Question

What mass 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 $15.0\mathrm{mL}$ of $0.200\mathrm{M}{\mathrm{AgNO}}_3$ solution, $0.2237\mathrm{g}$ of potassium chloride ($\mathrm{KCl}$) is needed.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between silver nitrate and potassium chloride to form a silver chloride precipitate and potassium nitrate. The balanced equation is: $AgN{O}_3+KCl\to AgC{l}_{(s)}+KN{O}_3$ Notice that the stoichiometric ratio between AgNO₃ and KCl is 1:1.

Calculate the moles of silver nitrate

We are given the volume (15.0 mL) and concentration (0.200 M) of the silver nitrate solution. In order to find the moles of silver nitrate, we can use the formula: Moles = Volume × Concentration. Convert the volume to liters and calculate the moles: Moles of $AgN{O}_3=15.0\mathrm{mL}\ast \frac{1\mathrm{L}}{1000\mathrm{mL}}\ast 0.200\mathrm{M}=0.00300\mathrm{mol}$

Calculate the moles of potassium chloride

We know that the stoichiometric ratio between silver nitrate and potassium chloride is 1:1 from the balanced chemical equation. Therefore, the moles of potassium chloride needed for the reaction will be the same as the moles of silver nitrate. Moles of KCl = Moles of $AgN{O}_3=0.00300\mathrm{mol}$

Calculate the mass of potassium chloride

To find the mass of potassium chloride (KCl), we will use the molecular weight of KCl and the moles calculated in the previous step. The molecular weight of KCl = 39.10 g/mol (K) + 35.45 g/mol (Cl) = 74.55 g/mol Now, calculate the mass of KCl: Mass of KCl = Moles of KCl × Molecular Weight of KCl $Mass\mathrm{\_}of\mathrm{\_}KCl=0.00300\mathrm{mol}\ast 74.55\frac{g}{\mathrm{mol}}=0.2237g$

Final answer

Therefore, 0.2237 grams of potassium chloride (KCl) is needed to completely precipitate the silver ions from 15.0 mL of 0.200 M silver nitrate (AgNO₃) solution.

Key concepts

Chemical Equations

Understanding the language of chemical reactions is fundamental in chemistry, and it starts with chemical equations. A chemical equation represents a chemical reaction where the reactants are listed on the left and the products on the right, separated by an arrow indicating the direction of the reaction. For instance, in the given exercise, the chemical equation is:
$$AgN{O}_3+KCl\to AgC{l}_{(s)}+KN{O}_3$$
This tells us that silver nitrate reacts with potassium chloride to yield silver chloride and potassium nitrate. Importantly, the equation must be balanced with the same number of atoms of each element on both sides, reflecting the conservation of mass. A balanced equation also allows us to understand the stoichiometry of the reaction - the quantitative relationship between reactants and products in a chemical reaction.

Molar Mass Calculation

When you're faced with a problem involving chemical compounds, it's critical to know how to calculate molar mass. Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule of the substance. For example:
$$MolarMassofKCl=(39.10g/mol(K))+(35.45g/mol(Cl))=74.55g/mol$$
You can find the atomic masses on the periodic table for each element comprising the compound. Knowing the molar mass is essential because it serves as a conversion factor between grams and moles, enabling you to perform crucial calculations related to the substance's quantity in a reaction.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where solutions containing different ions produce an insoluble solid, or a precipitate, when mixed. In the context of the exercise, silver nitrate and potassium chloride are aqueous solutions that react to form a precipitate of silver chloride:$$AgN{O}_3(aq)+KCl(aq)\to AgC{l}_{(s)}+KN{O}_3(aq)$$
The subscript 's' in AgCl_{(s)} indicates that silver chloride is the precipitate. Precipitation reactions are critical for understanding solubility, separating substances, and are widely used in various industries including water treatment, medicine, and mining. Recognizing what combinations of ions will form a precipitate is vital for predicting the course of a reaction and for the proper disposal of substances in environmental chemistry.