Question

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

Short answer

The minimum number of grams of \(I^-\) that must be present for \(\mathrm{PbI}_2(s)\) to form is approximately 0.0051 g.

Step-by-step solution

Find moles of \(\mathrm{Pb}(\mathrm{NO}_3)_2\)

Given 0.2 mL of 0.10 M \(\mathrm{Pb}(\mathrm{NO}_3)_2\), we can calculate the moles as follows: Moles = Molarity × Volume Moles of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) = 0.10 mol/L × 0.0002 L = 0.00002 mol

Use stoichiometry to determine moles of \(I^-\)

From the balanced equation, we know that 1 mole of \(\mathrm{Pb}(\mathrm{NO_3)_2}\) reacts with 2 moles of \(I^-\) to form 1 mole of \(\mathrm{PbI}_2\). Therefore: Moles of \(I^-\) = 2 × moles of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) Moles of \(I^-\) = 2 × 0.00002 mol = 0.00004 mol

Calculate the concentration of \(I^-\)

We are given a 10 mL sample of the solution, so the concentration of \(I^-\) can be obtained as follows: Concentration of \(I^-\) = moles of \(I^-\) / volume of solution Concentration of \(I^-\) = 0.00004 mol / 0.010 L = 0.004 M

Calculate the minimum number of grams of \(I^-\)

Now that we know the minimum concentration of \(I^-\) needed for the formation of \(\mathrm{PbI}_2\), we can calculate the minimum number of grams: Minimum grams of \(I^-\) = moles of \(I^-\) × molar mass of \(I^-\) The molar mass of \(I^-\) is 126.9 g/mol. Minimum grams of \(I^-\) = 0.00004 mol × 126.9 g/mol = 0.005076 g Therefore, the minimum number of grams of \(I^-\) that must be present for \(\mathrm{PbI}_2(s)\) to form is approximately 0.0051 g.

Key concepts

Molarity

Understanding molarity is crucial in stoichiometry as it is a measure of the concentration of a solution. Molarity, often represented by the symbol M, is defined as the number of moles of solute present in one liter of solution. It allows scientists and students to quantify how much of a certain solute is dissolved within a given volume.
For example, when we say a solution is 0.10 M, it means there are 0.10 moles of solute dissolved in each liter of the solution.
This concentration measurement helps us determine how much of a given reactant is involved in a chemical reaction.

When performing stoichiometric calculations, molarity is often used to find the number of moles of a substance in a given volume of solution. The formula to find moles from molarity is:

• Moles = Molarity × Volume

This calculation step is essential, as seen in the example practice where moles of ext{Pb(NO}_3)_2} was calculated using its molarity and the volume of the sample provided.

Balanced Chemical Equation

A balanced chemical equation is a tool chemists use to ensure the law of conservation of mass is followed in a chemical reaction. It provides a visual representation of the reactants turning into products while maintaining equal numbers of each type of atom on both sides.
For example, the reaction between ext{Pb(NO}_3)_2 and ext{I}^-} ions is governed by the equation:\[ ext{Pb(NO}_3)_2 + 2I^- ightarrow ext{PbI}_2 + 2NO_3^-\]This equation shows that one mole of ext{Pb(NO}_3)_2 reacts with two moles of ext{I}^-} to produce solid lead iodide ( ext{PbI}_2}).
By using stoichiometry alongside a balanced equation, you can easily calculate how much of each reactant is needed or how much of a product will be produced.
In our example, knowing that 1 mole of ext{Pb(NO}_3)_2 requires 2 moles of ext{I}^-} is crucial to deduce the amount of ext{I}^-} ions necessary for the reaction.

Concentration Calculation

Calculating concentration is a vital part of solution stoichiometry, which helps us understand how much solute is present in a specific volume of solution. In concentration calculations, we often begin by finding the number of moles of the solute in solution.
Using the example provided, we calculated the concentration of ext{I}^-} ions using:

• Concentration = Moles of Solute / Volume of Solution

In this case, by knowing that the solution volume was 10 mL (converted to 0.010 L for the calculation) and the moles of ext{I}^-} were 0.00004 mol, we determined the concentration to be 0.004 M.

Such calculations provide insights into the reactivity potential of a solution and are essential for knowing whether a precipitate will form or if a reaction will proceed under given conditions.
In this situation, these calculations helped to determine the minimum grams of ext{I}^-} necessary for the formation of lead iodide ( ext{PbI}_2}), confirming that sufficient concentration levels are met for the reaction to occur.