Question

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

Short answer

The final concentration of ammonium ions (\(NH_4^+\)) in the solution is \(0.622~M\), while the final concentration of sulfate ions (\(SO_4^{2-}\)) in the solution is \(0.311~M\).

Step-by-step solution

Find the initial concentration of the stock solution

First, we need to find the concentration of the stock solution (ammonium sulfate) in moles per liter (M). To do this, we will convert the mass of ammonium sulfate (10.8 g) to moles and then divide by the volume of the solution (100.0 mL) converted to liters. The molar mass of ammonium sulfate: \((NH_4)_2SO_4 = (2 \times (14.01 + 4 \times 1.01)) + (32.07 + 4 \times 16.00)\) \((NH_4)_2SO_4 = (2 \times (14.01 + 4.04)) + (32.07 + 64.00)\) Now, we can find the moles of ammonium sulfate: moles \(= \frac{mass}{molar~mass}\)

Find the concentration of the stock solution

Next, we need to find the concentration of the stock solution in moles per liter (M). To do this, divide the number of moles found in step 1 by the volume of the solution (converted to liters). Concentration \(= \frac{moles}{volume(in~liters)}\)

Calculate the concentration of ions in the final solution

For every one molecule of ammonium sulfate, there are two ammonium ions (\(NH_4^+\)) and one sulfate ion (\(SO_4^{2-}\)). Therefore, the concentrations of ammonium ions and sulfate ions in the stock solution will be twice and equal to the concentration of the ammonium sulfate, respectively. Now, to find the concentrations of both ions in the final solution, we will have to use the dilution formula: \(C_1V_1 = C_2V_2\) Where: \(C_1\) = initial concentration of the ion \(V_1\) = initial volume of the solution \(C_2\) = final concentration of the ion \(V_2\) = final volume of the solution

Calculate the final concentrations of ammonium ions and sulfate ions

Using the dilution formula from Step 3, we can find the final concentrations of the ammonium and sulfate ions in the final solution by substituting their initial concentrations and volumes. For ammonium ions: \(C_1 = 2 \times \text{concentration of ammonium sulfate}\) \(V_1 = 10.00~mL\) \(V_2 = 10.00 + 50.00~mL\) For sulfate ions: \(C_1 = \text{concentration of ammonium sulfate}\) \(V_1 = 10.00~mL\) \(V_2 = 10.00 + 50.00~mL\) The final concentrations can be determined by substituting these values into the dilution formula.

Key concepts

Ammonium Sulfate

Ammonium sulfate is a chemical compound with the formula \((NH_4)_2SO_4\). This compound is commonly used in fertilizers due to its high nitrogen content, which is beneficial for plant growth. When dissolved in water, it dissociates into its constituent ions: two ammonium ions \((NH_4^+)\) and one sulfate ion \((SO_4^{2-})\). Understanding its dissociation is key when analyzing solutions involving ammonium sulfate, particularly for calculating concentrations in chemistry problems. The behavior of these ions in solution is important for stoichiometry and reaction predictions in aqueous environments.

Dilution Formula

The dilution formula is a trusty tool in chemistry that helps us find the concentration of a solution after it has been diluted. This formula, \( C_1V_1 = C_2V_2 \), simply relates the concentration and volume before and after dilution. Here, \( C_1 \) is the initial concentration of the solution, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume. It works because the amount of solute remains constant before and after dilution. By using this formula, we can predict how the concentration of ions, like ammonium or sulfate, changes when water is added to a solution.

Molar Mass Calculation

Calculating the molar mass of a compound is like finding its weight in terms of atomic masses. For ammonium sulfate \((NH_4)_2SO_4\), you calculate it by adding the atomic masses of all atoms present in the compound. Here's how: calculate the total mass of two ammonium \((NH_4)^+\) groups and one sulfate \((SO_4)^{2-}\) group.

• The nitrogen (N) mass is 14.01 per atom, and hydrogen (H) is 1.01 per atom. So for \((NH_4)^+\), it’s \((14.01 + 4 \, \times \, 1.01) \times 2 \).
• The sulfur (S) has a mass of 32.07, and oxygen (O) is 16.00. In sulfate \((SO_4)^{2-}\), this provides \((32.07 + 4 \, \times \, 16.00)\).

By adding these together, you'll obtain the total molar mass, which is crucial for converting grams to moles – a common step when calculating concentrations from a given mass of solute. It brings theory to practical application, allowing for accurate preparation and analysis of chemical solutions.