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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: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.97QE: According to the Resource Conservation and Recovery Act (RCRA), waste material is classified as...
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Consider the balanced equation for the titration of an acid, HCl, with a base, NaOH: HC1(aq) + NaOH(aq)→NaCl(aq) + H2O(aq) According to the balanced equation, to neutralize one mole of the acid (HCl) you will have to add one mole of the base (NaOH). Based on the number of moles of HCl and NaOH at a particular stage of titration, the progress of the reaction with respect to the equivalence point can be decided, as shown in the following table: 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 For example, if you are titrating 50 mL of 0.1 M HCl with 25 mL of 0.1 M NaOH solution, the number of moles present in the solution can be determined from the molarity (M) of the solution and the volume of the solution in liters (L) by using the equation number of moles = molarity (mol/L) × volume (in liters) The number of moles of HCl present in 50 mL of 0.1 M HCl solution is 0.1 mol/L × 50 mL × 1 L 1000 mL 0.005 mol Based on similar calculations, the number of moles of NaOH present in 25 mL of 0.1 M NaOH solution is 0.0025 mol. Since the number of moles of NaOH is less than the number of moles of HCl, this condition occurs before the equivalence point.
Transcribed Image Text:Consider the balanced equation for the titration of an acid, HCl, with a base, NaOH: HC1(aq) + NaOH(aq)→NaCl(aq) + H2O(aq) According to the balanced equation, to neutralize one mole of the acid (HCl) you will have to add one mole of the base (NaOH). Based on the number of moles of HCl and NaOH at a particular stage of titration, the progress of the reaction with respect to the equivalence point can be decided, as shown in the following table: 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 For example, if you are titrating 50 mL of 0.1 M HCl with 25 mL of 0.1 M NaOH solution, the number of moles present in the solution can be determined from the molarity (M) of the solution and the volume of the solution in liters (L) by using the equation number of moles = molarity (mol/L) × volume (in liters) The number of moles of HCl present in 50 mL of 0.1 M HCl solution is 0.1 mol/L × 50 mL × 1 L 1000 mL 0.005 mol Based on similar calculations, the number of moles of NaOH present in 25 mL of 0.1 M NaOH solution is 0.0025 mol. Since the number of moles of NaOH is less than the number of moles of HCl, this condition occurs before the 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 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. • View Available Hint(s) Reset Help 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
Transcribed Image Text: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. • View Available Hint(s) Reset Help 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
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