   Chapter 15, Problem 65QAP ### 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

# 65. Many metal ions are precipitated from solution by the sulfide ion. As an example, consider treating a solution of copper(II) sulfate with sodium sulfide solution: CuSO 4 ( aq ) + Na 2 S ( aq ) → CuS ( s ) + Na 2 SO 4 ( aq ) What Volume of 0.105 M Na2S solution would be required to precipitate all of the copper( II) ion from 27.5 mL of 0.121 M CuSO4solution?

Interpretation Introduction

Interpretation:

The volume of 0.105M

Na2S solution which would be required to precipitate all the copper(II) ions from a given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolarmass

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The molarity and volume of CuSO4 solution is given to be 0.121M and 27.5mL respectively.

The conversion of units of 27.5mL into L is done as,

27.5mL=27.51000L=0.0275L

The number of moles of CuSO4 solution is calculated by the formula,

Numberofmoles=Molarity×Volumeofsolution        (1)

Substitute the values of molarity and volume of CuSO4 solution in the equation (1).

Numberofmoles=0.121M×0.0275L=0.0033275moles

The equation for precipitation reaction is shown below.

CuSO4aq+Na2SaqCuSs+Na2SO4aq

It indicates that 0

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