Problem 72 - Chemistry The Central Science In Si Units

Question

Suppose that a \(10-\mathrm{mL}\) sample of a solution is to be tested for \(1^{-}\) ion by addition of 1 drop \(\left(0.2 \mathrm{~mL}\right.\) ) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). What is the minimum number of grams of \(I^{-}\) that must be present for \(\mathrm{Pbl}_{2}(s)\) to form?

Step-by-Step Solution

Answer

The minimum amount of iodide required is 0.005076 grams.

Step 1: Understand the Reaction

The reaction involved here is the precipitation of lead(II) iodide, \( \text{PbI}_2(s) \). The chemical reaction can be written as: \[ \text{Pb}^{2+} + 2 \text{I}^- \rightarrow \text{PbI}_2(s) \] This means that 1 mole of \( \text{Pb}^{2+} \) reacts with 2 moles of \( \text{I}^- \) to form the solid \( \text{PbI}_2 \).

Step 2: Determine Moles of Pb(NO₃)₂ Added

We're adding 0.2 mL (or 0.0002 L) of \( 0.10 \text{ M} \) \( \text{Pb(NO}_3)_2 \). Calculate the number of moles of \( \text{Pb}^{2+} \) ions: \[ \text{Moles of } \text{Pb}^{2+} = 0.10 \, \text{mol/L} \times 0.0002 \, \text{L} = 2 \times 10^{-5} \, \text{mol} \]

Step 3: Calculate Moles of Iodide Required

From the reaction equation, 2 moles of \( \text{I}^- \) are needed for every mole of \( \text{Pb}^{2+} \). Therefore, \( 2 \times 2 \times 10^{-5} = 4 \times 10^{-5} \) moles of \( \text{I}^- \) are required.

Step 4: Find Mass of Iodide Required

The molar mass of \( \text{I}^- \) is approximately 126.9 g/mol. Calculate the mass needed: \[ \text{Mass of } \text{I}^- = 4 \times 10^{-5} \, \text{mol} \times 126.9 \, \text{g/mol} = 5.076 \times 10^{-3} \, \text{g} = 0.005076 \, \text{g} \]

Step 5: Conclusion

Therefore, the minimum amount of iodide ion required for the formation of a precipitate of \( \text{PbI}_2 \) is \( 0.005076 \) grams.

Key Concepts

Solubility

Solubility is a critical concept in chemistry that describes how substances dissolve in solvents. The solubility of a compound tells you the maximum amount of that compound that can dissolve in a certain amount of solvent at a specified temperature and pressure. In the reaction we're examining, lead(II) iodide \( \text{PbI}_2(s) \) is the precipitate formed. When certain ions in solution reach a concentration that exceeds the solubility limit, they form a solid that separates from the solution. This phenomenon is called a precipitation reaction.Understanding the solubility of products involved in a reaction is vital for predicting whether a precipitation reaction will occur. If the product of the reaction is insoluble or only slightly soluble, a precipitate will likely form. For example, lead(II) iodide is known to be poorly soluble in water, which is why it precipitates out when lead and iodide ions are present in sufficient quantities.

Mole Calculation

Mole calculation is a fundamental aspect of chemical reactions. The mole is a unit used to express amounts of a chemical substance, giving us a way to quantify particles and entities at the atomic or molecular scale with real-world amounts.In our exercise, we determine the number of moles of both reactants involved: lead(II) ions \(\text{Pb}^{2+}\) and iodide ions \(\text{I}^-\). Calculating moles allows chemists to understand how much of a substance it's reacting with. For instance, in the exercise, we start by calculating the moles of lead(II) ions in the solution using the formula:

• Moles = Molarity \(\times\) Volume

Here, a molarity of 0.10 M and a volume of 0.0002 L yields \(2 \times 10^{-5} \) moles of \(\text{Pb}^{2+}\). Such calculations help ensure that the quantities align with the chemical reaction's requirements.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the amount of reactants required or products formed in a reaction.In our chemical equation \( \text{Pb}^{2+} + 2 \text{I}^- \rightarrow \text{PbI}_2(s) \), stoichiometry informs us that one mole of \(\text{Pb}^{2+}\) requires two moles of \(\text{I}^-\) to produce one mole of \(\text{PbI}_2\). By applying the stoichiometric coefficients from the balanced equation, we calculate that for \(2 \times 10^{-5} \) moles of \(\text{Pb}^{2+}\), there must be \(4 \times 10^{-5} \) moles of \(\text{I}^-\). This ensures the reaction can proceed to completion, forming lead iodide precipitate.Stoichiometry is an essential tool for chemists, providing a mathematical method for converting between moles, mass, and number counts based on the balanced chemical equations they work with.