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Why does the amount of excess solid solute present in a solution not affect the amount of solute that ultimately dissolves in a given amount of solvent?

Short Answer

Expert verified
The amount of excess solid solute present in a solution does not affect the amount of solute that ultimately dissolves in a given amount of solvent because the solution is in a state of equilibrium. In a saturated solution, the rate of dissolution is equal to the rate of precipitation, maintaining a constant concentration of solute and solvent. Therefore, adding more solid solute does not alter this equilibrium or increase the amount of solute that ultimately dissolves, as long as temperature and pressure remain constant.

Step by step solution

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1. Understanding Solubility

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure. When a solute dissolves in a solvent, the solute particles interact with the solvent particles, and a homogeneous mixture is formed. Solubility can be affected by various factors, such as temperature, pressure, and the nature of the solute and solvent.
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2. Understanding Saturation

A solution is said to be saturated when the maximum amount of solute has dissolved in the solvent at a given temperature and pressure, and no more solute can dissolve in the solvent. In a saturated solution, the rate at which the solute particles dissolve in the solvent is equal to the rate at which they come out of the solution and form solid particles. This is known as equilibrium.
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3. Understanding Equilibrium

In a saturated solution, the dissolved solute particles are in a dynamic equilibrium with the undissolved solute particles. This means that the rates of dissolution and precipitation are equal, and the concentrations of solute and solvent in the solution remain constant over time. As long as the temperature and pressure remain constant, the amount of solute that dissolves in the solvent will not change, regardless of the amount of excess solid solute present.
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4. Excess Solid Solute and Equilibrium

When there is excess solid solute present in a solution, it does not affect the equilibrium between the dissolved and undissolved solute particles. This is because the rate of dissolution is equal to the rate of precipitation, so any solute particles that dissolve into the solvent will be matched by an equal number of solute particles coming out of the solution to form solid particles. Therefore, the amount of excess solid solute does not affect the amount of solute that ultimately dissolves in the solvent.
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5. Conclusion

The amount of excess solid solute present in a solution does not affect the amount of solute that ultimately dissolves in a given amount of solvent because the solution is in equilibrium. The rate of dissolution is equal to the rate of precipitation, and the concentrations of solute and solvent in the solution remain constant, as long as the temperature and pressure remain constant.

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Key Concepts

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

Saturation
Saturation refers to the point at which a solution can no longer dissolve additional solute at a given temperature and pressure. When a solution reaches this state, it is known as a saturated solution. In a saturated solution, a balance is maintained. This means:
  • The maximum amount of solute has dissolved in the solvent.
  • No more solute can be dissolved unless conditions change.
For instance, if more solute is added to a saturated solution, it will remain undissolved. This is because the solution has reached its capacity to hold dissolved particles.
One of the important aspects of saturation is that it sets the stage for equilibrium, a concept that links closely with saturation in maintaining stability in a solution.
Equilibrium
Equilibrium in the context of solutions refers to a state where the process of dissolution and the reverse process, precipitation, occur at the same rate. When a solution is saturated and in equilibrium, it means:
  • The rate of solute particles dissolving into the solvent is equal to the rate at which they are re-precipitating out of the solution.
  • The total amount of solute in the solution remains constant over time.
This balanced state is crucial because it ensures that any added excess solid solute, beyond saturation, doesn’t affect the equilibrium of the solution.
Factors such as temperature and pressure must remain constant to maintain this equilibrium. If these conditions change, the balance can shift, altering how much solute can dissolve.
Thus, understanding equilibrium helps explain why excess solute does not increase the solute concentration in a solution beyond its saturation point.
Dissolution
Dissolution is the process by which solute particles intermingle with solvent particles to form a homogeneous mixture, known as a solution. During dissolution:
  • Solute particles separate and disperse throughout the solvent.
  • The individual particles become surrounded by solvent molecules.
This process continues until the solution reaches saturation. At saturation, equilibrium between dissolution and precipitation is established, and any excess solute remains undissolved as solid.
The effectiveness of dissolution is influenced by factors such as:
  • Temperature, which can increase the dissolution rate for many solutes.
  • Pressure, particularly for gases dissolved in liquids.
  • The nature of both the solute and solvent, including their polarities.
By understanding dissolution, students can grasp why the amount of solute that dissolves is limited by solubility at a given condition, regardless of excess solute presence.

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Most popular questions from this chapter

Suppose that for a hypothetical reaction: $$ \mathrm{A}_{2}(g)+2 \mathrm{~B}(g) \rightleftharpoons 2 \mathrm{AB}(g) $$ It is determined that at a certain temperature the equilibrium concentrations are: \(\left[\mathrm{A}_{2}\right]=0.0090,[\mathrm{~B}]=0.940,\) and \([\mathrm{AB}]=5.3 \times 10^{-4} .\) Calculate the numerical value of \(K\) for the reaction.

Magnesium fluoride dissolves in water to the extent of \(8.0 \times 10^{-2} \mathrm{~g} / \mathrm{L}\) at \(25 \mathrm{C}\). Calculate the solubility of \(\mathrm{MgF}_{2}(s)\) in moles per liter, and calculate \(K_{\mathrm{sp}}\) for \(\mathrm{MgF}_{2}\) at 25 .

Plants synthesize the sugar dextrose according to the following reaction by absorbing radiant energy from the sun (photosynthesis). $$ 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ Will an increase in temperature tend to favor or discourage the production of $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) ? $$

Most hydroxides are not very soluble in water. For example, \(K_{\mathrm{sp}}\) for nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2},\) is \(2.0 \times 10^{-15}\) at 25 . \(\mathrm{C}\). How many grams of nickel(II) hydroxide dissolve per liter at 25 ' \(\mathrm{C}\) ?

True or false? The position of a heterogeneous equilibrium does not depend on the amounts of pure solid or pure liquid reactants or products present because the concentrations of pure solids and liquids cannot change. True False

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