Sodium Carbonate (Na2CO3) is a base and an accurately weighed sample can be used to standardize an acid. A sample of sodium carbonate (0.263 g) requires 28.35 mL of HCI for titration to reach the equivalence point. What is the concentration of HCI? (ans: 0.175 M) NazCO3 + 2HCI → 2 NaCI + H2O+ CO2
Sodium Carbonate (Na2CO3) is a base and an accurately weighed sample can be used to standardize an acid. A sample of sodium carbonate (0.263 g) requires 28.35 mL of HCI for titration to reach the equivalence point. What is the concentration of HCI? (ans: 0.175 M) NazCO3 + 2HCI → 2 NaCI + H2O+ CO2
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![**Educational Content on Titration and Standardization**
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**Experiment: Standardization of Hydrochloric Acid using Sodium Carbonate**
**Objective:**
To determine the concentration of hydrochloric acid (HCl) through titration with a known quantity of sodium carbonate (Na₂CO₃).
**Background:**
Sodium carbonate is a base that can be accurately weighed to standardize an acid. Through the process of titration, the equivalence point is achieved when the amount of acid equals the amount of base present in a solution, allowing for the calculation of the acid's concentration.
**Chemical Reaction:**
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
**Procedure Details:**
- A known mass of sodium carbonate (0.263 g) is used.
- The volume of HCl solution required to reach the equivalence point is 28.35 mL.
**Calculation Goal:**
Determine the concentration of the HCl solution. Given the answer: 0.175 M.
**Steps for Calculation (as shown visually in the image):**
1. **Given:**
- Mass of Na₂CO₃: 0.263 g
- Volume of HCl used: 28.35 mL
2. **Conversion to Liters:**
- Volume in liters: \( \frac{28.35 \, \text{mL}}{1000 \, \text{mL/L}} = 0.02835 \, \text{L} \)
3. **Use the balanced equation:**
- Na₂CO₃ reacts with 2 moles of HCl.
4. **Calculate moles of Na₂CO₃:**
- Molar mass of Na₂CO₃ is approximately 106 g/mol.
- Moles of Na₂CO₃ = \( \frac{0.263 \, \text{g}}{106 \, \text{g/mol}} \)
5. **Mole Ratio:**
- From the equation, 1 mole of Na₂CO₃ reacts with 2 moles of HCl.
6. **Find moles of HCl required:**
- Multiply the moles of Na](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F163dc8ba-851f-4764-b6aa-51728fdb757c%2F864ddb71-8c3d-4f65-a0f2-c73c1e484210%2F7qo2yf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content on Titration and Standardization**
---
**Experiment: Standardization of Hydrochloric Acid using Sodium Carbonate**
**Objective:**
To determine the concentration of hydrochloric acid (HCl) through titration with a known quantity of sodium carbonate (Na₂CO₃).
**Background:**
Sodium carbonate is a base that can be accurately weighed to standardize an acid. Through the process of titration, the equivalence point is achieved when the amount of acid equals the amount of base present in a solution, allowing for the calculation of the acid's concentration.
**Chemical Reaction:**
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
**Procedure Details:**
- A known mass of sodium carbonate (0.263 g) is used.
- The volume of HCl solution required to reach the equivalence point is 28.35 mL.
**Calculation Goal:**
Determine the concentration of the HCl solution. Given the answer: 0.175 M.
**Steps for Calculation (as shown visually in the image):**
1. **Given:**
- Mass of Na₂CO₃: 0.263 g
- Volume of HCl used: 28.35 mL
2. **Conversion to Liters:**
- Volume in liters: \( \frac{28.35 \, \text{mL}}{1000 \, \text{mL/L}} = 0.02835 \, \text{L} \)
3. **Use the balanced equation:**
- Na₂CO₃ reacts with 2 moles of HCl.
4. **Calculate moles of Na₂CO₃:**
- Molar mass of Na₂CO₃ is approximately 106 g/mol.
- Moles of Na₂CO₃ = \( \frac{0.263 \, \text{g}}{106 \, \text{g/mol}} \)
5. **Mole Ratio:**
- From the equation, 1 mole of Na₂CO₃ reacts with 2 moles of HCl.
6. **Find moles of HCl required:**
- Multiply the moles of Na
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