Problem 79 - Chemistry The Central Science In Si Units

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 \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) 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 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 \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution costs \(\$ 39.95\) for \(500 \mathrm{~mL}\) and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

Step-by-Step Solution

Answer

Use KCl; it's much cheaper than HCl.

Step 1 - Calculate moles of Ag+

Start by finding the moles of silver ions (\(\text{Ag}^+\)) in the \(15.0\ \text{mL}\) of \(0.200\ \text{M}\ \text{AgNO}_3\). \[ \text{moles of } \text{Ag}^+ = \text{Molarity} \times \text{Volume (L)} = 0.200\ \text{M} \times 0.0150\ \text{L} = 0.0030\ \text{ moles} \]

Step 2 - Determine volume of HCl needed

For reaction of \(\text{Ag}^+\) with \(\text{Cl}^-\): \(\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} (s)\). The moles of \(\text{Cl}^-\) required is equal to the moles of \(\text{Ag}^+\), which is \(0.0030\ \text{ moles}\). Using the molarity of \(\text{HCl}\), find the volume: \[ \text{Volume of HCl (L)} = \frac{\text{Moles of } \text{HCl}}{\text{Molarity of } \text{HCl}} = \frac{0.0030}{0.150} = 0.0200\ \text{L} \] Convert this to mL: \[ 0.0200\ \text{L} \times 1000\ \frac{\text{mL}}{\text{L}} = 20.0\ \text{mL} \]

Step 3 - Calculate mass of KCl needed

To calculate the mass of \(\text{KCl}\), determine first the moles of \(\text{KCl}\) needed, which is equal to the moles of \(\text{Cl}^-\): \[ \text{moles of KCl} = 0.0030\ \text{ moles}\]. Given the molar mass of \(\text{KCl}\) is approximately \(74.55\ \text{g/mol}\), find the mass: \[ \text{Mass of KCl} = 0.0030\ \text{ moles} \times 74.55\ \text{g/mol} = 0.224\ \text{g} \]

Step 4 - Compare costs for HCl and KCl

Calculate the cost for using each method. **HCl cost**: Given \(20.0\ \text{mL}\) of \(0.150\ \text{M} \text{HCl}\) costs: \[ \frac{39.95}{500\ \text{mL}} \times 20.0\ \text{mL} = 1.598\ \text{USD} \]**KCl cost**: Converting \(0.224\ \text{g}\) to metric tons: \[ 0.224\ \text{g} \approx 0.000224\ \text{kg} = 0.000000224\ \text{ton} \] Using \(10\ \text{USD/ton}\) cost, \[ 0.000000224\ \times 10 = 0.00000224\ \text{USD} \]

Step 5 - Determine the more cost-effective procedure

Now compare the costs, \(1.598\ \text{USD}\) for \(\text{HCl}\) and \(0.00000224\ \text{USD}\) for \(\text{KCl}\). Thus, using solid \(\text{KCl}\) is vastly more cost-effective than using \(\text{HCl}\).

Key Concepts

Molarity and Concentration

To fully understand how solutions are analyzed and utilized in chemistry, one must first grasp the concept of molarity. Molarity ( extbf{M}) is a way of expressing the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is: \[ M = \frac{n}{V} \] where

• \( M \) represents molarity in moles per liter (mol/L),
• \( n \) is the number of moles of solute, and
• \( V \) is the volume of the solution in liters.

In the given exercise, the molarity is used to determine how much \( \text{HCl}(aq) \) is needed to react with \( \text{Ag}^+ \) ions. Understanding molarity helps in precisely preparing solutions with desired concentrations, ensuring reactions proceed as expected.
It's essential to accurately measure the volume and accurately calculate moles to apply molarity correctly. This ensures the chemical reactions will have the proper reactants and proceed efficiently.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It relies on the principles of conservation of mass and the stoichiometric coefficients derived from balanced chemical equations. For instance, consider the simple reaction in the problem we are analyzing: \[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s) \]This indicates one mole of \( \text{Ag}^+ \) reacts with one mole of \( \text{Cl}^- \) to produce one mole of solid \( \text{AgCl} \). To perform stoichiometric calculations, you need to:

• Identify the balanced chemical equation for the reaction.
• Convert quantities of known reactants to moles.
• Use the mole ratio from the balanced equation to find moles of desired reactant or product.
• Convert moles back to other units such as mass or volume.

This approach was applied to find the amount of \( \text{HCl} \) and \( \text{KCl} \) needed to fully react with \( \text{Ag}^+ \) ions in the initial solution. Understanding stoichiometry ensures chemicals are used efficiently and safety protocols are followed.

Precipitation Reactions

Precipitation reactions are those in which solutions are mixed, resulting in the formation of an insoluble solid known as the precipitate. These reactions are crucial in analytical chemistry as they help in separating out specific components from a solution.Precipitation occurs when the product of certain ions in solution exceeds the solubility product (K_sp) of the compound formed. In the initial exercise, silver ions \( \text{Ag}^+ \) react with chloride ions \( \text{Cl}^- \) to form silver chloride \( \text{AgCl}(s) \), a classic example of a precipitation reaction.To predict and analyze such reactions, consider:

• The solubility rules, which predict the solubility of ionic compounds.
• Reactants’ concentrations and their ionic products in water.
• Determining when a solid forms from a solution mix.

Precipitation reactions are vital in fields like water treatment and analytical chemistry, where obtaining a solid from a solution is necessary for analysis or further reactions.

Cost Analysis in Chemistry

Cost analysis in chemistry involves evaluating the expenses associated with chemical processes and reactions. It helps in choosing the most efficient and economically favorable method for achieving desired results. The exercise compares the cost of two reagents: \( \text{HCl} \) and \( \text{KCl} \), used to precipitate \( \text{AgCl} \) from a silver nitrate solution.To perform a cost analysis, consider:

• The price of chemicals per unit of measurement (e.g., dollars per liter or per ton).
• The amount needed to fully react or process the desired quantity.
• Operational costs including labor and equipment, if applicable.

In our example, while \( \text{HCl} \) is more straightforward to use, \( \text{KCl} \) proved to be vastly more cost-effective. This is evident because the cost to use \( \text{KCl} \) was a fraction of a cent compared to several dollars for \( \text{HCl} \). Overall, a comprehensive cost analysis ensures that decisions made will be both economic and efficient, maximizing the use of resources.