Problem 72 - Chemistry: The Central Science in SI Units, Global Edition

Question

Suppose that a 10 -mL sample of a solution is to be tested for \(\mathrm{I}^{-}\) 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 \(\mathrm{I}^{-}\) that must be present for \(\mathrm{PbI}_{2}(s)\) to form?

Step-by-Step Solution

Answer

The minimum number of grams of $\mathrm{I}^{-}$ ions needed for $\mathrm{PbI}_{2}(s)$ to form when 0.2 mL of a 0.10 M $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ solution is added to a 10-mL sample is approximately \(5.80 \times 10^{-4}\) grams.

Step 1: Find the solubility product constant for PbI₂

From the given information, we have to find the solubility product constant (Ksp) of lead(II) iodide (PbI₂). With the help of a reference table or other data source, we find that the Ksp value for PbI₂ is 7.1 x 10⁻₉.

Step 2: Write the balanced equation for PbI₂(s) dissolution

The dissolution of PbI₂(s) in water can be represented as follows: \[PbI_{2}(s) \leftrightarrow Pb^{2+}(aq) + 2I^-(aq)\]

Step 3: Determine the initial concentrations

Since 0.2 mL of 0.10 M Pb(NO₃)₂ is added to the solution, we can convert the volume to liters and then determine the moles of Pb²⁺ ions introduced: Volume of Pb(NO₃)₂ = \(0.2\,mL \times \frac{1\,L}{1000\, mL} = 2 × 10^{-4} L\) Moles of Pb²⁺ ions = Molarity × Volume = \(0.10\,M \times 2 × 10^{-4} L = 2 × 10^{-5}\,mol\) Since the volume of the sample solution is 10 mL, the final volume of the mixture after adding Pb(NO₃)₂ is 10.2 mL. Hence, the initial concentration of Pb²⁺ ions in the mixture after addition is: Concentration of Pb²⁺ ions = \(\frac{2 × 10^{-5}\, mol}{10.2 × 10^{-3} L} = 1.96 × 10^{-3} M\) Considering the 1:2 stoichiometry in the chemical equation, Pb²⁺ and I⁻ have a 1:2 molar ratio.

Step 4: Evaluate the minimum concentration of I⁻ ions required using Ksp

Let x be the minimum concentration of I⁻ ions needed. The solubility product expression for the dissociation of PbI₂ can be written as follows: \[Ksp = [Pb^{2+}][I^-]^2 = (1.96 × 10^{-3} - x)(2x)^2\] Since the solubility product constant Ksp is given as 7.1 x 10⁻₉, we substitute this value in the equation and solve for x: \[7.1 \times 10^{-9} = (1.96 × 10^{-3} - x)(2x)^2\]

Step 5: Solve for x, the minimum concentration of I⁻ ions

By solving this equation for x, we can find the minimum concentration of I⁻ ions for PbI₂ to form. The equation will be challenging to solve algebraically, so we can use an iterative method or graphing calculator to find the roots. Upon solving, we get x ≈ \(4.57 × 10^{-4} M\) as the minimum concentration of I⁻ ions needed.

Step 6: Calculate the minimum grams of I⁻ ions present in the 10-mL sample

Now that we have the minimum concentration of I⁻ ions, we can determine the minimum grams of I⁻ ions present in the 10-mL sample. First, convert the concentration to moles of I⁻ ions in a 10-mL sample: Moles of I⁻ ions = Concentration × Volume = \(4.57 × 10^{-4} M \times 10 × 10^{-3} L = 4.57 × 10^{-6}\,mol\) Next, convert moles to grams using the molar mass of iodide ions (I⁻ = 127 g/mol): Minimum grams of I⁻ ions = Moles × Molar Mass = \(4.57 × 10^{-6} mol \times 127 \frac{g}{mol} = 5.80 × 10^{-4} g\) Thus, the minimum number of grams of I⁻ ions needed for PbI₂ to form is approximately \(5.80 × 10^{-4} g\).

Key Concepts

Iodide Ion Concentration

The iodide ion concentration is a key factor when considering the solubility of salts like lead(II) iodide (\(\text{PbI}_2\)). When you add a reagent such as lead nitrate (\(\text{Pb(NO}_3)_2\)) to a solution, it provides lead ions, which can combine with iodide ions to potentially form an insoluble compound, \(\text{PbI}_2\).
When dissolved, \(\text{PbI}_2\) dissociates according to the equation:\[\text{PbI}_2(s) \leftrightarrow \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)\]Understanding the balance between these ions is crucial because their high concentration can drive the precipitation of \(\text{PbI}_2\).
To determine the concentration, one must understand the relationship given by the solubility product constant, \(\text{Ksp}\). This constant reflects the maximum product of the ion concentrations that can exist in solution without precipitating. Practically, in our step-by-step solution, we used a \(\text{Ksp}\) value of 7.1 x 10\(^{-9}\), which helps us set up the equation to find the necessary iodide ion concentration.

Lead(II) Iodide Formation

The formation of lead(II) iodide involves a reaction where lead ions combine with iodide ions in solution to form a solid precipitate, \(\text{PbI}_2\). This compound is typically yellow and indicates the presence of certain concentrations of its constituent ions.
The solubility product constant, \(\text{Ksp}\), guides us in predicting whether a solid will form based on the ion concentrations:- \(\text{Ksp}\) is defined for slightly soluble salts. For \(\text{PbI}_2\), the constant is quite low (7.1 x 10\(^{-9}\)), reflecting its low solubility.- The equation \((\text{Ksp} = [\text{Pb}^{2+}][\text{I}^-]^2)\) helps in determining at what point the salt will begin to precipitate. When you reach the \(\text{Ksp}\) threshold, additional iodide ions can drive more lead ions to form \(\text{PbI}_2\), leading to its precipitation. Understanding when and how this occurs is essential for controlling the formation of \(\text{PbI}_2\) and similar compounds in chemical reactions or laboratory settings.

Chemical Stoichiometry

Chemical stoichiometry is the study of quantitative relationships in chemical reactions. It is crucial for determining how much of each reactant is needed to achieve a complete reaction without excess leftover. In this specific example, stoichiometry helps establish the right mix of lead ions and iodide ions for forming \(\text{PbI}_2\).
Let's break it down:- Stoichiometry provides the molar ratio of reactants and products in a balanced equation. - For the formation of \(\text{PbI}_2\), we use a 1:2 molar ratio from the equation:\[\text{Pb}^{2+} + 2\text{I}^- \rightarrow \text{PbI}_2\]- This ratio tells us that for every mole of \(\text{Pb}^{2+}\), two moles of \(\text{I}^-\) are required to form the compound. In practice, stoichiometry allows us to calculate how many grams or moles of \(\text{I}^-\) are necessary for a reaction. Using the molar mass of iodide ions (127 g/mol), we can convert between moles and grams, facilitating the prediction of the necessary conditions for reactions, which is especially useful in laboratory or industrial chemical procedures.