Chapter 9, Problem 55QAP

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

Chapter
Section

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
17 views

# A common method for determining how much chloride ion is present in a sample is to precipitate the chloride from an aqueous solution of the sample with silver nitrate solution and then to weigh the silver chloride that results. The balanced net ionic reaction is:math> Ag + ( a q ) + Cl − ( a q ) → AgCl ( s ) ppose a 5.45-g sample of pure sodium chloride is dissolved in water and is then treated with a solution containing 1.15 g of silver nitrate. Will this quantity of silver nitrate be capable of precipitating all the chloride ion from the sodium chloride sample?

Interpretation Introduction

Interpretation:

If the quantity of silver nitrate is capable of precipitating all the chloride ion from the sodium chloride sample should be calculated.

Concept Introduction:

The balanced chemical reaction is as follows:

Ag+(aq)+Cl(aq)AgCl(s)

Here, chloride ion precipitates out from the solution in the form of AgCl salt.

The mass of sodium chloride dissolved in water is 5.45 g, this is then treated with a solution containing 1.15 g of silver nitrate.

From the precipitation reaction, 1 mol of Ag+ ion reacts with 1 mol of Cl to form 1 mol of AgCl.

Explanation

The mass of sodium chloride is 5.45 g and its molar mass is 58.44 g/mol

Thus, its number of moles will be:

n=mM

Putting the values,

n=5.45 g58.44 g/mol=0.09325 mol

Thus, number of moles of chloride ion will be 0.09325 mol.

Similarly, mass of silver nitrate is 1.15 g and molar mass of silver nitrate is 169.87 g/mol thus, number of moles will be:

n=mM

Putting the values,

n=1

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