HCl(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2 \text{H}_2\text{O(l)} + \text{CaCl}_2\text{(aq)} \]
### Detailed Calculations:
To find the molarity of the HCl solution, follow these steps:
1. **Calculate the moles of Ca(OH)₂ used:**
\[
\text{Moles of Ca(OH)}_2 = \text{Molarity of Ca(OH)}_2 \times \text{Volume of Ca(OH)}_2
\]
Given:
- Molarity of Ca(OH)₂ = 0.450 M
- Volume of Ca(OH)₂ = 11.20 mL = 0.01120 L
\[
\text{Moles of Ca(OH)}_2 = 0.450 \, \text{M} \times 0.01120 \, \text{L} = 0.00504 \, \text{moles}
\]
2. **Use the stoichiometry of the reaction to find the moles of HCl:**
The balanced chemical equation shows that 2 moles of HCl react with 1 mole of Ca(OH)₂.
\[
\text{Moles of HCl} = 2 \times \text{Moles of Ca(OH)}_2
\]
\[
\text{Moles of HCl} = 2 \times 0.00504 \, \text{moles} = 0.01008 \, \text{moles}
\]
3. **Calculate the molarity of HCl:**
\[
\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F776f865c-aff7-4060-8b66-2ed96dd6121a%2F85f18e1c-909a-415c-9321-5a1d2ef73f9e%2F1rrutlq_processed.jpeg&w=3840&q=75)
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