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conjugate base conjugate acid 1:1 There is a ✓stochiometric ratio between NaOH and H₂CO3. Therefore, if we have 5 moles of NaOH, we need 5 moles of not neutralized H₂CO3 for neutralization. If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3, the acid is in excess and is therefore therefore neutralized If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M equal moles to the base H₂CO3, the acid is V the acid is and H₂O is the ✓ [Select] neutralized If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3, [Select] ✓and is therefore not neutralized V and is

conjugate base conjugate acid 1:1 There is a ✓stochiometric ratio between NaOH and H₂CO3. Therefore, if we have 5 moles of NaOH, we need 5 moles of not neutralized H₂CO3 for neutralization. If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3, the acid is in excess and is therefore therefore neutralized If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M equal moles to the base H₂CO3, the acid is V the acid is and H₂O is the ✓ [Select] neutralized If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3, [Select] ✓and is therefore not neutralized V and is

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Both are the same question, first image for context. The second part is asking about whether the acid will be in: excess, limiting reagent, or equal moles to base. The therefore part is referring to whether it will be neutralized or not. Can someone help with this? Thanks
--- **Neutralization of NaOH and \( \text{H}_2\text{CO}_3 \): Stoichiometry and Reaction Details** **Stoichiometry** There is a \( 1:1 \) stoichiometric ratio between NaOH (sodium hydroxide) and \( \text{H}_2\text{CO}_3 \) (carbonic acid). Therefore, for neutralization: - If we have 5 moles of NaOH, we need **5 moles** of \( \text{H}_2\text{CO}_3 \). **Experimental Scenarios** 1. **Scenario 1:** - **Solutions Used:** - 5 mL of 1 M NaOH - 2.5 mL of 2 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The acid is **in excess** and is therefore **not neutralized**. 2. **Scenario 2:** - **Solutions Used:** - 10 mL of 0.1 M NaOH - 16 mL of 0.05 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The acid is in **equal moles to the base** and is therefore **neutralized**. 3. **Scenario 3:** - **Solutions Used:** - 2 mL of 5 M NaOH - 5 mL of 3 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The conditions for whether the acid is in excess, less, or neutralized are to be selected. --- **Note:** Ensure appropriate safety and measurement practices in all experimental setups involving chemical reactions.
Transcribed Image Text:--- **Neutralization of NaOH and \( \text{H}_2\text{CO}_3 \): Stoichiometry and Reaction Details** **Stoichiometry** There is a \( 1:1 \) stoichiometric ratio between NaOH (sodium hydroxide) and \( \text{H}_2\text{CO}_3 \) (carbonic acid). Therefore, for neutralization: - If we have 5 moles of NaOH, we need **5 moles** of \( \text{H}_2\text{CO}_3 \). **Experimental Scenarios** 1. **Scenario 1:** - **Solutions Used:** - 5 mL of 1 M NaOH - 2.5 mL of 2 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The acid is **in excess** and is therefore **not neutralized**. 2. **Scenario 2:** - **Solutions Used:** - 10 mL of 0.1 M NaOH - 16 mL of 0.05 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The acid is in **equal moles to the base** and is therefore **neutralized**. 3. **Scenario 3:** - **Solutions Used:** - 2 mL of 5 M NaOH - 5 mL of 3 M \( \text{H}_2\text{CO}_3 \) - **Conclusion:** - The conditions for whether the acid is in excess, less, or neutralized are to be selected. --- **Note:** Ensure appropriate safety and measurement practices in all experimental setups involving chemical reactions.
**Text Transcription for Educational Website:** --- **Conjugate Base and Acid Relationships:** - Specify the conjugate base and the conjugate acid in the given reactions. **Stoichiometric Ratio and Neutralization:** - There is a 1:1 stoichiometric ratio between NaOH and H₂CO₃. Therefore, if we have 5 moles of NaOH, we need 5 moles of H₂CO₃ for neutralization. **Scenario Analysis:** 1. **Scenario 1:** - If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO₃, the acid is in excess and is therefore not neutralized. 2. **Scenario 2:** - If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M H₂CO₃, the acid is equal moles to the base and is therefore neutralized. 3. **Scenario 3:** - If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO₃, the result needs to be selected from either neutralized or not neutralized. **Note:** - Users can select the appropriate option from dropdown menus to determine whether the acid is neutralized or not based on the reaction conditions provided. --- This setup allows users to apply the concepts of stoichiometry and acid-base relationships in practical scenarios, enhancing their understanding of chemistry.
Transcribed Image Text:**Text Transcription for Educational Website:** --- **Conjugate Base and Acid Relationships:** - Specify the conjugate base and the conjugate acid in the given reactions. **Stoichiometric Ratio and Neutralization:** - There is a 1:1 stoichiometric ratio between NaOH and H₂CO₃. Therefore, if we have 5 moles of NaOH, we need 5 moles of H₂CO₃ for neutralization. **Scenario Analysis:** 1. **Scenario 1:** - If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO₃, the acid is in excess and is therefore not neutralized. 2. **Scenario 2:** - If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M H₂CO₃, the acid is equal moles to the base and is therefore neutralized. 3. **Scenario 3:** - If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO₃, the result needs to be selected from either neutralized or not neutralized. **Note:** - Users can select the appropriate option from dropdown menus to determine whether the acid is neutralized or not based on the reaction conditions provided. --- This setup allows users to apply the concepts of stoichiometry and acid-base relationships in practical scenarios, enhancing their understanding of chemistry.
Expert Solution
Step 1

Given : 

The stoichiometry of the reacting species is equal or 1 : 1  .

The neutralisation  reaction occuring between acetic acid and sodium hydroxide.

 The various concentration and volume of reacting species that are sodium hydroxide and acetic acid are given.

To find : 

Weather the reaction is completely neutralised or not 

steps

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