Question

In another experiment a small sample of pure ${\mathrm{PbI}}_2$ is shaken with water to produce a saturated solution, a. What must be the relationship between $\left[{\mathrm{Pb}}^{2+}\right]$ and $\left[{\mathrm{I}}^{-}\right]$ in that solution? (See Reaction 1.) b. The concentration of $\mathrm{I}$ - in the saturated solution is found to be $5.0\times {10}^{-3}\mathrm{M}$. What is the concentration of ${\mathrm{Pb}}^{2+}$ ion? __________ M c. Using the results of Part b, calculate a value for $K_{\mathrm{sp}}$ for ${\mathrm{PbI}}_2$. (Use Eq. 1.) __________

Short answer

a. The relationship between the concentrations of Pb2+ and I- ions is $[I^{-}]=2[P{b}^{2+}]$. b. The concentration of Pb2+ ions in the solution is $2.5\times {10}^{-3}M$. c. The solubility product constant (K_sp) for PbI2 is $6.25\times {10}^{-8}$.

Step-by-step solution

- Write the Dissolution Reaction of PbI2

Write out the dissolution reaction of PbI2 in water: $$Pb{I}_2(s)\rightleftharpoons P{b}^{2+}(aq)+2I^{-}(aq)$$ This is the reaction we will base our work on for the remainder of this exercise.

- Express Relationship Between Concentrations

As per the dissolution reaction, when one PbI2 molecule dissolves, it produces one Pb2+ ion and two I- ions. Therefore, we can write the relationship between the concentration of Pb2+ and I- ions as: $$[I^{-}]=2[P{b}^{2+}]$$ This equation shows the relationship between the concentration of I- ions and the concentration of Pb2+ ions.

- Calculate the Concentration of Pb2+

Now that we have the relationship between the concentrations of Pb2+ and I-, we can find the concentration of Pb2+ given that the concentration of I- is $5.0\times {10}^{-3}M$. Using the above relationship, we can write: $$[P{b}^{2+}]=\frac{[I^{-}]}{2}=\frac{5.0\times {10}^{-3}}{2}M$$ Now, calculate the concentration of Pb2+: $$[P{b}^{2+}]=2.5\times {10}^{-3}M$$ The concentration of Pb2+ ions in the solution is $2.5\times {10}^{-3}M$.

- Calculate the Solubility Product Constant (K_sp)

We can now calculate the solubility product constant (K_sp) for PbI2 using the concentrations obtained above based on the following equation: $$K_{sp}=[P{b}^{2+}][I^{-}{]}^2$$ Substitute the values of [Pb2+] and [I-] we obtained: $$K_{sp}=(2.5\times {10}^{-3})(5.0\times {10}^{-3}{)}^2$$ Now, calculate the K_sp value: $$K_{sp}=6.25\times {10}^{-8}$$ The solubility product constant (K_sp) for PbI2 is $6.25\times {10}^{-8}$.

Key concepts

Dissolution Reaction

The concept of a dissolution reaction is a cornerstone in understanding solubility and the behavior of substances in solutions. A dissolution reaction occurs when a solid solute dissolves in a solvent, forming a homogeneous mixture, or solution. In the context of our exercise, the dissolution reaction is represented by the equation:

$$\begin{array}{rlr}Pb{I}_2(s)& \rightleftharpoons & P{b}^{2+}(aq)+2I^{-}(aq)\end{array}$$This reaction is reversible, indicating that the solid lead(II) iodide () can dissolve in water to produce lead ions () and iodide ions () in the aqueous phase, but it can also precipitate back into solid form. In a saturated solution, the rate of dissolution is equal to the rate of precipitation, which is a prime example of dynamic equilibrium. The dissolution process is influenced by factors like temperature, the presence of other ions, and the solute-solvent interactions, highlighting the intricate nature of chemical processes in solution.

• In the provided exercise, understanding the dissolution reaction is imperative to grasp the resultant ion concentrations and the solubility product constant.
• The stoichiometry of the dissolution reaction is critical since it dictates the molar ratio of the ions in the solution, which is essential for subsequent calculations.

Ion Concentration

Ion concentration is a measure of the amount of ions present in a solution per unit volume, typically expressed in moles per liter (M or mol/L). It’s an important concept in understanding the strength of ionic compounds dissolved in solutions and plays a vital role in stoichiometry calculations in chemistry.

In our specific exercise involving , when the salt dissolves, the ions are released into the solution in a fixed stoichiometric ratio, directly derived from the coefficient of each ion in the balanced dissolution equation. The ratio tells us that for every one ion formed, two ions are released.

Relationship Between and Concentrations

By determining the ion concentration of one type of ion, we can use the stoichiometric relationship to calculate the concentration of the other. For example, in the equation:$$[I^{-}]=2[P{b}^{2+}]$$Knowing the concentration of enables us to find the concentration of by dividing by two. This relationship allows us to understand the proportions of ions in solution, which is crucial for predicting product formation and for finding the solubility product constant, which we will discuss next.

• Accurate measurement of ion concentration is significant for various applications including environmental monitoring, health, and industrial processes.
• The concepts of molarity and molality are used to express concentration in different scenarios based on the presence of a solute in a solvent.

Chemical Equilibrium

Chemical equilibrium is one of the most fundamental concepts in chemistry, occurring when the forward and reverse reactions in a chemical process occur at equal rates, resulting in no net change of reactants and products over time. It signifies a state of balance, not the absence of motion, where the reactants are converted to products, and the products are converted back to reactants at the same rate.

In the context of our problem, the dissolution of into its constituent ions and will reach a point where the rate of dissolution equals the rate of precipitation, establishing equilibrium in a saturated solution. This balance is quantitatively expressed by the solubility product constant ().

Understanding

$$K_{sp}=[P{b}^{2+}][I^{-}{]}^2$$The is a special equilibrium constant specific for the saturated solutions of sparingly soluble salts. It reflects the maximum amount of a compound that can dissolve in a solvent at a given temperature and pressure, and it’s calculated by taking the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced dissolution equation.

• Factors affecting chemical equilibrium include concentration, temperature, and pressure, each of which can shift the equilibrium position according to Le Chatelier’s Principle.
• In real-life applications, understanding chemical equilibrium is crucial for chemical manufacturing, pharmaceuticals, and analytical chemistry.

We've seen how this principle is applied in our exercise, demonstrating how equilibrium concepts are not merely theoretical but also practical and essential for predicting how reactions will behave in different conditions.