Question

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out \(\mathrm{AgCl}(s) .\) What volume of a \(0.150 \mathrm{MHCl}(a q)\) solution is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of a \(0.200 \mathrm{MAgNO}_{3}\) solution? (b) You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a \(0.150 \mathrm{MHCl}(a q)\) solution costs \(\$ 39.95\) for \(500 \mathrm{~mL}\) and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

Short answer

We need 20 mL of 0.150 M HCl solution or 0.22365 g of KCl to precipitate the silver ions from 15.0 mL of 0.200 M AgNO3 solution. The cost of using HCl solution is $1.598, while the cost of using KCl is $0.000002. Therefore, the more cost-effective procedure is to add solid KCl to the solution to precipitate the silver ions.

Step-by-step solution

Calculate moles of AgNO3

First, we will find the moles of AgNO3 present in the 15.0 mL of 0.200 M solution. We can calculate it using the following formula: Number of moles = Molarity × Volume //Convert volume to liters: \(15.0\:mL = 0.015\:L\) Moles of AgNO3 = \(0.200\:\cancel{Mol\:L^{-1}}\cdot0.015\:\cancel{L} = 0.003\:mol\)

Calculate moles of HCl needed

Since one mole of AgNO3 reacts with one mole of HCl to produce AgCl, the number of moles of HCl required is equal to the moles of AgNO3. So, we need 0.003 moles of HCl.

Calculate volume of 0.150 M HCl solution needed

Now, we will use the number of moles of HCl and its molarity to find the volume of the solution needed: Volume = Moles / Molarity Volume = \(0.003\:\cancel{mol}/0.150\:\cancel{Mol\:L^{-1}} = 0.02\:L\) Volume = \(0.02\:L = 20\:mL\) So, we need 20 mL of 0.150 M HCl solution to precipitate all the silver ions. #b) Calculating the mass of KCl needed#

Calculate moles of KCl needed

Again, one mole of AgNO3 reacts with one mole of KCl to produce AgCl, so we need 0.003 moles of KCl.

Calculate mass of KCl needed

Now, we'll use the molar mass of KCl to find the mass of KCl required: Mass of KCl = Moles × Molar mass Mass of KCl = \(0.003\:mol \times 74.55\:g/mol = 0.22365\:g\) So, we need 0.22365 g of KCl to precipitate all the silver ions. #c) Comparing costs of the two procedures#

Calculate the cost of HCl solution needed

To calculate the cost of the HCl solution, we need to find the cost of 20 mL of 0.150 M HCl solution. We are given that 500 mL of the solution costs $39.95, so the cost per milliliter is: Cost per mL = Total cost / Volume Cost per mL = \(39.95 / 500 mL = \)0.0799/mL Now, we can find the cost of 20 mL: Cost of 20 mL = 20 mL × \(0.0799/mL = \)1.598

Calculate the cost of KCl needed

In this case, we need to find the cost of 0.22365 g of KCl. We are given that 1 ton (= 1,000,000 g) of KCl costs $10, so the cost per gram is: Cost per g = Total cost / Mass Cost per g = \(10/ 1,000,000 g = \)0.00001/g Now, we can find the cost of 0.22365 g: Cost of 0.22365 g = \(0.00001/g × 0.22365 g = \)0.0000022365 ≈ $0.000002

Compare the costs

Now that we have the costs of both procedures: Cost of HCl solution: ≈ $1.598 Cost of KCl: ≈ $0.000002 The cost of using KCl is significantly lower than using HCl solution. Therefore, adding solid KCl to the solution to precipitate the silver ions is the more cost-effective procedure.

Key concepts

Molarity

Molarity is a fundamental concept in stoichiometry and chemistry that explains the concentration of a solute in a solution. It is specified in terms of the number of moles of solute per liter of solution. It's often represented with the symbol 'M' and provides a way to quantify how much of a particular substance is present in a given volume of solution. For instance, a solution with a molarity of 0.200 M of AgNO3 means there are 0.200 moles of silver nitrate in every liter of this solution. Understanding molarity helps in determining how much of a reactant is required or expected in a chemical reaction.

To find the molarity, use the formula:

• Molarity = \( \frac{\text{Number of Moles of Solute}}{\text{Volume of Solution in Liters}} \)

For example, in the original exercise, to ascertain how much HCl is needed to precipitate the silver ions, the reaction stoichiometry tells us that the moles of AgNO3 (calculated as 0.003 mol) must be equal to the moles of HCl needed, due to their one-to-one reaction ratio as governed by the reaction:\[ \text{AgNO}_{3(aq)} + \text{HCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{HNO}_{3(aq)} \]

Chemical Reactions

Chemical reactions form the core of chemistry as they reveal how substances interact with each other. They involve rearrangements of atoms and lead to the production of different substances. In our current context of analyzing silver nitrate, the primary reaction of interest includes the precipitation of silver chloride (AgCl) when reacting either with HCl or KCl.

Precipitation reactions entail the formation of a solid (precipitate) from a solution, and this reaction is exploited in the given exercise to separate silver ions. With

• HCl, the reaction is: \[ \text{AgNO}_{3(aq)} + \text{HCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{HNO}_{3(aq)} \]
• Similarly, with KCl: \[ \text{AgNO}_{3(aq)} + \text{KCl}_{(s)} \rightarrow \text{AgCl}_{(s)} + \text{KNO}_{3(aq)} \]

Each of these reactions is stoichiometrically balanced, which means that one mole of AgNO3 will react with one mole of HCl or KCl, demonstrating a simple 1:1 ratio in terms of moles of reactants.

Cost Analysis

Cost analysis in stoichiometry involves evaluating the financial implications of choosing particular reactants or methods in a chemical process. From an economic viewpoint, you need to consider not just the chemical feasibility, but also the costs involved. In the problem at hand, the goal is to determine which reagent, HCl solution or solid KCl, is more cost-effective for precipitating the silver ions from a solution.

The cost for precipitating using HCl is:

• HCl solution: 500 mL costs $39.95.
• Given a need for 20 mL, the cost per mL being $0.0799, the total is $1.598.

Alternatively, when calculating for KCl:

• KCl costs $10 per ton (1,000,000 g).
• For the required 0.22365 g, the cost of KCl ends up being negligibly small, at approximately $0.000002.

Thus, checking these costs highlights that using solid KCl is significantly less expensive than HCl solution, illustrating a more economical option for the purpose of precipitating silver ions.