# 7.11: Solubility: Calculations


Learning Objectives
• Apply a solubility conversion factor to calculate the amount of solute that can be dissolved in a specified quantity of solvent.  Then,
• determine whether the resultant solution is saturated or unsaturated and
• calculate the amount of excess solute that remains undissolved in the resultant solution.
• Use a solubility curve to determine whether a solution is saturated or unsaturated and to estimate the amount of excess solute that remains undissolved in the solution.

As stated previously, a solubility limit is expressed as a ratio that relates the maximum amount of a solute that can dissolve to a standardized 100.-gram, or 100.-milliliter, quantity of solvent.  Because of its fractional format, a solubility proportion can be applied as a conversion factor and utilized to determine the maximum amount of solute that can dissolve in a non-standard amount of solvent.  By comparing the resultant calculated value to the amount of solute that is used to prepare a solution, that solution can be classified as saturated or unsaturated, and the amount of excess solute that remains undissolved can be determined.

For example, a solution is prepared by mixing 1.75 grams of hydrogen sulfide and 650. grams of water at 20 degrees Celsius.  Use the solubility information that is presented in Table 7.9.1 to determine whether the resultant solution is saturated or unsaturated, and calculate the amount of excess solute that remains undissolved in this solution.

Before a solubility limit can be applied as a conversion factor, each substance that is referenced in the given problem must first be classified as a solute or a solvent.  As stated in Section 7.9, the solvent is the substance that is reported as a 100.-gram, or 100.-milliliter, quantity in the denominator of a solubility limit.  Since the chemical formula for water, H2O, is associated with the 100.-gram quantities in the denominators of the solubilities in Table 7.9.1, water, H2O, is the solvent in this solution, and the remaining substance, hydrogen sulfide, H2S, is the solute, "by default."

In order to determine whether this solution is saturated or unsaturated, the solubility of hydrogen sulfide, H2S, which has a reported value of 0.33 g/100. g H2O, must be used as a conversion factor to calculate the maximum amount of solute, hydrogen sulfide, H2S, that can dissolve in the given amount of solvent, 650. grams of water, H2O.  The chemical formula for hydrogen sulfide, H2S, must be incorporated into the numerator of this conversion, as shown below, in order to achieve the desired unit transformation.  Finally, because solubility limits correspond to the exact amount of solute that can dissolve in a corresponding amount of solvent, the quantities that are calculated using a solubility proportion should not be rounded.

$${\text {650.}}$$ $${\cancel{\rm{g} \; \rm{H_2O}}} \times$$ $$\dfrac{0.33 \; \rm{g} \; \rm{H_2S}}{100. \; \cancel{\rm{g} \; \rm{H_2O}}}$$ = $${\text {2.145}}$$ $${\rm{g} \; \rm{H_2S}}$$

Based on this calculated value, exactly 2.145 grams of hydrogen sulfide, H2S, can dissolve in 650. grams of water, H2O.  Since the solution that is described in the problem is prepared using 1.75 grams of hydrogen sulfide, H2S, which is less than this calculated quantity, the resultant solution is unsaturated.  Furthermore, because the amount of solute that is added to prepare this solution is less than the quantity of solute that can dissolve in 650. grams of water, H2O, all of the solute that is added to the solvent successfully dissolves, and no excess solute remains undissolved in this solution.

Example $$\PageIndex{1}$$

A solution is prepared by mixing 129 grams of ammonium iodide and 75.0 grams of water at 20 degrees Celsius.  Use the solubility information that is presented in Table 7.9.1 to determine whether the resultant solution is saturated or unsaturated, and calculate the amount of excess solute that remains undissolved in this solution.

Solution

Before a solubility limit can be applied as a conversion factor, each substance that is referenced in the given problem must first be classified as a solute or a solvent.  Since the chemical formula for water, H2O, is associated with the 100.-gram quantities in the denominators of the solubilities in Table 7.9.1, water, H2O, is the solvent in this solution, and the remaining substance, ammonium iodide, NH4I, is the solute, "by default."

In order to determine whether this solution is saturated or unsaturated, the solubility of ammonium iodide, NH4I, which has a reported value of 172 g/100. g H2O, must be used as a conversion factor to calculate the maximum amount of solute, ammonium iodide, NH4I, that can dissolve in the given amount of solvent, 75.0 grams of water, H2O.  The chemical formula for ammonium iodide, NH4I, must be incorporated into the numerator of this conversion, as shown below, in order to achieve the desired unit transformation.  Finally, because solubility limits correspond to the exact amount of solute that can dissolve in a corresponding amount of solvent, the quantities that are calculated using a solubility proportion should not be rounded.

$${\text {75.0}}$$ $${\cancel{\rm{g} \; \rm{H_2O}}} \times$$ $$\dfrac{172 \; \rm{g} \; \rm{NH_4I}}{100. \; \cancel{\rm{g} \; \rm{H_2O}}}$$ = $${\text {129}}$$ $${\rm{g} \; \rm{NH_4I}}$$

Based on this calculated value, exactly 129 grams of ammonium iodide, NH4I, can dissolve in 75.0 grams of water, H2O.  Since the solution that is described in the problem is prepared using 129 grams of ammonium iodide, NH4I, which is exactly equal to this calculated quantity, the resultant solution is saturated.  Furthermore, because the amount of solute that is added to prepare this solution is exactly equal to the quantity of solute that can dissolve in 75.0 grams of water, H2O, all of the solute that is added to the solvent successfully dissolves, and no excess solute remains undissolved in this solution.

Exercise $$\PageIndex{1}$$

A solution is prepared by mixing 34.45 grams of mercury (II) cyanide and 310. grams of water at 20 degrees Celsius.  Use the solubility information that is presented in Table 7.9.1 to determine whether the resultant solution is saturated or unsaturated, and calculate the amount of excess solute that remains undissolved in this solution.

Before a solubility limit can be applied as a conversion factor, each substance that is referenced in the given problem must first be classified as a solute or a solvent.  Since the chemical formula for water, H2O, is associated with the 100.-gram quantities in the denominators of the solubilities in Table 7.9.1, water, H2O, is the solvent in this solution, and the remaining substance, mercury (II) cyanide, Hg(CN)2, is the solute, "by default."

In order to determine whether this solution is saturated or unsaturated, the solubility of mercury (II) cyanide, Hg(CN)2, which has a reported value of 9.30 g/100. g H2O, must be used as a conversion factor to calculate the maximum amount of solute, mercury (II) cyanide, Hg(CN)2, that can dissolve in the given amount of solvent, 310. grams of water, H2O.  The chemical formula for mercury (II) cyanide, Hg(CN)2, must be incorporated into the numerator of this conversion, as shown below, in order to achieve the desired unit transformation.  Finally, because solubility limits correspond to the exact amount of solute that can dissolve in a corresponding amount of solvent, the quantities that are calculated using a solubility proportion should not be rounded.

$${\text {310.}}$$ $${\cancel{\rm{g} \; \rm{H_2O}}} \times$$ $$\dfrac{9.30 \; \rm{g} \; \rm{Hg(CN)_2}}{100. \; \cancel{\rm{g} \; \rm{H_2O}}}$$ = $${\text {28.83}}$$ $${\rm{g} \; \rm{Hg(CN)_2}}$$

Based on this calculated value, exactly 28.83 grams of mercury (II) cyanide, Hg(CN)2, can dissolve in 310. grams of water, H2O.  The solution that is described in the problem is prepared using 34.45 grams of mercury (II) cyanide, Hg(CN)2, which is more than this calculated quantity.  A solution that contains more than the maximum amount of solute that should dissolve in a given quantity of solvent should be supersaturated.  However, supersaturated solutions are only created by manipulating the temperature of a solution.  Since the given problem does not indicate altering the temperature of the solution in any way, this solution cannot be supersaturated and, therefore, only the amount of solute that can dissolve at the given temperature does dissolve.  Therefore, since exactly 28.83 grams of mercury (II) cyanide, Hg(CN)2, can dissolve in 310. grams of water, H2O, at 20 degrees Celsius, exactly this quantity of solute does dissolve, and the resultant solution is saturated.  Furthermore, because the amount of solute that is added to prepare this solution, 34.45 grams of mercury (II) cyanide, Hg(CN)2, is more than the quantity of solute that can dissolve in 310. grams of water, 28.83 grams of mercury (II) cyanide, Hg(CN)2, the difference between these values, 5.62 g Hg(CN)2, remains undissolved in this solution.

A solubility curve, which graphs the experimentally-determined solubility of a solute as a function of its temperature, can also be used to determine whether a given solution is saturated or unsaturated.

For example, a solution is prepared by mixing 85 grams of nitrogen trihydride and 100. grams of water at 20 degrees Celsius.  Use the solubility information that is presented in Figure 7.9.1 to determine whether the resultant solution is saturated or unsaturated, and estimate the amount of excess solute that remains undissolved in this solution.

In order to apply a solubility curve to determine whether a solution is saturated or unsaturated, each substance that is referenced in the given problem must first be classified as a solute or a solvent.  Since the chemical formula for water, H2O, is associated with the 100.-gram quantity in the denominator of the solubility proportion in Figure 7.9.1, water, H2O, is the solvent in this solution, and the remaining substance, nitrogen trihydride, NH3, is the solute, "by default."

Based on the solubility curve for nitrogen trihydride, NH3, that is shown in Figure 7.9.1, the solubility of this solute has an estimated value of 55 g/100. g H2O at 20 degrees Celsius.  Therefore, because the solubility limit of a solute corresponds to the maximum amount of that chemical that can dissolve in a given amount of solvent, approximately 55 grams of nitrogen trihydride, NH3can dissolve in 100. grams of water at 20 degrees Celsius.  The solution that is described in the problem is prepared using 85 grams of nitrogen trihydride, NH3, which is more than this estimated quantity.  A solution that contains more than the maximum amount of solute that should dissolve in a given quantity of solvent should be supersaturated.  However, supersaturated solutions are only created by manipulating the temperature of a solution.  Since the given problem does not indicate altering the temperature of the solution in any way, this solution cannot be supersaturated and, therefore, only the amount of solute that can dissolve at the given temperature does dissolve.  Therefore, since approximately 55 grams of nitrogen trihydride, NH3can dissolve in 100. grams of water, H2O, at 20 degrees Celsius, this quantity of solute does dissolve, and the resultant solution is saturated.  Furthermore, because the amount of solute that is added to prepare this solution, 85 grams of nitrogen trihydride, NH3, is more than the quantity of solute that can dissolve in 100. grams of water, 55 grams of nitrogen trihydride, NH3, the difference between these values, 30 g NH3, remains undissolved in this solution.

Exercise $$\PageIndex{2}$$

A solution is prepared by mixing 22 grams of sodium chloride and 100. grams of water at 80 degrees Celsius.  Use the solubility information that is presented in Figure 7.9.1 to determine whether the resultant solution is saturated or unsaturated, and estimate the amount of excess solute that remains undissolved in this solution.