### Titration of HCl with NaOH: Understanding the Reaction**Balanced Equation for Titration:**- The chemical reaction for the titration of hydrochloric acid (HCl) with sodium hydroxide (NaOH) is as follows: \[ \text{HCl(aq) + NaOH(aq) → NaCl(aq) + H}_2\text{O(aq)} \]**Neutralization Process:**- To neutralize one mole of HCl, one mole of NaOH is required. The progress during titration is determined by comparing moles of HCl and NaOH:| Condition | Progress of the Reaction ||------------------------------------------------|---------------------------------|| number of moles of HCl > number of moles of NaOH | before the equivalence point || number of moles of HCl = number of moles of NaOH | at the equivalence point || number of moles of HCl < number of moles of NaOH | after the equivalence point |**Example Calculation:**- **Scenario:** Titrating 50 mL of 0.1 M HCl with 25 mL of 0.1 M NaOH.1. **Calculate Moles of HCl:** - Use the formula: \[ \text{number of moles} = \text{molarity (mol/L)} \times \text{volume (in liters)} \] - For 50 mL of 0.1 M HCl: \[ 0.1 \, \text{mol/L} \times 50 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.005 \, \text{mol} \]2. **Calculate Moles of NaOH:** - For 25 mL of 0.1 M NaOH: \[ 0.1 \, \text{mol/L} \times 25 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0025 \, \text{mol} \]**Conclusion:**- Since the number A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.**Drag the appropriate items to their respective bins.**- **Before equivalence point** - **At equivalence point** - **After equivalence point** **Options to Drag:**- 150 mL of 1.00 M NaOH- 10.0 mL of 1.00 M NaOH- 5.00 mL of 1.00 M NaOH- 100 mL of 1.00 M NaOH- 50.0 mL of 1.00 M NaOH- 200 mL of 1.00 M NaOH**Interactive Elements:**- View Available Hint(s)- Reset- Help(Note: This is part of an interactive educational activity where students classify given volumes of NaOH as occurring before, at, or after the equivalence point in a titration setup.)

The neutralization reaction between HCl and NaOH is NaOH +HCl ---->> NaCl + H2O So same…

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to ther flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence poi Drag the appropriate items to their respective bins. • View Available Hint(s) Reset Hel 150 mL of 1.00 M NaOH 10.0 mL of 1.00 M NAOH 5.00 mL of 1.00 M NaOH 100 mL of 1.00 M NaOH 200 mL of 1.00 M NaOH 50.0 mL of 1.00 M NaOH Before equivalence point At equivalence point After equivalence point

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to ther flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence poi Drag the appropriate items to their respective bins. • View Available Hint(s) Reset Hel 150 mL of 1.00 M NaOH 10.0 mL of 1.00 M NAOH 5.00 mL of 1.00 M NaOH 100 mL of 1.00 M NaOH 200 mL of 1.00 M NaOH 50.0 mL of 1.00 M NaOH Before equivalence point At equivalence point After equivalence point

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Only first 3 parts please A ,B,C
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