   Chapter 15, Problem 127AP ### 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
8 views

# How many grams of Ba (NO3)2are required to precipitate all the sulfate ion present in 15.3 mL of 0.139 M H2SO4solution? Ba ( NO 3 ) 2 ( aq ) + H 2 SO 4 ( aq ) → BaSO 3 ( s ) + 2HNO 3 ( aq )

Interpretation Introduction

Interpretation:

The mass of BaNO32 required to precipitate all the sulfate ion present in the given solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:

M=nV

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

The relation between number of moles and mass of a substance is given as:

n=mMm

Where,

• m represents the mass of the substance.
• Mm represents the molar mass of the substance.
Explanation

The molarity of the given H2SO4 solution is 0.139M.

The volume of the 0.139MH2SO4 solution is 15.3mL.

The molar mass of BaNO32 is 261.337gmol1.

The molarity of a solution is given as:

M=nV

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

Rearrange the above equation for the value of n.

n=MV

Substitute the value of molarity and volume of H2SO4 solution in above expression.

n=0.139M1molL11M15.3mL1L1000mL=2.127×103mol

The number of moles of H2SO4 present in 0.139MH2SO4 is 2.127×103mol.

One mole of SO42 ion is produced when one mole of H2SO4 is dissolved in water. Hence the number of moles of SO42 ion present in the 0.139MH2SO4 is 2.127×103mol

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