   Chapter 15.7, Problem 15.9SC ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
5 views

# ercise 15.9 Calculate the volume of 0.10 M HNO3 needed to neutralize 125 mL of 0.050 M KOH.

Interpretation Introduction

Interpretation:

The volume of 0.100 M HNO3 which is required to neutralize 125 mL of 0.050 M KOH should be calculated.

Concept Introduction:

Molarity of the solution is equal to the ratio of number of moles of solute to the volume of solution in litres.

The mathematical expression is given by:

Molarity = number of moles of solute volume of solution in litres

Number of moles = given massmolar mass.

Explanation

Since, nitric acid is a strong acid, thus it completely dissociates into ions. The reaction is shown below:

HNO3H++NO3

And, KOH dissociates into K+ and OH. On mixing these two solutions, H+ ion and OH ion reacts and form H2O.

The reaction is shown below:

H+(aq)+OH(aq)H2O(l)

According to the reaction, 1 mole of H+ ion and 1 mole OH ion reacts to form 1 mole of water. Thus, both ions present in 1:1.

Now, calculate the number of moles of OH present in 125 mL of 0.050 M KOH

Convert volume in litres:

Since, 1 L = 1000 mL

Volume in L = 125 mL×1 L1000 mL=0

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