   # The vanadium in a sample of ore is converted to VO 2+ . The VO 2+ ion is subsequently titrated with MnO 4 − in acidic solution to form V(OH) 4 + and manganese(II) ion. The unbalanced titration reaction is M n O 4 − ( a q ) + V O 2 + + ( a q ) + H 2 O ( l ) → V ( O H ) 4 + ( a q ) + M n 2 + ( a q ) + H + ( a q ) To titrate the solution, 26.45 mL of 0.02250 M MnO 4 − was required. If the mass percent of vanadium in the ore was 58.1 %, what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method. ### 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 144IP
Textbook Problem
20 views

## The vanadium in a sample of ore is converted to VO2+. The VO2+ ion is subsequently titrated with MnO4− in acidic solution to form V(OH)4+ and manganese(II) ion. The unbalanced titration reaction is M n O 4 − ( a q )   +   V O 2 + + ( a q ) +   H 2 O ( l )   →   V ( O H ) 4 + ( a q )   + M n 2 + ( a q ) + H + ( a q ) To titrate the solution, 26.45 mL of 0.02250 M MnO4− was required. If the mass percent of vanadium in the ore was 58.1 %, what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

Interpretation Introduction

Interpretation:

The mass of the ore sample has to be calculated.

Concept Introduction:

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

Massofcompound(ing)=Moles(ing)Masspercent(in%)

### Explanation of Solution

Explanation

To calculate the mass of ore sample

To balance the reaction by oxidation state method.

The chemical reaction can be given as,

MnO4-(aq)+VO2+(aq)+H2O(l)V(OH)4+(aq)+Mn2+(aq)+H+(aq)

The oxidation states for various ions are,

VO2+:O,-2V,x+(-2)=+2x=+4

MnO4-:O,-2Mn,x+4(-2)=-1x=+4

V(OH)4+:O,-2,H,+1V,x+4(-2)+4+(+1)=+1x=+5

Mn2+:Mn,+2

The oxidation state of vanadium increases from +4 to +5 in V(OH)4+.

The oxidation state of manganese decreases from +7 to +2 in Mn2+.

Thus balancing the equation, we get

MnO4-(aq)+5VO2+(aq)+11H2O(l)5V(OH)4+(aq)+Mn2+(aq)+22H+(aq)

The unbalanced chemical equation is balanced with the help of oxidation method. It was found that the oxidation state of vanadium increases in V(OH)4+ and oxidation state of manganese decreases in Mn2+

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Cells a. are self-contained, living units. b. serve the bodys needs but have few needs of their own. c. remain ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What types of forces give rise to quaternary structure?

Chemistry for Today: General, Organic, and Biochemistry

6.11 Define the term photon.

Chemistry for Engineering Students

How can errors in the cell cycle lead to cancer in humans?

Human Heredity: Principles and Issues (MindTap Course List)

Distinguish between acids and bases. What are their properties?

Biology: The Dynamic Science (MindTap Course List)

The string shown in Figure P16.11 is driven at a frequency of 5.00 Hz. The amplitude of the motion is A = 12.0 ...

Physics for Scientists and Engineers, Technology Update (No access codes included) 