   Chapter 20, Problem 50IL

Chapter
Section
Textbook Problem

The mineral claudetite contains the element arsenic in the form of arsenic(III) oxide, As2O3. The As2O3 in a 0.562-g sample of the impure mineral was converted first to H3AsO3 and then titrated with a 0.0480 M solution of I3−, which reacts with H3AsO3 according to the following balanced net ionic equationH3AsO3(aq) + 3 H2O(ℓ) + I3−(aq) → H3AsO4(aq) + 2 H3O+(aq) + 3 I−(aq)If the titration required 45.7 mL of the I3− solution, what is the percentage of As2O3 in the mineral sample?

Interpretation Introduction

Interpretation: The percentage of As2O3 in the mineral sample should be determined.

Concept introduction:

Concentration of a solution can be expressed in molarity.

Molarity=NumberofmolesVolumeofsolutioninlitres

Numberofmoles=massmolarmass

Explanation

Given:

H3AsO3+3H2O+I3-H3AsO4+2H3O++3I-

Massofsample=0.562g

Given that 0.0480MI3- reacts with H3AsO3

From the equation it is clear that one mole of H3AsO3 reacts with one mole of I3- to give the product.

Molarity=NumberofmolesVolumeofsolutioninlitres

Numberofmoles=Molarity×Volumeofsolutioninlitres

NumberofmolesofI3-=Molarity×Volumeofsolutioninlitres=0

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