Calculating Solubility - AP Chemistry

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by , and fluoride ions increase by .

End: The equilibrium expression for this salt is , which will allow us to calculate the solubility based on the change in ion concentration.

The solubility for barium fluoride is .

Definition of :

for .

For :

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

Solve for the unknown variable:

Multiply times the given volume:

Multiply times the molar mass of :

The correct answer is

First of all, we know this is an acid-base reaction. is the reactant base and is the reactant acid. We know that these acid-base reactions create a water and a salt. Also, we know that sulfates are soluble, except for those of calcium, strontium, and barium. As a result, is a precipitate in the reaction and water is produced.

Most salts containing halogens are soluble. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble except for cations containing alkaline earth metals and hydroxides of calcium, strontium, and barium, which are slightly soluble.

Start by writing the equation for the dissolution of calcium hydroxide.

Next, set up the following table to show the equilibrium concentrations of the ions:

| | \[ \] | \[\] | | | -------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------- | ---- | | Initial | 0.00 | 0.00 | | Change | +S | +2S | | Equilibrium | S | 2S |

Now substitute in the values of the concentrations of the ions into the expression to find .

Now, plug in the given value and solve for , the molar solubility.

For this question, we're told that an aqueous solution of magnesium fluoride is saturated. We're then told that the fluoride concentration in the solution is reduced by a factor of two, and we're asked to find how the magnesium concentration would need to change in order to keep the solution saturated.

First, it's important to recall what saturation means. Some substances cannot dissolve in water, whereas others can dissolve readily. In other words, different substances will dissolve to differing degrees in water. The dissolution of a compound in a solvent such as water can be represented by an equilibrium expression. When there is a relatively small amount of solute added, the solution is said to be unsaturated. This means that all of the added solute will dissolve. As more and more solute is added to the water, there will eventually reach a point at which so much solute is present that it can no longer dissolve. When this happens, any additional solute will not dissolve and will instead form a precipitate in the solution. This condition is referred to as supersaturated. Saturation is the "sweet spot" so to speak; it is the point in between unsaturated and supersaturated where the maximum amount of solute has been added to the solution where all of the solute can be in the dissolved form. In other words, saturation refers to the maximum concentration of added solute where there is NO precipitation.

In order to set up an equilibrium expression, we can first write out the reaction in which magnesium fluoride dissolves.

Knowing the reaction, we can now write an equilibrium expression. Remember that pure solids and liquids don't appear in equilibrium expressions! Thus, this expression will only contain the products of the above reaction.

In the above expression, refers to the solubility product constant, which is just a quantitative way of expressing the degree to which a given compound can dissolve within a given solvent.

Since the equilibrium expression tells us the concentrations necessary to have a saturated solution, we can determine how a change in fluoride concentration would affect the equilibrium. Then, we can determine what changes to magnesium are needed.

If the fluoride concentration were to be cut in half, the value would be decreased by a factor of . This means that in order to maintain the value, we would have to increase the magnesium concentration by a factor of in order to compensate for the loss of fluoride.

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by , and fluoride ions increase by .

End: The equilibrium expression for this salt is , which will allow us to calculate the solubility based on the change in ion concentration.

The solubility for barium fluoride is .

Definition of :

for .

For :

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

Solve for the unknown variable:

Multiply times the given volume:

Multiply times the molar mass of :

The correct answer is

First of all, we know this is an acid-base reaction. is the reactant base and is the reactant acid. We know that these acid-base reactions create a water and a salt. Also, we know that sulfates are soluble, except for those of calcium, strontium, and barium. As a result, is a precipitate in the reaction and water is produced.

Most salts containing halogens are soluble. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble except for cations containing alkaline earth metals and hydroxides of calcium, strontium, and barium, which are slightly soluble.

Start by writing the equation for the dissolution of calcium hydroxide.

Next, set up the following table to show the equilibrium concentrations of the ions:

| | \[ \] | \[\] | | | -------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------- | ---- | | Initial | 0.00 | 0.00 | | Change | +S | +2S | | Equilibrium | S | 2S |

Now substitute in the values of the concentrations of the ions into the expression to find .

Now, plug in the given value and solve for , the molar solubility.

For this question, we're told that an aqueous solution of magnesium fluoride is saturated. We're then told that the fluoride concentration in the solution is reduced by a factor of two, and we're asked to find how the magnesium concentration would need to change in order to keep the solution saturated.

First, it's important to recall what saturation means. Some substances cannot dissolve in water, whereas others can dissolve readily. In other words, different substances will dissolve to differing degrees in water. The dissolution of a compound in a solvent such as water can be represented by an equilibrium expression. When there is a relatively small amount of solute added, the solution is said to be unsaturated. This means that all of the added solute will dissolve. As more and more solute is added to the water, there will eventually reach a point at which so much solute is present that it can no longer dissolve. When this happens, any additional solute will not dissolve and will instead form a precipitate in the solution. This condition is referred to as supersaturated. Saturation is the "sweet spot" so to speak; it is the point in between unsaturated and supersaturated where the maximum amount of solute has been added to the solution where all of the solute can be in the dissolved form. In other words, saturation refers to the maximum concentration of added solute where there is NO precipitation.

In order to set up an equilibrium expression, we can first write out the reaction in which magnesium fluoride dissolves.

Knowing the reaction, we can now write an equilibrium expression. Remember that pure solids and liquids don't appear in equilibrium expressions! Thus, this expression will only contain the products of the above reaction.

In the above expression, refers to the solubility product constant, which is just a quantitative way of expressing the degree to which a given compound can dissolve within a given solvent.

Since the equilibrium expression tells us the concentrations necessary to have a saturated solution, we can determine how a change in fluoride concentration would affect the equilibrium. Then, we can determine what changes to magnesium are needed.

If the fluoride concentration were to be cut in half, the value would be decreased by a factor of . This means that in order to maintain the value, we would have to increase the magnesium concentration by a factor of in order to compensate for the loss of fluoride.

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by , and fluoride ions increase by .

End: The equilibrium expression for this salt is , which will allow us to calculate the solubility based on the change in ion concentration.

The solubility for barium fluoride is .

Definition of :

for .

For :

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

Solve for the unknown variable:

Multiply times the given volume:

Multiply times the molar mass of :

The correct answer is

First of all, we know this is an acid-base reaction. is the reactant base and is the reactant acid. We know that these acid-base reactions create a water and a salt. Also, we know that sulfates are soluble, except for those of calcium, strontium, and barium. As a result, is a precipitate in the reaction and water is produced.

Most salts containing halogens are soluble. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble except for cations containing alkaline earth metals and hydroxides of calcium, strontium, and barium, which are slightly soluble.

Start by writing the equation for the dissolution of calcium hydroxide.

Next, set up the following table to show the equilibrium concentrations of the ions:

| | \[ \] | \[\] | | | -------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------- | ---- | | Initial | 0.00 | 0.00 | | Change | +S | +2S | | Equilibrium | S | 2S |

Now substitute in the values of the concentrations of the ions into the expression to find .

Now, plug in the given value and solve for , the molar solubility.

For this question, we're told that an aqueous solution of magnesium fluoride is saturated. We're then told that the fluoride concentration in the solution is reduced by a factor of two, and we're asked to find how the magnesium concentration would need to change in order to keep the solution saturated.

First, it's important to recall what saturation means. Some substances cannot dissolve in water, whereas others can dissolve readily. In other words, different substances will dissolve to differing degrees in water. The dissolution of a compound in a solvent such as water can be represented by an equilibrium expression. When there is a relatively small amount of solute added, the solution is said to be unsaturated. This means that all of the added solute will dissolve. As more and more solute is added to the water, there will eventually reach a point at which so much solute is present that it can no longer dissolve. When this happens, any additional solute will not dissolve and will instead form a precipitate in the solution. This condition is referred to as supersaturated. Saturation is the "sweet spot" so to speak; it is the point in between unsaturated and supersaturated where the maximum amount of solute has been added to the solution where all of the solute can be in the dissolved form. In other words, saturation refers to the maximum concentration of added solute where there is NO precipitation.

In order to set up an equilibrium expression, we can first write out the reaction in which magnesium fluoride dissolves.

Knowing the reaction, we can now write an equilibrium expression. Remember that pure solids and liquids don't appear in equilibrium expressions! Thus, this expression will only contain the products of the above reaction.

In the above expression, refers to the solubility product constant, which is just a quantitative way of expressing the degree to which a given compound can dissolve within a given solvent.

Since the equilibrium expression tells us the concentrations necessary to have a saturated solution, we can determine how a change in fluoride concentration would affect the equilibrium. Then, we can determine what changes to magnesium are needed.

If the fluoride concentration were to be cut in half, the value would be decreased by a factor of . This means that in order to maintain the value, we would have to increase the magnesium concentration by a factor of in order to compensate for the loss of fluoride.

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by , and fluoride ions increase by .

End: The equilibrium expression for this salt is , which will allow us to calculate the solubility based on the change in ion concentration.

The solubility for barium fluoride is .

Definition of :

for .

For :

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

Solve for the unknown variable:

Multiply times the given volume:

Multiply times the molar mass of :