Problem 18 - Analytical Chemistry

Question

The chloride in a 0.12 -g sample of \(95 \%\) pure \(\mathrm{MgCl}_{2}\) is to be precipitated as \(\mathrm{AgCl}\). Calculate the volume of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution required to precipitate the chloride and give a \(10 \%\) excess.

Step-by-Step Solution

Answer

26.334 mL of 0.100 M AgNO3 solution is required.

Step 1: Calculate Amount of MgCl2

Since the sample is 95% pure, calculate the amount of \( \mathrm{MgCl}_{2} \) present: \(0.12 \, \text{g} \times 0.95 = 0.114 \, \text{g of} \, \mathrm{MgCl}_{2}.\)

Step 2: Calculate Moles of MgCl2

Determine the moles of \( \mathrm{MgCl}_{2} \) using its molar mass of \( \text{95.21 g/mol}: \)\(\frac{0.114 \, \text{g}}{95.21 \, \text{g/mol}} = 0.001197 \, \text{mol MgCl}_{2}.\)

Step 3: Determine Moles of Chloride Ions

Since \( \mathrm{MgCl}_{2} \) produces 2 moles of \( \mathrm{Cl}^- \) ions per mole, calculate the moles of chloride:\( \)\(0.001197 \, \text{mol MgCl}_{2} \times 2 = 0.002394 \, \text{mol Cl}^-.\)

Step 4: Calculate Moles of Required AgNO3

\( \mathrm{AgNO}_3 \) reacts with \( \mathrm{Cl}^- \) in a 1:1 ratio to form \( \mathrm{AgCl} \). Therefore, we need the same moles of \( \mathrm{AgNO}_3 \):\( \)\(0.002394 \, \text{mol AgNO}_3.\)

Step 5: Add 10% Excess of AgNO3

Calculate a 10% excess of \( \mathrm{AgNO}_3 \):\[0.002394 \, \text{mol} \times 1.10 = 0.0026334 \, \text{mol AgNO}_3.\]

Step 6: Calculate Volume of AgNO3 Solution

Use the molarity equation \( C = \frac{n}{V} \) to find the volume of \( 0.100 \, \mathrm{M} \) \( \mathrm{AgNO}_3 \):\[V = \frac{0.0026334 \, \text{mol}}{0.100 \, \text{mol/L}} = 0.026334 \, \text{L} = 26.334 \, \text{mL}.\]

Step 7: Final Volume for Precipitation with 10% Excess

Thus, you need \(26.334 \, \text{mL}\) of \(0.100 \, \mathrm{M} \, \mathrm{AgNO}_{3} \) solution to completely precipitate the chloride with a 10% excess.

Key Concepts

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, called the precipitate. In this exercise, chloride ions from the magnesium chloride (\(\text{MgCl}_2\)) in the sample react with silver nitrate (\(\text{AgNO}_3\)) to form silver chloride (\(\text{AgCl}\)), which is an insoluble white solid. Such reactions are often used to remove ions from solution or to produce pure salts.
In a general sense, precipitation reactions can be predicted using solubility rules.
For instance, most silver salts are insoluble except for \(\text{AgF}\) and a few others. This characteristic allows chemists to selectively precipitate silver chloride in a mixture. Precipitation reactions are crucial in chemical analysis and purification processes.
In our specific reaction, the balanced chemical equation is:\[ \text{AgNO}_3 (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) + \text{NO}_3^- (aq)\]. The nitrate ion (\(\text{NO}_3^-\)) remains in solution and does not participate in the precipitation event.

Molarity Calculation

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In this exercise, we use the concept of molarity to calculate the volume of the \(0.100 \, \text{M} \, \text{AgNO}_3\) solution required to react with the chloride ions.
Molarity is calculated using the formula:\[C = \frac{n}{V}\], where \(C\) is molarity, \(n\) is the number of moles, and \(V\) is the volume in liters.
In the exercise, once the moles of \(\text{AgNO}_3\) required are determined, they are multiplied by \(1.10\) to account for a \(10\%\) excess. This is especially important in practical lab settings to ensure complete reactions. The final calculation provides the necessary volume in milliliters.

Chemical Reactions

Chemical reactions represent the process by which reactants transform into products. They're characterized by changes in chemical bonds and the rearrangement of atoms. Here, magnesium chloride and silver nitrate participate in a reaction that also involves principles of stoichiometry.
Stoichiometry plays a vital role in predicting the outcomes of chemical reactions. It involves the use of balanced chemical equations to determine the quantitative relationships between reactants and products. For example, in this problem, knowing the stoichiometry helps in calculating the exact amounts of reactants necessary to achieve complete conversion to products.
Key characteristics of chemical reactions include changes in energy, observed as heat or light, formation of gases or precipitates, and color changes in the reacting solutions. Understanding these reactions is fundamental to grasping concepts from balancing equations to reaction mechanisms, making it a cornerstone of chemistry studies.