Chapter 4, Problem 139SCQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Two students titrate different samples of the same solution of HCl using 0.100 M NaOH solution and phenolphthalein indicator (Figure 4.12). The first student pipets 20.0 mL of the HCl solution into a flask, adds 20 mL of distilled water and a few drops of phenolphthalein solution, and titrates until a lasting pink color appears. The second student pipets 20.0 mL of the HCl solution into a flask, adds 60 mL of distilled water and a few drops of phenolphthalein solution, and titrates to the first lasting pink color. Each student correctly calculates the molarity of an HCl solution. What will the second student’s result be? (a) four times less than the first student’s result (b) four times greater than the first student’s result (c) two times less than the first student’s result (d) two times greater than the first student’s result (e) the same as the first student’s result

Interpretation Introduction

Interpretation:

The molarity of HCl solution used for titration by two students has to be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

Molarity=MassperlitreMolecular mass

• A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
• Concentrationofsubstance=Amountof substancevolumeofthesubstance
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
Explanation

Reason for correct option:

Given that two students titrate different samples of same solution of HCl using 0.100â€‰Mâ€‰â€‰NaOH.

HClâ€‰+â€‰NaOHâ†’â€‰NaClâ€‰+â€‰H2O

Consider the case of first student:

Amount of HCl in the 20â€‰mL solution can be calculated as follows,

Amountâ€‰ofâ€‰HClâ€‰=â€‰CHClÃ—â€‰VHCl

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  =â€‰0.100â€‰â€‰molâ€‰HClLâ€‰Ã—â€‰0.02â€‰Lâ€‰â€‰=â€‰0.002â€‰â€‰molâ€‰HCl

0.002â€‰â€‰molâ€‰HCl is in 20â€‰mL, when it is diluted to 40.00â€‰mL the concentration will be,

Concentrationâ€‰ofâ€‰HClâ€‰=Amountâ€‰ofÂ HClvolumeâ€‰ofâ€‰theâ€‰HClâ€‰â€‰

Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â Â Â Â 0.002â€‰â€‰mol0.04â€‰Lâ€‰â€‰=â€‰0.05â€‰â€‰M

Consider the case of second student:

Amount of HCl in the 20â€‰mL solution can be calculated as follows,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Amountâ€‰ofâ€‰HClâ€‰=â€‰CHClÃ—â€‰VHCl

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  =â€‰0

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