Chapter 4, Problem 111GQ

### 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

# One half liter (500. mL) of 2.50 M HCl is mixed with 250. mL of 3.75 M HCl. Assuming the total solution volume after mixing is 750. mL, what is the concentration of hydrochloric acid in the resulting solution? What is its pH?

Interpretation Introduction

Interpretation:

pH and concentration of HCl solution formed by mixing of two HCl solutions of different concentrations has to be determined.

Concept introduction:

• Strong acids dissociates completely into ions in solution but weak acids do not.
• pH of a solution is the negative of the base -10 logarithm of the hydronium ion concentration.

pH=-log[H3O+]

• Concentration of hydronium ion [H3O+]=10-pH
• Amountofsubstance=Concnetrationofthesubstance×Volume
• Concnetrationofthesubstance=AmountofsubstanceVolume
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
•  For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
Explanation

Here one half liter of 2.50â€‰Mâ€‰HCl is mixed with 250.0â€‰mLâ€‰ofâ€‰â€‰3.75â€‰Mâ€‰HCl. Â Total volume of the solution after mixing is 750â€‰mL.

The concentration of the resulting solution can be determined by calculating the amount of HCl in two different concentrations given.

From the concentration of HCl and its volume amount of HCl calculated as follows,

Amountâ€‰â€‰ofâ€‰â€‰HCl=â€‰â€‰Concnetrationâ€‰HClâ€‰Ã—â€‰Volumeâ€‰HCl

Amount of HCl in one half liter of 2.50â€‰Mâ€‰HCl is,

â€‚Â 2.50â€‰molLÃ—0.5â€‰Lâ€‰â€‰=â€‰1.25â€‰â€‰molâ€‰HCl

Amount of HCl in 250.0â€‰mLâ€‰ofâ€‰â€‰3.75â€‰Mâ€‰HCl is,

â€‚Â 3.75â€‰molLÃ—0.25â€‰Lâ€‰â€‰=â€‰0.9375â€‰â€‰molâ€‰HCl

Amount of HCl in 750.0â€‰mL solution is,

Â Â Â Â 0.9375â€‰â€‰molâ€‰HClâ€‰â€‰+â€‰â€‰1

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