Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.151P
Interpretation Introduction

Interpretation:

The molarities of NH4+ and PO43 are to be determined.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount(mol)ofsolutevolume (L)ofsolution (1)

Expert Solution & Answer
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Answer to Problem 13.151P

The molarities of NH4+ and PO43 are 7.98×103M and 1.48×103M.

Explanation of Solution

The dissociation of NH4NO3 occurs as follows:

NH4NO3NH4++NO3

One mole of NH4NO3 dissociates to give one mole of NH4+ and one mole of NO3.

The dissociation of (NH4)3PO4 occurs as follows:

(NH4)3PO43NH4++PO43

One mole of (NH4)3PO4 dissociates to give three moles of NH4+ and one mole of PO43.

The formula to calculate the moles of NH4+ in NH4NO3 is as follows:

MolesofNH4+=(Given mass ofNH4NO3Molar mass ofNH4NO3)(1molofNH4+1molofNH4NO3) (2)

Substitute 5.56g for the given mass of NH4NO3 and 80.05g/mol for the molar mass of NH4NO3 in equation (2).

MolesofNH4+=(5.56g)(1mol80.05g)(1molofNH4+1molofNH4NO3)=7.07058×102mol

The formula to calculate the moles of NH4+ in (NH4)3PO4 is as follows:

MolesofNH4+=(Given mass of(NH4)3PO4Molar mass of(NH4)3PO4)(3molofNH4+1molof(NH4)3PO4) (3)

Substitute 4.42g for the given mass of (NH4)3PO4 and 149.10g/mol for the molar mass of (NH4)3PO4 in equation (3).

MolesofNH4+=(4.42g)(1mol149.10g)(3molofNH4+1molof(NH4)3PO4)=8.89336×102mol

The total moles of NH4+ is calculated as follows:

MolesofNH4+=MolesofNH4+inNH4NO3+MolesofNH4+in(NH4)3PO4 (4)

Substitute 7.07058×102mol for the moles of NH4NO3 and 8.89336×102mol for the moles of (NH4)3PO4 in equation (4).

MolesofNH4+=7.07058×102mol+8.89336×102mol=15.96394×102mol

Substitute 15.96394×102mol for the amount of solute and 20L for the volume of the solution in equation (1) to calculate the molarity of NH4+.

Molarity ofNH4+=15.96394×102mol20L=7.98197×103M=7.98×103M

The formula to calculate the moles of PO43 in (NH4)3PO4 is as follows:

MolesofPO43=(Given mass of(NH4)3PO4Molar mass of(NH4)3PO4)(1molofPO431molof(NH4)3PO4) (5)

Substitute 4.42g for the given mass of (NH4)3PO4 and 149.10g/mol for the molar mass of (NH4)3PO4 in equation (5).

MolesofPO43=(4.42g)(1mol149.10g)(1molofPO431molof(NH4)3PO4)=2.96445×102mol

Substitute 2.96445×102mol for the amount of solute and 20L for the volume of the solution in equation (1) to calculate the molarity of PO43.

Molarity ofPO43=2.96445×102mol20L=1.482225×103M=1.48×103M

Conclusion

The molarities of NH4+ and PO43 are 7.98×103M and 1.48×103M.

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Chapter 13 Solutions

Chemistry: The Molecular Nature of Matter and Change

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