Solutions

Solutions in Chemistry

Definition of a Solution

  • A solution is defined as a homogeneous mixture.

    • Solvent: The substance present in the largest amount in the solution.

    • Solutes: Other substances that are dissolved within the solvent.

    • Aqueous solution: A specific type of solution where water is the solvent.

Types of Solutions

  • Ionic Substances: These substances break up into their constituent cations and anions when dissolved in a solvent.

  • Solubility of Ionic Substances:

    • Polar water molecules interact with the positive and negative ions in the ionic compounds.

Solubility of Polar Substances

  • An example of a polar substance that is soluble in water is ethanol, which has a polar OH group.

  • Sugar and Water: The question arises, why is solid sugar soluble in water?

Characteristics of Polar and Nonpolar Molecules

  • For a molecule to be classified as polar, two conditions must be met:

    1. The molecule must contain polar bonds.

    2. The spatial arrangement of these bonds must lead to a separation of charge within the molecule.

Interaction of Polar and Nonpolar Substances

  • Nonpolar substances, such as oil, do not interact with polar solvents like water.

    • Water-water hydrogen bonds prevent the mixing of nonpolar molecules with water.

  • To dissolve a solute in water, a "hole" must be created in the water structure for each solute particle, leading to replacing the lost water-water interactions with new water-solute interactions based on the principle: "like dissolves like".

Limitations of Solubility

  • Solubility of a solute is limited:

    • Saturated solution: Contains the maximum amount of solute that can dissolve at a given temperature.

    • Unsaturated solution: Has not reached the solubility limit; more solute can still be added.

    • Supersaturated solution: This occurs when a saturated solution is cooled after allowing more solute to dissolve, remaining in solution. This state is unstable; adding a crystal can cause precipitation.

Concentration of Solutions

  • A solution can vary in concentration based on the amounts of solute and solvent:

    • To describe concentration qualitatively:

    • Concentrated solution: Contains a relatively large amount of solute.

    • Dilute solution: Contains a relatively small amount of solute.

Mass Percent Concentration
  • The formula for calculating mass percent concentration is as follows:

    • \text{Mass Percent} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100\%

    • Alternatively: \text{Mass Percent} = \left( \frac{\text{grams of solute}}{\text{grams of solute} + \text{grams of solvent}} \right) \times 100\%

Example Calculation: Mass Percent
  • To calculate the percent by mass of glucose in a solution:

    • If 5.5 g of glucose is dissolved in 78.2 g of water:

    • \text{Percent by mass} = \left( \frac{5.5 g}{5.5 g + 78.2 g} \right) \times 100 = 6.6\%

Molarity
  • The formula for calculating molarity (M) is defined as:

    • M = \frac{\text{moles of solute}}{\text{liters of solution}}

  • Example:

    • If 6 moles of HCl is dissolved to give 2 liters of solution, then:

    • M = \frac{6\text{ moles}}{2\text{liters}} = 3 M

Example Calculation: Molarity
  • If provided with 1.00 mol of sugar in 125.0 mL solution:

    • Convert mL to L: 125.0 mL = 0.1250 L

    • Calculate molarity:

    • M = \frac{1.00 mol}{0.1250 L} = 8.00 M

Exercise on Molarity
  • If you dissolve 500.0 g of potassium phosphate and make 1.50 L of solution:

    • Convert 500.0 g to moles: 500.0 g \div 212.27 g/mol = 2.355 mol

    • Calculate molarity:

    • M = \frac{2.355 mol}{1.50 L} = 1.57 M

Further Practice
  • Determine the volume needed of a 10.0 M sugar solution to obtain 2.00 mol of sugar:

    • \text{Volume} = \frac{2.00 mol}{10.0 M} = 0.200 L

Concentration of Ions in Solutions
  • For a 0.25 M CaCl2 solution:

    • Dissociation occurs:

    • CaCl2 \rightarrow Ca^{2+} + 2Cl^{-}

    • Concentrations of ions:

      • [Ca^{2+}] = 1 \times 0.25 M = 0.25 M

      • [Cl^{-}] = 2 \times 0.25 M = 0.50 M

Compare Solutions with Different Concentrations
  • Given two HCl solutions (Solution A and Solution B), with Solution A having a higher concentration:

    • If equal volumes are compared, Solution B must contain more moles of HCl.

    • If comparing equal moles, Solution B must have a greater volume.

    • To achieve equal concentrations, water must be added to Solution B.

    • Adding more moles of HCl to both will increase their concentration.

Ion Concentrations Practice Exercise
  • Determine which of the following contains the greatest number of ions:

    • a) 400.0 mL of 0.10 M NaCl.

    • b) 300.0 mL of 0.10 M CaCl2.

    • c) 200.0 mL of 0.10 M FeCl3.

    • d) 800.0 mL of 0.10 M sucrose.

In-Class Exercise
  • Calculate the molarity of a solution prepared by dissolving 0.185 mol Na2SO4 in enough water to make 450 mL of solution.