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. A \(10.00\) 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

In the final solution, the concentration of ammonium ions (NH₄⁺) is 0.272 M and the concentration of sulfate ions (SO₄²⁻) is 0.136 M.

Step-by-step solution

Calculate the moles of Ammonium Sulfate (NH4)2SO4

To calculate the moles of (NH4)2SO4, we can use the given mass and the molar mass of (NH4)2SO4: Molar mass of (NH4)2SO4 = 2*(14.01 + 4*1.01) + 32.07 + 4*16.00 = 132.14 g/mol Now we can calculate the moles of (NH4)2SO4: moles = mass/molar_mass moles = 10.8 g / 132.14 g/mol ≈ 0.0817 mol

Calculate the Molarity of the Stock Solution

Molarity is defined as moles of solute per liter of solution: M₁ = moles / volume_of_stock_solution The volume of the stock solution is 100.0 mL, which we must convert to liters: volume_in_liters = 100.0 mL * (1 L / 1000 mL) = 0.100 L Now, we can find the molarity of (NH4)2SO4 in the stock solution: M₁ = 0.0817 mol / 0.100 L = 0.817 M

Calculate the Moles of Ions in the 10 mL of Stock Solution

Now we will find the moles of ions in the 10 mL of stock solution using the molarity: moles_of_ions = M₁ * volume_in_liters The volume of 10 mL must be converted to liters: volume_10mL_in_liters = 10 mL * (1 L / 1000 mL) = 0.010 L Now we can find the moles of ions in the 10 mL of stock solution: moles_of_NH4 = moles_of_SO4 = M₁ * volume_10mL_in_liters = 0.817 M * 0.010 L = 0.00817 mol

Calculate the Concentration of Ions in the Final Solution

Finally, we will find the concentration of ions in the final solution: M₂ = moles_of_ions / volume_of_final_solution The volume of the final solution is 50 mL + 10 mL which must be converted to liters: volume_final_solution_in_liters = 60 mL * (1 L / 1000 mL) = 0.060 L Now, we can find the concentration of ions in the final solution: M₂ = 0.00817 mol / 0.060 L ≈ 0.136 M Since there are 2 moles of ammonium ions and 1 mole of sulfate ions for every mole of (NH4)2SO4, the concentration of ammonium ions will be twice the concentration of sulfate ions in the final solution. Concentration of NH₄⁺ ions = 2 * 0.136 M = 0.272 M Concentration of SO₄²⁻ ions = 0.136 M In conclusion, the concentration of ammonium ions (NH₄⁺) in the final solution is 0.272 M and the concentration of sulfate ions (SO₄²⁻) is 0.136 M.

Key concepts

Molarity

When we talk about solution concentration, one of the most commonly used terms is molarity. It refers to the number of moles of a solute present in one liter of solution. To express it mathematically, molarity (\textbf{M}) is calculated using the formula: \[\begin{equation}M = \frac{\text{moles of solute}}{\text{liters of solution}}\end{equation}\]It's a straightforward measure of concentration because it tells you directly how many moles of a substance are in each liter, which can be especially helpful when you're mixing solutions or reacting them in stoichiometric calculations. In our original exercise, the molarity of the ammonium sulfate in the stock solution was calculated, helping us to find the concentration of ions in the final solution.

Understanding molarity is crucial for laboratory work and scientific research as it is the basis for dilution calculations and directly influences the outcomes of chemical reactions. Whenever you're dealing with solutes and solutions, make sure you're clear on molarity, as it's the centerpiece of many chemical calculations.

Molality

Molality is another measure of solution concentration, but unlike molarity, it does not depend on the volume of the solution, which can be affected by temperature changes. Instead, molality (\textbf{m}) is defined as the number of moles of solute per kilogram of solvent. \[\begin{equation}m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\end{equation}\]This makes molality particularly useful when dealing with experiments where temperature variations are common since it remains unchanged with temperature fluctuations. However, in our exercise, molality is not used because we're focused on the volume-based concentration (molarity) instead of weight-based concentration. It's important to distinguish when to use molarity or molality since each can be crucially important depending on the scientific context or laboratory conditions.

Dilution

Dilution is a process in which more solvent is added to a solution, thereby reducing the solute's concentration. The underlying principle is conservation of the amount of solute, meaning that no solute is added or removed during the dilution process. The equation below denotes this relationship and is vital for dilution calculations: \[\begin{equation}C_1V_1 = C_2V_2\end{equation}\]where - \(C_1\) is the initial concentration,- \(V_1\) is the initial volume,- \(C_2\) is the final concentration,- \(V_2\) is the final volume.

In the context of our original exercise, we applied dilution when water was added to the stock solution of ammonium sulfate. This step was crucial to calculating the concentration of ammonium and sulfate ions in the final solution correctly. Understanding dilution is essential not only in chemistry but also in everyday applications like preparing food recipes, medicinal doses, or even in ecology when considering pollutant dispersion.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It's akin to a recipe that dictates how much of each ingredient we need for the reaction to proceed. In stoichiometry, the molar ratios provided by the balanced chemical equation are the 'recipe' for the reaction. To apply stoichiometry, we must write a balanced chemical equation and then convert the measured quantity (usually mass or volume) to moles using the molar mass or molarity. This is what we performed in the exercise when we found the molar mass of ammonium sulfate to calculate the number of moles. Stoichiometry isn't only used for reactions but any time amounts of substances are involved, such as finding out the concentration of ions from a solute as demonstrated with our problem.

Mastering stoichiometry is fundamental for success in chemistry, as it helps in predicting product yields, understanding reaction mechanisms, and is essential for any future work in the field whether in research, industry, or education.