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Chapter 26: Problem 1

State in words the meaning of the solubility product equation for \(\mathrm{PbI}_{2}\) : $$ K_{\mathrm{xp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$

Short Answer

Expert verified
The solubility product equation for PbI鈧�, \( K_{\mathrm{xp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} \), represents the relationship between the concentrations of lead (II) cations (Pb虏鈦�) and iodide anions (I鈦�) in a saturated solution of PbI鈧�. It signifies the equilibrium constant for the solubility of PbI鈧� in water, with higher values indicating greater solubility.

Step by step solution

01

Dissociation of lead (II) iodide (PbI鈧�)

When lead (II) iodide is placed in water, it begins to dissociate into its individual ions. This dissolution can be represented by the following equilibrium chemical equation: $$ \mathrm{PbI_{2}\rightleftharpoons Pb^{2+} + 2I^{-}} $$
02

Understanding the solubility product (K鈧撯倸)

The solubility product (K鈧撯倸) is an equilibrium constant that represents the degree to which a solute will dissolve in a solvent (in this case, water). It is the product of the equilibrium concentrations of the ions, each raised to the power of their stoichiometric coefficients. For the PbI鈧� system, the solubility product equation is: $$ K_{\mathrm{xp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$
03

Relation between K鈧撯倸 and ion concentrations

The equilibrium constant K鈧撯倸 provides information about the solubility of a solute in a solvent. For a given solute, a larger K鈧撯倸 value indicates a higher level of solubility. In the case of PbI鈧�, the concentration of Pb虏鈦� ions and I鈦� ions in the solution can be represented by [Pb虏鈦篯 and [I鈦籡 respectively. The K鈧撯倸 equation then allows us to relate these concentrations to the solubility of PbI鈧�: $$ K_{\mathrm{xp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$ To summarize, the solubility product equation for PbI鈧� represents the relationship between the concentrations of the lead (II) cations (Pb虏鈦�) and iodide anions (I鈦�) in a saturated solution of PbI鈧�. The equilibrium constant, K鈧撯倸, indicates the solubility of PbI鈧� in water, with higher values signifying greater solubility.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in no overall change in the concentrations of reactants and products over time. It is essential to understand that this doesn't mean the reactants and products are at equal concentrations, but that their rates of formation are stable.

For the dissolution of lead (II) iodide, chemical equilibrium is achieved when the rate at which the solid PbI鈧� turns into its ions is the same as the rate at which the ions combine to form the solid. The solubility product equation comes into play to quantify this particular equilibrium state.
Dissolution of Lead (II) Iodide
The dissolution of lead (II) iodide (PbI鈧�) in water is an interesting process that involves the solid salt dividing into its constituent ions. This occurs as PbI鈧� interacts with water molecules, leading to the separation of lead (II) cations (Pb虏鈦�) and iodide anions (I鈦�).

The dissolution can be described by a simple chemical equation: \[\mathrm{PbI_{2}\rightleftharpoons Pb^{2+} + 2I^{-}}\] Each molecule of PbI鈧� separates into one Pb虏鈦� ion and two I鈦� ions. This reaction reaches a state of equilibrium, where the rate of dissolution equals the rate of precipitation, meaning solid formation.
Equilibrium Constant Calculation
The equilibrium constant calculation for the dissolution of a salt like PbI鈧� is represented by the solubility product constant (K鈧撯倸). It describes the level at which the salt can dissolve in a solvent at a particular temperature.

To calculate the equilibrium constant, you look at the concentrations of the dissolved ions at equilibrium. For lead (II) iodide, the equation is: \[K_{\mathrm{xp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\] The exponents correspond to the stoichiometric coefficients of the ions in the dissolution equation. Understanding this relationship helps predict the behavior of the salt in solution and calculate concentrations effectively.
Ion Concentration and Solubility
The ion concentration in a solution is paramount to determining the solubility of a substance. Ion concentration refers to the molarity, or the number of moles of an ion per liter of solution. Solubility is a measure of how much of a particular substance can be dissolved in a solvent, and it's directly influenced by the ion concentration in the solution at equilibrium.

For PbI鈧�, understanding the solubility product equation allows you to calculate the solubility of the salt. A high K鈧撯倸 value signifies that more lead (II) iodide can dissolve in the water, indicating a higher solubility. Conversely, a lower K鈧撯倸 implies lesser solubility. Thus, the equilibrium constant serves as a valuable tool in predicting the extent of a substance's solubility in various conditions.

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Most popular questions from this chapter

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.) __________

When \(5.0 \mathrm{mL}\) of \(0.012 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) are mixed with \(5.0 \mathrm{mL}\) of \(0.030 \mathrm{M} \mathrm{KI},\) a yellow precipitate of \(\mathrm{PbI}_{2}(\mathrm{s})\) forms. a. How many moles of \(\mathrm{Pb}^{2+}\) are initially present?\(\quad\quad\) __________ moles b. How many moles of \(I^{-}\) are originally present?\(\quad\quad\) __________ moles c. In a colorimeter the cquilibrium solution is analyzed for \(\mathrm{I}^{-}\), and its concentration is found to be \(7 \times\) \(10^{-3}\) mole/liter. How many moles of \(\mathbf{I}^{-}\) are present in the solution ( \(10 \mathrm{ml}\) )? __________ moles d. How many moles of I - precipitated?\(\quad\quad\) __________ moles e. How many moles of \(\mathrm{Pb}^{2+}\) precipitated?\(\quad\quad\) __________ moles f. How many moles of \(\mathrm{Pb}^{2+}\) are left in solution?\(\quad\quad\) __________ moles g. What is the concentration of \(\mathrm{Pb}^{2+}\) in the equilibrium solution?\(\quad\quad\) __________ moles/liter h. Find a value for \(K_{\mathrm{sp}}\) of \(\mathrm{PbI}_{2}\) from these data.\(\quad\quad\) __________
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