Problem 131 - Chemistry An Atoms First Approach

Question

You made \(100.0 \mathrm{mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 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 g. What was the concentration of the original lead(II)

Step-by-Step Solution

Answer

The concentration of the original lead(II) nitrate solution was 4.9 M.

Step 1: Find moles of PbCl₂ formed

Firstly, we need to calculate the moles of PbCl₂ formed in the reaction between the lead(II) nitrate solution and sodium chloride solution. We can do this using the molar mass of PbCl₂ and the mass of the solid obtained. Moles of PbCl₂ = \( \frac{Mass\ of\ PbCl_2}{Molar\ mass\ of\ PbCl_2}\) Moles of PbCl₂ = \( \frac{3.407\ g}{278.10\ g/mol} = 0.01225\ mol\)

Step 2: Find moles of Pb(NO₃)₂ in the remaining solution

Since 2.00 mL of the remaining lead(II) nitrate solution was used in the reaction, and we are assuming that the reaction was quantitative, the moles of Pb(NO₃)₂ in that sample would be equal to the moles of PbCl₂ formed. So, we have 0.01225 mol of Pb(NO₃)₂ in the 2.00 mL sample.

Step 3: Calculate moles of Pb(NO₃)₂ in the entire remaining solution

We can now use the ratio of the sample volume to the remaining solution volume to find the total initial moles of Pb(NO₃)₂ in the entire (80.0 mL) solution. Ratio = \( \frac{Volume\ of\ sample}{Total\ remaining\ volume}\) Ratio = \( \frac{2.00\ mL}{80.0\ mL} = 0.025\) Since there were 0.01225 mol of Pb(NO₃)₂ in the 2.00 mL sample, the total initial moles of Pb(NO₃)₂ in the entire 80.0 mL solution would be: \(0.01225\ mol\ \div\ 0.025 = 0.49\ mol\)

Step 4: Calculate the concentration of the original lead(II) nitrate solution

Now we have the moles of Pb(NO₃)₂ in the remaining (80.0 mL) solution. To find the concentration of the original solution, we need to consider the original volume, which was 100.0 mL. Concentration = \( \frac{Moles\ of\ Pb(NO_3)_2}{Original\ volume}\) Concentration = \( \frac{0.49\ mol}{100.0\ mL} = 0.0049\ mol/mL\) Since the question asks for the concentration in molarity (M), we will convert the concentration in mol/mL to mol/L: Concentration = \(0.0049\ mol/mL \times 1000\ mL/L = 4.9\ M\) So, the concentration of the original lead(II) nitrate solution was 4.9 M.

Key Concepts

Molar Mass

Understanding the concept of molar mass is critical when dealing with chemical reactions and calculations. At its essence, the molar mass is the weight of one mole (or 6.022 x 10²³ particles) of a substance, usually expressed in grams per mole (g/mol). This value is foundational for converting between the weight of a substance and the number of moles, allowing chemists to quantify substances in reactions.

Consider the calculations involving the compound PbCl₂ involved in the exercise. The molar mass of PbCl₂ is found by adding together the atomic masses of lead (Pb), chlorine (Cl), and evaluating them relative to their stoichiometric coefficients within the compound. The resulting molar mass then aids in determining the number of moles when the mass of the compound is known, as was used to solve the initial part of the exercise provided.

Stoichiometry

Stoichiometry represents the heart of chemical reaction calculations, allowing us to meticulously work out the quantitative relationships between reactants and products in a chemical equation. The principles of stoichiometry are applied by using the mole ratio from a balanced chemical equation to convert between moles of one chemical to the moles of another. During this process, we ensure that the Law of Conservation of Mass is adhered to, meaning the mass of reactants used equals the mass of products formed.

In the provided exercise, stoichiometry was used to derive the amount of lead(II) nitrate from the generated lead(II) chloride. This process required understanding the 1:1 molar ratio between the two substances based on the chemical equation of their reaction. Accurate stoichiometric calculations are pivotal for deducing the concentration of the original solution.

Quantitative Analysis

The field of quantitative analysis encompasses a variety of techniques and calculations used to determine the amount or concentration of a substance within a given mixture. It involves precise measurements and calculations, as seen in the provided exercise where the mass of the precipitated PbCl₂ is used to ultimately calculate the original concentration of Pb(NO₃)₂.

More specifically, in quantitative analysis, after a reaction is conducted and the product is measured, as in the formation of 3.407 g of PbCl₂, we can backtrack through stoichiometric relationships to reveal the amount of the original reactants. This process underscores the importance of accurate measurements and an understanding of reaction stoichiometry for identifying unknown concentrations in solutions—skills that are essential in a laboratory setting and industries requiring precise chemical quantification.

Molarity

Molarity, symbolized as M, is a standard unit of concentration in chemistry, defined as the number of moles of a solute divided by the volume of the solution in liters. This concept is crucial for preparing solutions in a lab and for conducting experiments that require precise concentrations. The calculation of molarity is a direct application of the relationship between moles, volume, and concentration. As demonstrated in the exercise, once the total moles of a solute are established, one can calculate the molarity by relating these moles to the original volume of the solution.

To elucidate, the final steps of the exercise involved converting the found moles of lead(II) nitrate in the reduced volume of 80.0 mL to the molarity of the initial 100.0 mL solution. Moreover, molarity is vital for predicting the outcome of reactions and is the cornerstone of dilutions and concentration adjustments in many chemical processes.