   # A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess H + or OH − ions left in solution. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 74E
Textbook Problem
1 views

## A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess H+ or OH− ions left in solution.

Interpretation Introduction

Interpretation: The concentration of H+ and OH- should be determined for the given mixture also the nature (acid or base) of the mixture to be determined.

Concept Introduction: The neutralization of acids and bases results in the formation of salts.

The concentration of solution can be defined in terms of molarity as moles of solute per volume of solution in litres. It is given by the equation,

Molarity=Molesofsolute(in moles)Volumeofsolution(inlitres)

### Explanation of Solution

Explanation

Record the data from the question.

Volume and Molarity of hydrochloric acid = 50 ml and 0.100M

Volume and Molarity of nitric acid         =100 ml and 0.200M

Volume and Molarity of calcium hydroxide=500ml and 0.0100M

Volume and Molarity of rubidium hydroxide=200ml and 0.100M

The volume and molarity of two different acids and two different bases are given and their data are recorded as shown above.

• Calculate the moles of acids and bases.

Moles of acids =0.05000.100molHClL×1molH+molHCl+0.1000L×0.200molHNO3L×1molH+molHNO3

=0.00500+0.0200

= 0.0250molH+

Moles of base =0.50000.0100molCa(OH)2L×2molOH-molCa(OH)2+0.2000L×0.100molRbOHL×1molOH-molRbOH

=0.0100+0.0200=0.0300molOH-

The given acids and bases are strong in nature. By plugging in the values of volumes and molarity, the moles of acids and bases were found to be 0.0250molH+and0

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