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Chapter 9: Problem 73

The traditional method of analysis for the amount of chloride ion prescnt in a sample is to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample. Suppose a \(1.054-g\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Short Answer

Expert verified
To completely precipitate the chloride ions from the 1.054-g sample with 10.3% chloride ions, 0.520140 g of silver nitrate must be used. This will result in 0.438851 g of silver chloride obtained.

Step by step solution

01

Calculate the mass of chloride ions in the sample

Given that the sample contains 10.3% chloride ions, we can find the mass of chloride ions (Cl-) in the 1.054-g sample as follows: Mass of Cl- = (10.3 / 100) * 1.054 g = 0.108562 g
02

Calculate the moles of chloride ions in the sample

Now, find the moles of Cl- in the sample using the molar mass of Cl (35.45 g/mol): Moles of Cl- = Mass of Cl- / Molar Mass of Cl = 0.108562 g / 35.45 g/mol = 0.003063 mol
03

Determine the moles of silver nitrate required

From the stoichiometry of the reaction between AgNO3 (silver nitrate) and Cl- (chloride ion) to form AgCl (silver chloride), we know that one mole of Cl- will react with one mole of AgNO3: AgNO3 + Cl- 鈫� AgCl + NO3- Thus, the moles of AgNO3 required is equal to the moles of Cl-: Moles of AgNO3 = 0.003063 mol
04

Calculate the mass of silver nitrate required

Using the molar mass of AgNO3 (169.87 g/mol), we can calculate the mass of AgNO3 required to precipitate all Cl- from the sample: Mass of AgNO3 = Moles of AgNO3 * Molar Mass of AgNO3 = 0.003063 mol * 169.87 g/mol = 0.520140 g
05

Calculate the mass of silver chloride obtained

Again, from the stoichiometry of the reaction, we know that one mole of Cl- will react with one mole of AgNO3 to form one mole of AgCl: Moles of AgCl = Moles of Cl- = 0.003063 mol Now, using the molar mass of AgCl (143.32 g/mol), we can determine the mass of AgCl obtained: Mass of AgCl = Moles of AgCl * Molar Mass of AgCl = 0.003063 mol * 143.32 g/mol = 0.438851 g The mass of silver nitrate required is 0.520140 g and the mass of silver chloride obtained is 0.438851 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Chloride Ion Determination
Chloride ion determination is a crucial step in quantitative chemical analysis. For a sample containing chloride ions, determining their amount involves precise measurements and calculations. To start, you need to know the sample's total mass and the percentage of chloride ions present.
The formula to find the mass of chloride ions is:
  • Mass of Cl- = (Percentage of chloride ions / 100) * Total sample mass
This formula gives you the mass of chloride ions in your sample. In this exercise, it's given that a sample contains 10.3% chloride ions. By plugging the numbers into the formula, you determine the mass of chloride ions.
Ace your understanding by practicing with different sample percentages and masses to enhance your calculation skills.
Exploring the Silver Nitrate Reaction
The silver nitrate reaction is a classic method to identify and quantify chloride ions. When silver nitrate ( AgNO_3 ) interacts with chloride ions ( Cl^- ), it forms silver chloride ( AgCl ) and nitrate ions ( NO_3^- ). This is a classic precipitation reaction.
The balanced chemical equation for this reaction is:
  • AgNO鈧� + Cl鈦� 鈫� AgCl + NO鈧冣伝
Here, every chloride ion will react with a silver nitrate molecule to form one molecule of silver chloride. This 1:1 stoichiometric ratio is the essence of the process. Stoichiometry helps you determine how much of each reactant is needed and what products form. Here, the moles of silver nitrate required equals the moles of chloride ions.
As you calculate, consider the reaction's stoichiometric balance to find the maximum yield of silver chloride you can achieve. Knowing stoichiometry helps in ensuring precise reactions to avoid excess or shortages.
The Process of Silver Chloride Precipitation
Precipitation involves the solid formation from a solution. In this exercise, silver chloride ( AgCl ) precipitates when silver nitrate is added to a solution with chloride ions. This is a perfect example of a precipitation reaction where an insoluble solid is formed and separates from the remaining liquid.
The moment silver chloride forms, you might observe a cloudy appearance in the solution. To calculate how much silver chloride you obtain, use the moles of chloride ion, as they directly convert to moles of silver chloride:
  • Moles of AgCl = Moles of Cl鈦�
Now, to find the mass of this solid, employ the formula:
  • Mass of AgCl = Moles of AgCl * Molar mass of AgCl
The precipitation of silver chloride quantifies the initial chloride ions by mass. This technique is beneficial in analytical chemistry for diverse applications. Understanding this principle enables you to accurately calculate product masses and hone laboratory skills.

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Most popular questions from this chapter

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