Question

If a solution of \(\mathrm{AgNO}_{3}\) is added to an \(\mathrm{HCl}\) solution, insoluble \(\mathrm{AgCl}\) will precipitate: $$ \mathrm{AgNO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{HNO}_{3}(a q) $$ (a) What volume of a \(1.20 \mathrm{M} \mathrm{Ag} \mathrm{NO}_{3}\) solution must be added to \(250.0 \mathrm{~mL}\) of a \(0.100 M \mathrm{HCl}\) solution to precipitate all the chloride ion? (b) What mass of \(\mathrm{AgCl}\) should precipitate?

Short answer

a) 20.8 mL b) 3.583 g

Step-by-step solution

Write the balanced Chemical equation

Before starting any calculations, the balanced chemical equation must be confirmed: \[ \mathrm{AgNO}_3(aq) + \mathrm{HCl}(aq) \longrightarrow \mathrm{AgCl}(s) + \mathrm{HNO}_3(aq) \] This formula will be used in the molar calculations in the steps below.

Calculate the moles of HCl

Now need to calculate the number of moles of HCl. The formula \(Moles = Molarity \times Volume\) can be used to find the moles of HCl. Given \(Molarity_{HCl}=0.100M\) and \(Volume_{HCl}=250.0mL\), we get: \[Moles_{HCl} = 0.100 M \times 250.0 mL \times 1L/1000mL = 0.025 moles\]

Calculate the volume of AgNO3 solution needed

From the balanced equation, it is evident that one mole of AgNO3 reacts with one mole of HCl to produce one mole of AgCl. Therefore, the moles of AgNO3 required are equal to the moles of HCl (0.025 moles). Now, to convert this into volume, we rearrange \(Moles = Molarity \times Volume\) to give \(Volume = Moles / Molarity\). Substituting values we get: \[Volume_{AgNO3} = 0.025 moles / 1.20 M = 0.0208 L or 20.8 mL \]

Calculate the mass of AgCl precipitated

From the stoichiometric coefficients in the balanced equation, we observe that the mole of AgCl precipitated equals the mole of HCl. We know the molar mass of AgCl is 143.32 g/mol, so multiply our moles of AgCl (which equals our moles of HCl, 0.025 moles) with the molar mass to find the mass of AgCl precipitated: \[Mass_{AgCl} = 0.025 moles \times 143.32 g/mol = 3.583 g \]

Key concepts

Stoichiometry

Stoichiometry involves the calculation of the quantities of reactants and products involved in a chemical reaction. It's based on the law of conservation of mass and the principle that, in a balanced chemical equation, the ratio of the moles of any two substances is equal to the ratio of their coefficients.

To begin with any stoichiometric problem, you must first have a balanced chemical equation, which provides the proportion of reactants that combine and the amount of products formed. Once the equation is balanced, as in the step-by-step solution provided, you can then use mole ratios to convert between the amounts of reactants and products.

To effectively master stoichiometry, it's essential to:

• Understand the concept of the mole and Avogadro's number,
• Be comfortable with converting between mass, moles, and the number of particles,
• Have a good grasp of molar mass and its significance in these conversions.

Following these points ensures that you can accurately determine the quantity of one chemical substance that reacts with or is produced from another substance in a chemical reaction.

Molarity Calculations

Molarity is a measure of concentration in chemistry, defining the number of moles of solute per liter of solution. To perform molarity calculations, we use the following equation: \( Molarity = \frac{Moles \, of \, solute}{Volume \, of \, solution \, in \, liters} \).

When you're given the molarity and volume of a solution, as seen in the exercise, you can calculate the number of moles of the solute using the formula. Conversely, knowing the moles and the molarity, you can solve for the volume, which is essential for dilution problems or to determine how much to administer to achieve a desired reaction outcome.

In the example provided, understanding molarity is fundamental to figuring out not only the volume of \(\mathrm{AgNO}_3\) needed but also crucial for the subsequent calculation of how much silver chloride (\(\mathrm{AgCl}\)) will precipitate, thus bridging the stoichiometry to the practical outcome of the chemical reaction.

Silver Chloride Precipitation

Silver chloride precipitation is a classic example of a precipitation reaction, a type of chemical reaction in which an insoluble salt forms and separates from the solution. The equation \(\mathrm{AgNO}_3(aq) + \mathrm{HCl}(aq) \longrightarrow \mathrm{AgCl}(s) + \mathrm{HNO}_3(aq)\) illustrates the reaction between aqueous solutions of silver nitrate and hydrochloric acid to form solid silver chloride and aqueous nitric acid.

In practical terms, when you mix these two clear solutions, you can expect to see the appearance of a white solid - silver chloride. Precipitation reactions like this one are not only important in laboratory analysis and synthesis but also have applications in waste treatment and resource recovery. Understanding the stoichiometry and molarity calculations for such reactions ensures that one can predict and control the amount of precipitate formed—a crucial skill in various chemical industries.

To predict the amount of precipitate formed:

• Identify the precipitating ions,
• Use the solubility rules to determine if any insoluble compounds form,
• Apply stoichiometry to calculate the theoretical yield of the precipitate.

By mastering these steps, you can efficiently tackle questions about silver chloride precipitation and similar processes.