Understanding Equivalence Point and Concentration Calculation

Understanding Equivalence Point and Concentration Calculation

Assessment

Interactive Video

Chemistry, Science

10th - 12th Grade

Hard

Created by

Jackson Turner

FREE Resource

The video tutorial explains the concept of the equivalence point in titration, where the amount of titrant added is stoichiometrically equivalent to the substance being titrated. It demonstrates how to calculate the moles of hydrochloric acid used in a titration and convert units from milliliters to liters. The tutorial further explains how to determine the concentration of potassium sorbate in a stock solution by using the calculated moles of hydrochloric acid and the initial volume of the solution.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial volume of hydrochloric acid used in the problem?

29.95 mL

45.00 mL

50.00 mL

60.00 mL

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the equivalence point signify in a titration?

The point where the reaction starts

The point where the solution changes color

The point where all reactants are consumed

The point where the number of moles of titrant equals the number of moles of analyte

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the volume of hydrochloric acid converted to liters?

By multiplying by 1000

By dividing by 1000

By adding 1000

By subtracting 1000

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the molarity of the hydrochloric acid used in the titration?

2.00 M

1.00 M

1.25 M

1.50 M

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How many moles of hydrochloric acid are used in the reaction?

0.0500 moles

0.0450 moles

0.0374 moles

0.0600 moles

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial volume of the potassium sorbate stock solution?

29.95 mL

45.00 mL

60.00 mL

50.00 mL

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the concentration of potassium sorbate calculated?

By dividing moles of potassium sorbate by volume in liters

By multiplying moles of potassium sorbate by volume in liters

By adding moles of potassium sorbate to volume in liters

By subtracting moles of potassium sorbate from volume in liters

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