Problem 119 - General Chemistry: Principles and Modern Applications

Question

Silver nitrate is a very expensive chemical. For a particular experiment, you need \(100.0 \mathrm{mL}\) of \(0.0750 \mathrm{M}\) \(\mathrm{AgNO}_{3},\) but only \(60 \mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{AgNO}_{3}\) is available. You decide to pipet exactly \(50.00 \mathrm{mL}\) of the solution into a \(100.0 \mathrm{mL}\) flask, add an appropriate mass of \(\mathrm{AgNO}_{3},\) and then dilute the resulting solution to exactly \(100.0 \mathrm{mL}\). What mass of \(\mathrm{AgNO}_{3}\) must you use?

Step-by-Step Solution

Answer

You must use 0.849 g of \(AgNO_{3}\)

Step 1: Calculate moles needed for final solution

First, calculate the number of moles needed for the final solution using the equation \(M = n/v\), where \(M\) is the molarity, \(n\) is the number of moles, and \(v\) is the volume (in litres). The final solution has a volume of 0.100 litres (or 100.0 mL) and a molarity of 0.075 mol/L, so you need \(n = M*v = 0.075 \times 0.100 = 0.0075\) mol of \(AgNO_{3}\)

Step 2: Calculate moles in initial solution

Next, calculate the number of moles in the initial solution (again using \(M = n/v\)), which has a volume of 0.050 L and a molarity of 0.050 mol/L. This gives \(n = M*v = 0.050 \times 0.050 = 0.0025\) mol of \(AgNO_{3}\)

Step 3: Calculate moles of \(AgNO_{3}\) to be added

Now, subtract the number of moles in the initial solution from the number of moles needed for the final solution. This gives \(0.0075 - 0.0025 = 0.0050\) mol.

Step 4: Calculate mass of \(AgNO_{3}\) to be added

Finally, convert the number of moles to be added into mass using the molar mass of \(AgNO_{3}\), which is 169.87 g/mol. This gives the mass as \(m = n \times mw = 0.0050 \times 169.87 = 0.849 g\).

Key Concepts

Molarity calculations

Molarity is a concept used to describe how concentrated a solution is. Specifically, it is expressed as the number of moles of solute (the substance being dissolved) per liter of solution. In mathematical terms, it's represented by the formula: \( M = \frac{n}{V} \). Here, \( M \) stands for molarity, \( n \) is the number of moles of the solute, and \( V \) is the volume of the solution in liters.
To determine how many moles of silver nitrate (\( \text{AgNO}_3 \)) you need for your experiment, you first need to find the desired molarity and volume of your final solution. In this exercise, you’re aiming for a solution that is 0.075 M in a total volume of 0.1 L. Simply plug these values into the equation to find out how many moles you need: \( n = M \times V = 0.075 \times 0.100 = 0.0075 \) moles.
Molarity calculations are crucial when preparing solutions with precise concentrations, especially in chemistry experiments where the concentration of reactants affects reaction rates and outcomes.

Mass calculation

Once the number of moles required for the final solution is determined, you need to figure out how much extra silver nitrate you need to add. In this particular exercise, you start with a certain amount already in a 0.050 L solution at 0.050 M. Calculating the initial moles using \( M = \frac{n}{V} \), you get: \( n = 0.050 \times 0.050 = 0.0025 \) moles of \( \text{AgNO}_3 \).
The difference between the moles needed (0.0075 moles) and what you already have (0.0025 moles) tells you exactly how much more is required: 0.0075 - 0.0025 = 0.005 moles.
To convert this mole requirement into mass, multiply by the molar mass of \( \text{AgNO}_3 \), which is 169.87 g/mol. Thus, 0.005 moles \( \times 169.87 \text{ g/mol} = 0.849 \text{ g} \).
Performing mass calculations accurately ensures that the correct amount of a substance is used, which is essential for the reliability of experimental results.

Solution preparation

Preparing a solution accurately involves a series of calculated steps to ensure that the final outcome meets the required specifications. Here’s how to approach this task from the exercise:

• Start by measuring out 50.00 mL of the existing 0.050 M \( \text{AgNO}_3 \) solution. This will provide an initial amount of 0.0025 moles of \( \text{AgNO}_3 \).
• Next, calculate and weigh out 0.849 g of additional \( \text{AgNO}_3 \), to achieve the desired moles for your final concentration.
• Add this mass to the measured 50.00 mL of solution.
• Finally, dilute the entire mixture to reach the total volume of 100.0 mL, ensuring uniform concentration throughout the solution.

Each step helps to create the precise conditions needed for the experiment. Solution preparation involves attention to detail, accuracy, and understanding each component's role in achieving the overall desired concentration.