   # A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete. ### 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 63E
Textbook Problem
417 views

## A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate.a. Write a balanced chemical equation for any reaction that occurs.b. What precipitate forms?c. What mass of precipitate is produced?d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation has to be written for the reaction that occurs.

Concept Introduction:

Formula equation contains all the ions together. The chemical equation contains the formula of all those compounds taking part in the reaction. The left hand side of the reaction has reactants and the right hand side of the reaction has the products.

To write: The balanced formula equation of the given reaction.

### Explanation of Solution

Potassium hydroxide on reaction magnesium nitrate gives a precipitate of potassium hydroxide with potassium nitrate. Potassium nitrate remains as spectator ions. The chemical equation for this reaction can be written as,

KOH(aq)+Mg(NO3)2(aq)Mg(OH)2(s)+KNO3(aq)

This is unbalanced chemical reaction since the moles of reactant and product are not equal

(b)

Interpretation Introduction

Interpretation:

The precipitate that is formed in the given equation has to be given.

Concept Introduction:

When two solutions containing soluble salts are mixed together, the product obtained is an insoluble salt called as precipitate. Such chemical reactions are called as precipitation reactions.

(c)

Interpretation Introduction

Interpretation:

The amount of precipitate that is produced in the given reaction has to be calculated.

Concept Introduction:

The mass of a compound can be calculated using the molar mass of the compound to the given moles. It can be given by the equation,

Massofcompound(ing)=Moles(ing)×Molarmass(ing)

(d)

Interpretation Introduction

Interpretation:

The concentration of the ions present in the remaining solution has to be calculated.

Concept Introduction:

Concentration of solution can be defined in terms of molarity as moles of solute to the total volume of the solution. It can be given as,

Molarity(inM)=Massofsolute(ing)Volumeofsolution(inL)

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