Question

A solution is prepared by diluting \(0.7850 \mathrm{~L}\) of \(1.262 \mathrm{M}\) potassium sulfide solution with water to a final volume of \(2.000 \mathrm{~L}\). (a) How many grams of potassium sulfide were dissolved to give the original solution? (b) What are the molarities of the potassium sulfide, potassium ions, and sulfide ions in the diluted solution?

Short answer

Answer: The mass of potassium sulfide in the original solution is 109.277 g. In the diluted solution, the molarities are as follows: K2S has a molarity of 0.497125 M, K+ ions have a molarity of 0.99425 M, and S^2- ions have a molarity of 0.497125 M.

Step-by-step solution

Calculate the mass of K2S in the original solution

To find the mass of K2S in the original solution, we'll use the formula mass = molarity × volume × molar mass. Molarity (M1) = 1.262 M Volume (V1) = 0.7850 L Molar mass of K2S = 39.098 + 39.098 + 32.07 = 110.266 g/mol Mass = (1.262 mol/L) × (0.7850 L) × (110.266 g/mol) = 109.277 g There are 109.277 grams of potassium sulfide dissolved in the original solution.

Calculate the molarity of K2S in the diluted solution

To find the molarity of K2S in the diluted solution, we'll use the dilution formula M1V1 = M2V2. Final volume (V2) = 2.000 L M1 = 1.262 M V1 = 0.7850 L M2 = (M1 × V1) / V2 = (1.262 mol/L × 0.7850 L) / 2.000 L= 0.497125 M The molarity of K2S in the diluted solution is 0.497125 M.

Calculate the molarities of K+ and S^2- ions

As each molecule of K2S dissociates into two potassium ions (K+) and one sulfide ion (S^2-): Molarity of K+ ions = 2 × Molarity of K2S = 2 × 0.497125 M = 0.99425 M Molarity of S^2- ions = Molarity of K2S = 0.497125 M In the diluted solution: Molarity of potassium sulfide (K2S) = 0.497125 M Molarity of potassium ions (K+) = 0.99425 M Molarity of sulfide ions (S^2-) = 0.497125 M

Key concepts

Dilution Calculation

Dilution is an essential process in chemistry where a solution is made less concentrated by adding more solvent. In the context of the exercise, diluting a potassium sulfide solution involves increasing its volume by adding water. This decreases the molarity of the solution.To perform dilution calculations, we use the formula: \( M_1V_1 = M_2V_2 \), where:

• \( M_1 \) is the initial molarity before dilution.
• \( V_1 \) is the initial volume of the solution.
• \( M_2 \) is the new molarity after dilution.
• \( V_2 \) is the final volume of the solution.

For the given problem:- \( M_1 = 1.262 \) M- \( V_1 = 0.7850 \) L- \( V_2 = 2.000 \) LBy substituting these values into the formula, we find that the molarity of the diluted solution, \( M_2 \), is \( 0.497125 \) M. This consistent use of the dilution formula helps maintain concentration ratios accurately.

Molarity

Molarity is a crucial concept when discussing solutions in chemistry. It quantifies the concentration of a solute in a solution, expressed in moles per liter (mol/L). This measurement allows chemists to understand how much of a substance is present in a given solution.In practice:- To calculate molarity, you divide the number of moles of solute by the volume of the solution in liters.- The formula for molarity (\( M \)) is: \( M = \frac{n}{V} \), where \( n \) is the number of moles, and \( V \) is the volume of the solution.In the original exercise, potassium sulfide's initial molarity was calculated by knowing the volume of the solution (0.7850 L) and its concentration expressed as 1.262 M. By configuration of molarity in calculations, you can find how concentration changes with the volume.Understanding molarity helps analyze and predict the outcomes of chemical reactions, as the precise amount of reactants needed can be calculated.

Potassium Sulfide

Potassium sulfide (\( K_2S \)) is a chemical compound composed of two potassium (K) ions and one sulfide (S) ion. As an ionic compound, it dissolves in water to dissociate into its respective ions, contributing to the conductivity of the solution.Calculating the mass of potassium sulfide in a solution involves multiplying the molarity of \( K_2S \), the solution's volume, and the molar mass of \( K_2S \) which is 110.266 g/mol. In the exercise, the original solution had a molarity of 1.262 M and a volume of 0.7850 L, which weighed approximately 109.277 g.Understanding the properties of potassium sulfide is vital for applications that involve sulfur compounds or need strong bases. Knowing how to calculate and manipulate its concentration is essential in various fields such as materials science and chemistry.

Ions in Solution

When ionic compounds like potassium sulfide are dissolved in water, they dissociate into their respective ions. This ionic dissociation is crucial for processes such as electrical conductivity in solutions and in many chemical reactions.In the case of potassium sulfide (\( K_2S \)), each molecule separates into:

• Two potassium ions (\( K^+ \))
• One sulfide ion (\( S^{2-} \))

In a diluted solution with molarity \( 0.497125 \) M for \( K_2S \), the individual ion concentrations are:- Potassium ion (\( K^+ \)) concentration: 2 × 0.497125 M = 0.99425 M- Sulfide ion (\( S^{2-} \)) concentration: 0.497125 MThis concept is especially significant in analytical chemistry and electrochemistry, where ion concentration impacts the behavior of solutions. Understanding the dissociation of compounds into ions helps predict how they will interact in a given chemical environment. Properly managing ionic concentrations is essential in fields ranging from environmental science to pharmacology.