Question

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

Short answer

The volume of \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution needed to precipitate the silver ions is \(20.0\,\text{mL}\), and the mass of solid \(\mathrm{KCl}\) needed is \(0.224\,\text{g}\). Comparing the costs of the two procedures, the solid \(\mathrm{KCl}\) procedure (\(\$0.000224\)) is more cost-effective than the \(\mathrm{HCl}(a q)\) procedure (\(\$1.598\)).

Step-by-step solution

Calculate the moles of AgNO₃ in the given solution.

To find the moles of \(\mathrm{AgNO}_3\), multiply the volume of the solution by its concentration: $\text{moles of } \mathrm{AgNO}_3 = \text{volume of solution} \times \text{concentration} \\ = 15.0\,\text{mL} \times 0.200\,\text{M} \\ = \frac{15.0\,\text{mL}}{1000\,\text{mL/L}} \times 0.200\,\text{M} \\ = 0.003\,\text{mol}$ ##Step 2: Determine the moles of HCl needed##

Use stoichiometry to find the moles of HCl required.

For every mole of \(\mathrm{AgNO}_3\), one mole of \(\mathrm{HCl}\) is required to precipitate \(\mathrm{AgCl}\). So, the moles of \(\mathrm{HCl}\) required will be equal to the moles of \(\mathrm{AgNO}_3\). \(\text{moles of HCl} = \text{moles of AgNO}_3 = 0.003\,\text{mol}\) ##Step 3: Calculate the volume of HCl solution needed##

Find the volume of the HCl solution required to precipitate AgCl.

To find the volume of \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution needed, divide the moles of \(\mathrm{HCl}\) by its concentration: $\text{volume of HCl solution} = \frac{\text{moles of HCl}}{\text{concentration}} \\ = \frac{0.003\,\text{mol}}{0.150\,\text{M}} \\ = 0.020\,\text{L} \\ = 20.0\,\text{mL}$ ##Step 4: Determine the moles of KCl needed##

Use stoichiometry to find the moles of KCl required.

For every mole of \(\mathrm{AgNO}_3\), one mole of \(\mathrm{KCl}\) is required to precipitate \(\mathrm{AgCl}\). So, the moles of \(\mathrm{KCl}\) required will be equal to the moles of \(\mathrm{AgNO}_3\). \(\text{moles of KCl} = \text{moles of AgNO}_3 = 0.003\,\text{mol}\) ##Step 5: Calculate the mass of KCl needed##

Find the mass of solid KCl required to precipitate AgCl.

To find the mass of \(\mathrm{KCl}\) needed, multiply the moles of \(\mathrm{KCl}\) by its molar mass (\(74.551\,\text{g/mol}\) for \(\mathrm{KCl}\)): $\text{mass of KCl} = \text{moles of KCl} \times \text{molar mass} \\ = 0.003\,\text{mol} \times 74.551\,\text{g/mol} \\ = 0.224\,\text{g}$ ##Step 6: Calculate the costs of both procedures##

Calculate the cost of each analysis procedure.

For the \(\mathrm{HCl}(a q)\) procedure: $\text{cost} = \frac{\$39.95}{500\,\text{mL}} \times 20.0\,\text{mL} \\ = \$1.598$ For the solid \(\mathrm{KCl}\) procedure: $\text{cost} = \frac{\$10}{1000\,\text{kg}} \times 0.000224\,\text{kg} \\ = \$0.000224$ ##Step 7: Compare the costs and determine the most cost-effective procedure##

Determine which analysis procedure is more cost-effective.

Since the cost of the \(\mathrm{HCl}(a q)\) procedure (\(\$1.598\)) is greater than the cost of the solid \(\mathrm{KCl}\) procedure (\(\$0.000224\)), the solid \(\mathrm{KCl}\) procedure is more cost-effective.

Key concepts

Mole Calculations

Moles are the fundamental unit in chemistry that helps us count the number of atoms, molecules, or ions in a given sample. Understanding mole calculations is crucial in solving problems related to stoichiometry and chemical reactions.

**Converting Volume to Moles**
To find the number of moles in a liquid solution, you use the formula: \[ \text{moles} = \text{volume (L)} \times \text{concentration (M)} \] For example, if you have a 15.0 mL solution of \(\mathrm{AgNO}_3\) at a concentration of 0.200 M, first convert the volume to liters (which becomes 0.015 L). Then multiply by the concentration: \[ 0.015 \text{ L} \times 0.200 \text{ M} = 0.003 \text{ mol} \]

**Equivalence in Reactions**
In the reaction between \(\mathrm{AgNO}_3\) and \(\mathrm{HCl}\), the stoichiometry shows a 1:1 mole ratio. This means for every mole of \(\mathrm{AgNO}_3\), one mole of \(\mathrm{HCl}\) is required to completely react.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two soluble salts react in solution to form an insoluble solid, or precipitate. In the case of \(\mathrm{AgNO}_3\) and \(\mathrm{HCl}\), the reaction produces a white solid, \(\mathrm{AgCl}\), which precipitates out of the solution.

**Reaction Equation**
The chemical equation for this reaction is:\[\mathrm{AgNO}_3(aq) + \mathrm{HCl}(aq) \rightarrow \mathrm{AgCl}(s) + \mathrm{HNO}_3(aq)\]This illustrates how the silver ions (\(\mathrm{Ag}^+\)) and chloride ions (\(\mathrm{Cl}^-\)) come together to form \(\mathrm{AgCl}(s)\).

**Using Stoichiometry**
Since the ratio of \(\mathrm{AgNO}_3\) to \(\mathrm{KCl}\) or \(\mathrm{HCl}\) is 1:1 for forming \(\mathrm{AgCl}\), the moles of each reactant involved are equal. Knowing the amount of \(\mathrm{AgNO}_3\) allows you to determine how much \(\mathrm{HCl}\) or \(\mathrm{KCl}\) you need to add for complete precipitation.

Cost Analysis in Chemistry

Cost analysis in chemistry is essential when choosing between different chemical procedures, especially on an industrial scale. It involves comparing the expense of using various reactants to determine the most cost-effective process.

**Analyzing Costs for AgCl Precipitation**
Consider the cost of the chemicals used for a procedure. For precipitation reactions:

• The cost for \(\mathrm{HCl}\) at \\(39.95 for 500 mL might seem high, but you calculate the actual cost needed by determining how much volume you require. For instance, if you need 20.0 mL, then the cost becomes approximately \\)1.598.
• In contrast, \(\mathrm{KCl}\), which costs \\(10 per ton (or 1000 kg), when used in a small amount like 0.224 g, translates to a negligible cost of around \\)0.000224.

**Choosing the Economical Option**
Given the vast cost difference, opting for \(\mathrm{KCl}\) over \(\mathrm{HCl}\) results in considerable savings, making it the more economically favorable method for precipitating \(\mathrm{AgCl}\) from a solution.