Chapter 4, Problem 42PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# What is the mass of solute, in grams, in 125 mL of a 1.023 × 10–3 M solution of Na3PO4? What is the molar concentration of the Na+ and PO43– ion?

Interpretation Introduction

Interpretation:

The mass of solute in grams, in 125mL of a 1.023×10-3M solution of Na3PO4 and the molar concentration of Na+andPO43- ions should be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one litre of the solution.

Molarity=MassperlitreMolecular mass

• A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

Numberofmole=GivenmassofthesubstanceMolarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
Explanation

Â Molar concentration of sodium ion [Na+]= 3Ã—0.0012â€‰â€‰Mâ€‰â€‰Na+(aq)â€‰â€‰=â€‰â€‰0.0036â€‰â€‰M

Â Molar concentration of phosphate ion [PO43-â€‰] =0.0012â€‰â€‰M

Â Molar concentration of given Na3PO4 solution is 0.001023â€‰â€‰M and the volume of the solution in litre is .125â€‰L.

Â Before to find the mass of solute in gram, the amount of Na3PO4 required to make 0.001023â€‰â€‰M solution has to be determined.

Â Thus,

Â Amount of Na3PO4 required = .125â€‰Lâ€‰Ã—0.001023â€‰molâ€‰ofâ€‰â€‰Na3PO4â€‰1â€‰â€‰Lâ€‰solutionâ€‰â€‰=â€‰0.0001278molâ€‰

Â The mass of solute (Na3PO4) in grams can be calculated as follows,

0.0001278molâ€‰Na3PO4â€‰Ã—163.94â€‰gâ€‰Na3PO4â€‰â€‰1â€‰molâ€‰Na3PO4â€‰â€‰â€‰â€‰=â€‰0.021â€‰gâ€‰Na3PO4

Â Therefore the mass of solute, in grams, in .125â€‰L of a 0.001023â€‰â€‰M solution of Na3PO4 is 0.021â€‰g.

Â The molar concentrations of Na+â€‰andâ€‰PO43-â€‰ is,

Â Amount of Na3PO4 (in mol) can be calculated as follows,

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