Problem 112 - Chemistry The Central Science In Si Units

Question

A \(3.50 \mathrm{~g}\) of an alloy which contains only lead and tin is dissolved in hot \(\mathrm{HNO}_{3} .\) Excess sulfuric acid is added to this solution and \(1.57 g\) of \(\mathrm{PbSO}_{4}(s)\) is obtained. (a) Write the net ionic equation for the formation of \(\mathrm{PbSO}_{4}\). (b) Assuming all the lead in the alloy reacted to form \(\mathrm{PbSO}_{4}\), what was the amount, in grams, of lead and tin in the alloy respectively?

Step-by-Step Solution

Answer

Pb net ionic equation: \( \text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 \). Lead: 1.073 g, Tin: 2.427 g.

Step 1: Write Net Ionic Equation for PbSO4 Formation

In this step, we will write the net ionic equation for the formation of lead sulfate (PbSO4). The reaction between lead ions and sulfate ions forms solid lead sulfate. The net ionic equation is:\[\text{Pb}^{2+} (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{PbSO}_4 (s)\]

Step 2: Understanding the Conversion of Pb2+ to PbSO4

We know that the entire mass of lead forms lead sulfate. Given its molar mass of PbSO4 (303.26 g/mol), determine the moles of Pb in the PbSO4. Use the formula:\[moles = \frac{\text{given mass}}{\text{molar mass}}\]\[moles \ of \ PbSO_4 = \frac{1.57 \ g}{303.26 \ g/mol} \approx 0.00518 \text{ mol}\]

Step 3: Determine Mass of Lead in the Alloy

Recognizing that 1 mole of PbSO4 contains 1 mole of Pb, we can say the moles of Pb in PbSO4 equals the moles of Pb in the alloy. Calculate the mass of lead using its molar mass (207.2 g/mol):\[\text{mass} = moles \times \text{molar mass} = 0.00518 \times 207.2 \approx 1.073\text{ g}\]

Step 4: Calculate Mass of Tin

The initial total mass of the alloy is 3.50 g. After determining the mass of lead, subtract from the total to find the mass of tin:\[\text{mass of tin} = 3.50 \text{ g} - 1.073 \text{ g} \approx 2.427 \text{ g}\]

Step 5: Conclusion

We have determined the mass of lead and tin in the alloy. The mass of lead is approximately 1.073 g, and the mass of tin is approximately 2.427 g.

Key Concepts

Net Ionic Equation

Net ionic equations are simplified chemical equations that only show the species directly involved in a chemical reaction. These equations omit spectator ions, which are present in the solution but don't participate in the reaction. In the context of the alloy composition problem, the net ionic equation represents the formation of lead sulfate from lead and sulfate ions.

Specifically, when nitride ions in a solution interact with sulfate ions, solid lead sulfate (\[\text{PbSO}_4(s)\]) precipitates. This is because lead ions (\[\text{Pb}^{2+}(aq)\]) directly combine with sulfate ions (\[\text{SO}_4^{2-}(aq)\]) to form lead sulfate. The net ionic equation captures this reaction as:

• \[\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s)\]

This reaction underscores the basic principle of double displacement and precipitation reactions in chemistry, where two ions in solution form an insoluble compound.

Lead Sulfate Formation

Lead sulfate (\[\text{PbSO}_4\]) is an important chemical compound resulting from the reaction of lead ions with sulfate ions. It is typically formed as a solid precipitate in aqueous solutions. In these reactions, any soluble lead source can react with a soluble sulfate source, like sulfuric acid, to form this naturally occurring mineral.

This process is significant in various applications, such as in the lead-acid batteries or certain industrial processes. Importantly, during the chemical reaction, the stoichiometry or the quantitative relation between constituent compounds must be calculated accurately to predict the amount of product formed. In this scenario, we obtain 1.57 g of \(\text{PbSO}_4\), which directly indicates the amount of lead originally present in the alloy.

Such reactions are a prime example of using precipitation to isolate specific components from a mixture, which is fundamental in analytical chemistry practices.

Stoichiometry

Stoichiometry is the study of the quantitative aspects of chemical reactions, focusing on the measurement and calculation of reactants and products in a chemical equation. It requires the use of molar ratios, derived from the balanced chemical equation, to determine the amounts of various substances involved in a reaction.

In solving the alloy composition problem, stoichiometry helps us identify the amount of lead in 1.57 g of \(\text{PbSO}_4\) formed. Knowing that 1 mole of \(\text{PbSO}_4\) contains 1 mole of lead ions, we calculate the moles of lead using the formula:

• Given Mass of \(\text{PbSO}_4 = 1.57\) g
• Molar Mass of \(\text{PbSO}_4 = 303.26\) g/mol

By dividing the mass of lead sulfate by its molar mass, we find the moles of lead sulfate, and consequently the moles of lead, allowing us to calculate the mass of lead in the alloy. This kind of calculation is central to predicting and understanding chemical yields and compositions.

Solution Chemistry

Solution chemistry involves the study of the interactions and reactions that occur in solutions. It's essential for understanding how substances dissolve, react, and precipitate in a solvent like water. When the alloy was dissolved in nitric acid (\(\text{HNO}_3\)), the metals within the alloy dissociated into ions.

The introduction of an excess amount of sulfuric acid (\(\text{H}_2\text{SO}_4\)) caused lead ions to react with sulfate ions, forming the insoluble solid, lead sulfate. Understanding the solubility rules helps predict which compounds will dissolve or form a precipitate.

• Lead sulfate is one of those compounds that is poorly soluble in water, thus it precipitates out of solution when formed.
• This principle is used to separate compounds in mixtures or to identify specific ions in a solution.

Solution chemistry enables chemists to design reactions that efficiently convert reactants into desired products by controlling conditions such as concentration and temperature. This knowledge is invaluable in both academic research and industrial applications.