Question

Whatmass of \(\mathrm{KCl}\) is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution?

Short answer

To precipitate the silver ions from 15.0 mL of 0.200 M AgNO₃ solution, 0.224 g of KCl is needed.

Step-by-step solution

Calculate the moles of AgNO₃ in the solution.

To find the moles of AgNO₃, we will use the molarity and volume of the solution. Molarity (M) is defined as the number of moles of solute per liter of solution. So, we can calculate the moles of AgNO₃ using: Moles of AgNO₃ = Molarity × Volume The volume should be in liters, so we'll convert 15.0 mL to liters by dividing by 1000: 15.0 mL = 0.015 L Now, we will plug in the given values to find the moles of AgNO₃: Moles of AgNO₃ = 0.200 M × 0.015 L = 0.003 moles

Determine the stoichiometry of the reaction.

The reaction between AgNO₃ and KCl is a precipitation reaction, in which silver ions and chloride ions form solid silver chloride (AgCl), while potassium and nitrate ions remain in solution as spectator ions: AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq) The stoichiometry of the reaction is 1:1 for AgNO₃ and KCl, meaning that each mole of AgNO₃ will react with one mole of KCl.

Calculate the moles of KCl needed.

Since the stoichiometry of the reaction between AgNO₃ and KCl is 1:1, the moles of KCl needed to precipitate all the silver ions will be the same as the moles of AgNO₃ calculated in step 1: Moles of KCl = 0.003 moles

Convert the moles of KCl to mass.

Now that we know the moles of KCl needed, we can calculate the mass by using the molar mass of KCl. The molar mass of KCl is: K: 39.1 g/mol Cl: 35.5 g/mol Molar mass of KCl = 39.1 + 35.5 = 74.6 g/mol So, the mass of KCl can be found using: Mass of KCl = Moles of KCl × Molar mass of KCl Mass of KCl = 0.003 moles × 74.6 g/mol = 0.224 g Therefore, 0.224 g of KCl is needed to precipitate the silver ions from 15.0 mL of 0.200 M AgNO₃ solution.

Key concepts

Precipitation Reaction

In chemistry, a precipitation reaction is a type of chemical reaction where two solutions are mixed together, and an insoluble solid, called a precipitate, is formed. In the case of the reaction between silver nitrate (\( \mathrm{AgNO}_3 \)) and potassium chloride (\( \mathrm{KCl} \)), silver ions (\( \mathrm{Ag}^+ \)) react with chloride ions (\( \mathrm{Cl}^- \)) to form silver chloride (\( \mathrm{AgCl} \)), a solid precipitate. - The general equation for a precipitation reaction is: \( \text{AB (aq) + CD (aq) } \rightarrow \text{ AD (s) + CB (aq)} \) - Here: - \( \mathrm{AgNO}_3 \) (aq) and \( \mathrm{KCl} \) (aq) are in aqueous form, meaning they are dissolved in water. - \( \mathrm{AgCl} \) is the solid that forms, making it the precipitate. The remaining ions, \( \mathrm{K}^+ \) and \( \mathrm{NO}_3^- \), remain in the solution and do not participate in forming a solid.Precipitation reactions are fundamental in chemistry because they are often used to isolate certain elements from a solution or to detect the presence of specific ions.

Molar Mass

Molar mass is a crucial concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is important for converting between the mass of a substance and the amount in moles. For instance, to calculate the amount of potassium chloride (\( \mathrm{KCl} \)) needed in a reaction, we utilize its molar mass.To determine the molar mass of a compound: - Add up the atomic masses of each element present within the compound based on their occurrences. - For \( \mathrm{KCl} \): - Potassium (\( \mathrm{K} \)) has an atomic mass of approximately 39.1 g/mol. - Chlorine (\( \mathrm{Cl} \)) has an atomic mass of approximately 35.5 g/mol.The resultant molar mass is: - \( 39.1 + 35.5 = 74.6 \) g/mol. This molar mass helps in converting moles of \( \mathrm{KCl} \) to grams, which can subsequently allow us to determine the precise mass required for a reaction.

Moles Calculation

Moles calculation is an essential skill in solution chemistry, allowing one to determine the amount of a substance in a given solution. The concept of moles connects the number of individual particles to a measurable mass, which is crucial for performing chemical reactions accurately.To find the moles of a solute in a solution, you need to know its molarity and volume: - **Molarity** (\( \mathrm{M} \)) is defined as the number of moles of solute per liter of solution. - **Volume** must be in liters (for accuracy in molarity calculations).For example, in the problem, we're dealing with a 0.200 M \( \mathrm{AgNO_3} \) solution: - Convert 15.0 mL to liters: \( 15.0 \) mL = \( 0.015 \) L- The formula to calculate moles is: \[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]Here, the moles of \( \mathrm{AgNO_3} \) would be: \[ 0.200 \times 0.015 = 0.003 \text{ moles} \]Understanding this calculation is foundational for tackling more complex chemical equations and reactions.

Solution Chemistry

Solution chemistry is a branch of chemistry that focuses on the properties and behaviors of solutions and their solutes. Solutions contain a solute, which is the dissolved substance, and a solvent, which is the medium that dissolves the solute. In solution chemistry, the concentration of solutions is often discussed in terms of molarity.Important Aspects of Solution Chemistry:- **Molarity (\( \mathrm{M} \))**: This expresses the concentration in moles of solute per liter of solution, critical for determining reactant amounts.- **Dissolution Process**: Understanding how solutes dissolve and interact with the solvent helps in predicting reaction outcomes.- **Spectator Ions**: These are ions present in a solution that do not participate in the actual chemical reaction, such as \( \mathrm{K}^+ \) and \( \mathrm{NO}_3^- \) in the precipitation reaction we observed.Solution chemistry is fundamental for comprehending various laboratory and industrial processes, as it influences rates of reaction, the ability to form precipitates, and the general equilibrium of chemical systems.