A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?

Short Answer

Expert verified
The system is in equilibrium because the dissolution and precipitation of the electrolyte occur at equal rates, with no net change over time. It's called heterogeneous because it involves more than one phase – solid and liquid.

Step by step solution

01

Understanding the Definitions

A saturated solution is a solution in which the maximum amount of solute has been dissolved at a particular temperature. If no more solute can dissolve, and some remains as solid, this indicates a state of dynamic equilibrium between the dissolution and precipitation processes of the solute.
02

Explaining the Equilibrium

The system is at equilibrium because the rate at which the solid electrolyte dissolves into the solution is equal to the rate at which the dissolved electrolyte re-deposits back into the solid form. This means there is no net change in the concentration of the dissolved substance over time.
03

Defining Heterogeneous Equilibrium

Such a system is called a heterogeneous equilibrium because it involves multiple phases: the solid phase (the undissolved electrolyte) and the liquid phase (the saturated solution). The term 'heterogeneous' refers to a system containing different phases in contrast to a homogeneous equilibrium, which would involve a single phase.

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Key Concepts

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

Saturated Solution
Imagine adding sugar to a cup of tea until no more sugar can dissolve; you've just created a saturated solution. In scientific terms, a saturated solution is one where the solute, like salt or sugar, has been dissolved in the solvent, such as water, to its maximum capacity at a given temperature.

Think of this scenario as a crowded dance floor where no more dancers can fit. Just as some partygoers might wait for a chance to hop onto the dance floor, the excess solute waits in solid form, ready to enter the solution when space becomes available. This state represents a delicate balance where the same number of dissolved particles that leave the solution and revert to solid form are replaced by an equal number of solid particles dissolving into the liquid. This is where our concept of equilibrium comes into play.
Dynamic Equilibrium
In a party, the flow of guests in and out of the door remains constant, and the number of people inside stays the same; this reflects a dynamic equilibrium. Applying this to saturated solutions, it occurs when the dissolving solute and the precipitating solute occur at equal rates, and as a result, there is no net change in the number of dissolved particles over time.

The interesting part is the 'dynamic' aspect, which implies that there is constant motion; particles are continuously dissolving and re-crystallizing. However, to an observer, the saturated solution seems unchanged because the overall concentration of the dissolved substance remains stable. This 'in-motion yet constant' scenario provides learners with an excellent example of the dynamic nature of chemical processes at the microscopic level.
Solubility
Solubility is a measure determining how much solute you can dissolve in a solvent before reaching saturation. It's like a limit to how many people can comfortably fit in an elevator; once you reach that number, no one else can get in without someone else getting out.

In a chemical context, solubility is dictated by temperature, pressure, and the nature of the solute and solvent. Factors such as temperature can prompt a solute to dissolve more or less. For instance, as the temperature of water increases, more sugar can dissolve within it. The solubility of a substance is a critical factor for many industries, from pharmaceuticals to food production, influencing how compounds are managed and utilized. Understanding solubility and its variable nature allows us to predict and manipulate the behavior of solutions in practical applications.

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Most popular questions from this chapter

What is the effect on the amount of CaHPO \(_{4}\) that dissolves and the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HPO}_{4}^{-}\) -when each of the following are added to a mixture of solid \(\mathrm{CaHPO}_{4}\) and water at equilibrium? (a) \(\mathrm{CaCl}_{2}\) (b) HCl (c) \(\mathrm{KClO}_{4}\) (d) \(\mathrm{NaOH}\) (e) \(\mathrm{CaHPO}_{4}\)

Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for \(K_{\mathrm{sp}}\) values.) (a) \(\mathrm{KClO}_{4}:\left[\mathrm{K}^{+}\right]=0.01 \mathrm{M},\left[\mathrm{ClO}_{4}^{-}\right]=0.01 \mathrm{M}\) (b) \(\mathrm{K}_{2} \mathrm{Pt} \mathrm{Cl}_{6}:\left[\mathrm{K}^{+}\right]=0.01 M,\left[\mathrm{PtCl}_{6}^{2-}\right]=0.01 \mathrm{M}\) (c) \(\mathrm{PbI}_{2}:\left[\mathrm{Pb}^{2+}\right]=0.003 \mathrm{M},\left[\mathrm{I}^{-}\right]=1.3 \times 10^{-3} \mathrm{M}\) (d) \(\mathrm{Ag}_{2} \mathrm{S}:\left[\mathrm{Ag}^{+}\right]=1 \times 10^{-10} \mathrm{M},\left[\mathrm{S}^{2-}\right]=1 \times 10^{-13} \mathrm{M}\)

What is the molar solubility of BaSO \(_{4}\) in a \(0.250-M\) solution of \(\mathrm{NaHSO}_{4} ? K_{\mathrm{a}}\) for \(\mathrm{HSO}_{4}^{-}=1.2 \times 10^{-2}\).

How many grams of Milk of Magnesia, Mg(OH) \(_{2}(s)\) (58.3 g/mol), would be soluble in 200 mL of water. \(K_{\mathrm{sp}}=\) \(7.1 \times 10^{-12} .\) Include the ionic reaction and the expression for \(K_{\mathrm{sp}}\) in your answer. \(\left(K_{\mathrm{w}}=1 \times 10^{-14}=\right.\) \(\left.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\right)\)

Calculate the molar solubility of AgBr in \(0.035 M \mathrm{NaBr}\left(K_{\mathrm{sp}}=5 \times 10^{-13}\right)\).

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