Question

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Short answer

The concentration of ammonium ions in the final solution is given by \( \frac{2 \times (\text{Concentration of ammonium sulfate}) \times 0.010 \,\text{L}}{0.060 \,\text{L}} \), and the concentration of sulfate ions is given by \( \frac{(\text{Concentration of ammonium sulfate}) \times 0.010 \,\text{L}}{0.060 \,\text{L}} \). Substitute the calculated concentration of ammonium sulfate from step 2 to find the final concentrations for both ions.

Step-by-step solution

Calculate the moles of ammonium sulfate in the stock solution

First, we need to find the moles of ammonium sulfate in the stock solution. The formula of ammonium sulfate is (NH4)2SO4, and its molar mass is 132.14 g/mol. The mass given is 10.8 g. To find the moles, we will use the formula: Moles = mass (g) / molar mass (g/mol) Moles of ammonium sulfate = \( \frac{10.8 \,\text{g}}{132.14 \,\text{g/mol}} \)

Calculate the concentration of ammonium sulfate in the stock solution

We are given that the stock solution has a volume of 100.0 mL. To find the concentration of ammonium sulfate in the stock solution, we can use the formula: Concentration = Moles / Volume (L) The volume should be converted to liters: Volume = 100.0 mL * \( \frac{1 \,\text{L}}{1000 \,\text{mL}} \) = 0.100 L Concentration of ammonium sulfate = \( \frac{\text{Moles of ammonium sulfate}}{0.100 \,\text{L}} \)

Calculate the moles of ammonium and sulfate ions in the stock solution sample

A 10.00 mL sample of the stock solution is taken. Since the concentration of a solution is uniform, we can use the same process as in step 2 to find the moles of ammonium sulfate in this sample: Moles of ammonium sulfate = Concentration of ammonium sulfate * Volume of the sample (L) Volume of the sample = 10.00 mL * \( \frac{1 \,\text{L}}{1000 \,\text{mL}} \) = 0.010 L Moles of ammonium sulfate in the sample = Concentration of ammonium sulfate * 0.010 L Since there are two ammonium ions (NH4+) and one sulfate ion (SO4^2-) in each ammonium sulfate molecule, the moles of ammonium ions in the sample will be twice the moles of ammonium sulfate, and the moles of sulfate ions will be equal to the moles of ammonium sulfate. Moles of ammonium ions = 2 * Moles of ammonium sulfate in the sample Moles of sulfate ions = Moles of ammonium sulfate in the sample

Calculate the moles of ammonium and sulfate ions in the final solution

We are given that the sample is added to 50.00 mL of water. Since the volumes are additive, the total volume of the final solution is 50.00 + 10.00 = 60.00 mL. Since the moles of ammonium and sulfate ions in the final solution are equal to those in the sample, we can move on to step 5.

Calculate the concentration of ammonium and sulfate ions in the final solution

The final solution has a combined volume of 60.00 mL. To find the concentration of ammonium and sulfate ions in the final solution, we can use the formula: Concentration = Moles / Volume (L) Volume of the final solution = 60.00 mL * \( \frac{1 \,\text{L}}{1000 \,\text{mL}} \) = 0.060 L Concentration of ammonium ions = \( \frac{\text{Moles of ammonium ions}}{0.060 \,\text{L}} \) Concentration of sulfate ions = \( \frac{\text{Moles of sulfate ions}}{0.060 \,\text{L}} \) Now, we can plug in the values from the previous steps to find the final concentrations for ammonium and sulfate ions in the final solution.

Key concepts

Ammonium Sulfate

Ammonium sulfate, \(\text{(NH}_4\text{)}_2\text{SO}_4\), is a solid compound that is commonly used in fertilizers for its nitrogen content. This compound is composed of two ammonium ions, \(\text{NH}_4^+\), and one sulfate ion, \(\text{SO}_4^{2-}\). Understanding the chemical structure of ammonium sulfate is key to understanding how it behaves in solutions, especially when discussing concentration calculations. When dissolved in water, ammonium sulfate separates into its constituent ions, contributing to the overall concentration of ions in solution. The dissociation can be represented as:

• \(\text{(NH}_4\text{)}_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ + \text{SO}_4^{2-}\)

This dissociation is crucial in predicting the behavior of the compound in a given solution, especially when calculating the concentrations of these ions individually later on.

Molar Mass

The molar mass is a critical concept when it comes to calculating the amount of a substance in terms of moles. For ammonium sulfate, \(\text{(NH}_4\text{)}_2\text{SO}_4\), the molar mass is calculated by summing the atomic masses of all atoms in the formula:

• Nitrogen (N) = 14g/mol, Hydrogen (H) = 1g/mol
• Sulfur (S) = 32g/mol, Oxygen (O) = 16g/mol

The calculation follows: \[ \text{Molar mass of } \text{(NH}_4\text{)}_2\text{SO}_4 = 2\times(1\times4 + 14) + 32 + 4\times16 \]This results in a total molar mass of 132.14 g/mol for ammonium sulfate. The molar mass serves as a conversion factor from grams to moles, helping to calculate the quantity of substances involved in reactions and solutions.

Dilution

Dilution refers to the process of reducing the concentration of a solute in a solution, usually by adding more solvent. In our problem, a sample from the stock solution is further diluted by adding distilled water. This process affects the concentration of ions in the final solution. When conducting a dilution, the number of moles of solute remains constant, but the volume increases, which decreases the concentration.For example, by adding 50.00 mL of water to a 10.00 mL sample of stock solution, the total volume of the solution becomes 60.00 mL. It is important to understand the dilution equation often used here, \( C_1V_1 = C_2V_2 \), to determine the new concentration:

• \(C_1\) and \(V_1\) are the concentration and volume before dilution \(\text{(initial stock solution)}\)
• \(C_2\) and \(V_2\) are the concentration and volume after dilution \(\text{(final solution)}\)

This equation helps ensure the concentrations are calculated accurately after dilution steps have been done.

Stock Solution

A stock solution is a concentrated solution that can be diluted to create a set number of new solutions with varied concentrations. In this exercise, the stock solution of ammonium sulfate was made by dissolving 10.8 grams of the compound in water up to a total volume of 100.0 mL. This initial preparation is crucial as it sets the basis for further experiments or reactions involving this solution. With stock solutions, it's easier to dilute to precise concentrations, maintaining consistency in experiments. The concept of a stock solution allows chemists to work efficiently, taking small volumes of the concentrated solution and diluting them to desired concentrations for specific needs. As shown in the previous sections, understanding the concentration of a stock solution directly affects the calculations ensuing from further dilutions or sample removals.