Question

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out 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 \(\mathrm{MgNO}_{3}\) 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 \(\mathrm{KCl}\) costs \(\$ 10 / \mathrm{ton},\) which analysis procedure is more cost-effective?

Short answer

To precipitate silver ions from 15.0 mL of 0.200 M AgNO₃ solution, 20 mL of 0.150 M HCl solution or 0.224 g of KCl are needed. The cost of using HCl solution is \$1.598, while the cost of using KCl is only \$0.00000224. Therefore, using KCl is more cost-effective for this analysis.

Step-by-step solution

(Step 1: Moles of Ag+ in the solution)

(First, we need to find the moles of Ag+ in the 15.0 mL of 0.200 M AgNO3 solution. We can do this using the formula: Moles of Ag+ = Molarity of AgNO3 * Volume of AgNO3 solution Moles of Ag+ = 0.200 mol/L * 15.0 mL * (1 L / 1000 mL) Moles of Ag+ = 0.003 mol)

(Step 2: Moles of HCl needed to precipitate Ag+)

(Next, we need to find the moles of HCl needed to precipitate all the silver ions. The balanced chemical equation for the reaction is: AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) From the balanced equation, we can see that 1 mole of HCl is required for each mole of AgNO3. Moles of HCl needed = Moles of Ag+ Moles of HCl needed = 0.003 mol)

(Step 3: Volume of 0.150 M HCl required)

(Now we can find the volume of the 0.150 M HCl solution needed to precipitate the silver ions: Volume of HCl = Moles of HCl / Molarity of HCl Volume of HCl = 0.003 mol / 0.150 mol/L Volume of HCl = 0.02 L or 20 mL)

(Step 4: Moles of KCl needed to precipitate Ag+)

(The balanced chemical equation for the reaction of AgNO3 with KCl is: AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) From the balanced equation, we can see that 1 mole of KCl is required for each mole of AgNO3. Moles of KCl needed = Moles of Ag+ Moles of KCl needed = 0.003 mol)

(Step 5: Mass of KCl needed)

(Now we will find the mass of KCl needed to precipitate the silver ions: Mass of KCl = Moles of KCl * Molar mass of KCl Mass of KCl = 0.003 mol * 74.55 g/mol Mass of KCl = 0.224 g)

(Step 6: Cost comparison)

(First, we calculate the cost of using 0.150 M HCl solution: Cost of HCl = (Volume of HCl used / Total volume of HCl) * Cost of HCl solution Cost of HCl = (20 mL / 500 mL) * $39.95 Cost of HCl = $1.598 Next, we calculate the cost of using KCl: Cost of KCl = (Mass of KCl used / Total mass of KCl) * Cost of KCl Cost of KCl = (0.224 g / 1,000,000 g) * $10 Cost of KCl = $0.00000224 Now, we can compare the costs: Cost for HCl: $1.598 Cost for KCl: $0.00000224 Clearly, using KCl is more cost-effective for this analysis.)

Key concepts

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two aqueous solutions combine to form an insoluble solid, known as a precipitate. These reactions are essential in various applications, such as in chemical analysis and separation processes.
In our exercise, when mixing silver nitrate (\( \text{AgNO}_3 \)) with hydrochloric acid (\( \text{HCl} \)), silver chloride (\( \text{AgCl} \)) forms as a white precipitate, leaving behind nitric acid (\( \text{HNO}_3 \)) in the solution. The balanced equation representing this reaction is:
\[ \text{AgNO}_3(aq) + \text{HCl}(aq) \rightarrow \text{AgCl}(s) + \text{HNO}_3(aq) \]
This balanced equation shows a 1:1 molar ratio between \( \text{AgNO}_3 \) and \( \text{HCl} \), indicating that one mole of hydrochloric acid is needed to precipitate one mole of silver ions.
By understanding this fundamental principle of precipitation reactions, we can gracefully predict and control the formation of precipitates during chemical reactions, which is crucial for laboratory experiments and industrial processes.

Molarity Calculations

Molarity plays a key role in quantifying the concentration of solutions in chemistry. It is defined as the number of moles of solute per liter of solution. The unit of molarity is moles per liter (\( \text{mol/L} \)). Calculating molarity helps us determine how concentrated a solution is.
To solve our exercise, we first calculated the moles of silver ions in the 15.0 mL of \( 0.200 \) \( \text{M} \) \( \text{AgNO}_3 \) solution using:

• Moles of \( \text{Ag}^+ \) = Molarity \( \times \) Volume

Subsequently, we used the molarity of \( \text{HCl} \) to find the required volume:

• Volume of \( \text{HCl} \) = Moles of \( \text{HCl} \) \( / \) Molarity of \( \text{HCl} \)
• Volume of \( \text{HCl} \) = \( 0.003 \text{ mol} / 0.150 \text{ mol/L} \)
• Volume is 20 mL.

These simple calculations allowed us to determine precisely how much of a reagent is needed to complete the reaction, which is vital for careful preparation and accurate experimental results.

Cost Analysis

Cost analysis is crucial in selecting the most economical procedure for chemical reactions. It involves comparing the expenses associated with different methods to achieve a task or outcome.
In our given scenario, we have two options for precipitating silver ions: using \( \text{HCl} \) solution or solid \( \text{KCl} \).

• The cost of \( \text{HCl} \) was calculated based on the volume needed and the price per 500 mL.
• Conversely, the cost of solid \( \text{KCl} \) was determined based on the mass required and its price per ton.

With the cost for \( \text{HCl} \) coming to \(1.598, and only \)0.00000224 for \( \text{KCl} \), it is evident that \( \text{KCl} \) is the more cost-effective option.
Conducting a cost analysis like this aids in making informed decisions, ensuring that procedures are not only effective but also economically viable. This is especially significant in fields like chemical engineering and industrial production, where cost savings can lead to substantial financial benefits.