You made \(100.0 \mathrm{~mL}\) of a lead(II) \(\mathrm{nm}\) olution for lab but forgot to cap it. The next lab session you noticed that there was only \(80.0 \mathrm{~mL}\) left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{~mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of \(3.407 \mathrm{~g} .\) What was the concentration of the original lead(II) nitrate solution?

Stoichiometry is a foundational concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. When you have a balanced chemical equation, stoichiometry allows you to predict how much product will form from a given amount of reactant and vice versa. It's essential when you're trying to determine the concentration of a solution, as in our textbook example with the lead(II) nitrate and sodium chloride reaction. A balanced equation tells us that one mole of lead(II) nitrate reacts with two moles of sodium chloride to yield one mole of lead(II) chloride. This 1:1:2 ratio underpins the calculations needed to find the concentration of the original lead(II) nitrate solution. Understanding stoichiometry can help students connect the theoretical aspects of chemical reactions with practical laboratory scenarios.

For exercises like these, ensure to identify the molar ratios between reactants and products in a balanced chemical equation. From these ratios, you can convert between masses, volumes, and moles of different substances involved in a reaction.

The molar mass of a compound is calculated by summing the atomic masses of all the atoms in its chemical formula. In our example, the compound lead(II) chloride (PbCl₂) has a molar mass that is the sum of one lead atom's mass and two chlorine atoms' masses. Finding the molar mass is crucial when converting between moles and grams of a substance, a common task in chemistry problems. With the correct molar mass, as provided in the step-by-step solution, we can calculate how many moles of lead(II) chloride were present in our precipitate, leading us one step closer to the final concentration of our lead(II) nitrate solution.

Remember that molar mass is expressed in grams per mole (g/mol), and when solving problems involving molar mass, always ensure that you use accurate atomic masses for each element as listed on the periodic table.

Balancing chemical reactions is an essential practice in stoichiometry. It ensures that the number of atoms for each element is the same on both sides of a chemical equation, aligning with the law of conservation of mass. The problem involves the reaction between lead(II) nitrate and sodium chloride, which produces lead(II) chloride and sodium nitrate. The chemical equation needs to be balanced so that you can accurately use stoichiometry to find out how much of each reactant and product is involved. In our situation, we end up with a balanced equation indicating that one mole of lead(II) nitrate reacts with two moles of sodium chloride, forming one mole of lead(II) chloride and two moles of sodium nitrate, a direct application of balanced reaction in resolving the concentration of the initial solution.

The concentration of a solution is a measure of how much solute is dissolved in a given quantity of solvent. It's typically expressed in molarity (M), which is moles of solute per liter of solution. In our example, we needed to determine the molarity of the lead(II) nitrate solution after evaporation had decreased its volume. By calculating the molarity of the reduced solution, we deduced how concentrated the solute had become. Knowing how to calculate and interpret solution concentration is pivotal for preparing solutions of precise molarities in laboratory settings and is a frequent task in chemistry homework assignments.

To compute concentration, divide the number of moles of solute by the volume of the solution in liters. This is exactly what was demonstrated in the step-by-step solution with the reduced volume after evaporation.

Dilution is the process of decreasing the concentration of a solution by adding more solvent. The dilution principle states that the amount of solute in the solution remains the same before and after dilution; only the volume changes. This principle is mathematically expressed as C₁V₁ = C₂V₂, where C₁ and V₁ are the concentration and volume of the initial solution, and C₂ and V₂ are the concentration and volume after dilution. In our exercise, by applying the dilution principle, we were able to find the concentration of the original lead(II) nitrate solution before evaporation occurred.

When faced with problems involving dilution, it's important to remember that while the concentrations before and after dilution are different, the total number of moles of the solute remains constant. This principle helps in a range of applications, from preparing laboratory reagents to issues regarding environmental pollution.