Properties of Solutions and Concentration Calculations

Properties of Solutions

  • Mixtures:

    • Defined as a combination of two or more non-reacting pure substances.

    • Common examples include:

    • Saltwater (water and NaCl).

    • Air (oxygen, nitrogen, and trace gases).

    • Milk (water, sugar, fat, proteins, vitamins, calcium).

    • Components can be separated into pure substances; e.g., boiling saltwater to retrieve salt.

    • Mixtures can be classified as:

    • Heterogeneous Mixtures:

      • Composition varies throughout the sample (e.g., rocks, salad dressing).

      • Properties differ from one region to another; may require magnification to observe differences.

    • Homogeneous Mixtures:

      • Uniform composition and properties throughout (e.g., sugar water, air).

      • Also known as solutions.

Solution Components

  • A solution consists of:

    1. Solute:

    • The substance being dissolved; usually lesser in amount.

    1. Solvent:

    • The substance that dissolves the solute; usually greater in amount.

    • If the solvent is water (H2O), it’s termed an aqueous solution.

  • Solutions can vary in state:

    • Examples of different types of solutions include:

    • Gas in liquid: Club soda (CO2 in water).

    • Liquid in liquid: Antique in radiators (ethylene glycol in water).

    • Solid in liquid: Sugar in water.

    • Solid in solid: Solder (tin in lead).

Factors Affecting Solubility

  • Effect of Chemical Structure:

    • “Like dissolves like” principle:

    • Ionic and polar solutes dissolve in polar solvents.

    • Non-polar solutes dissolve in non-polar solvents.

    • Examples:

    • NaCl is soluble in water (polar) but not in CCl4 (non-polar).

    • CH4 is soluble in non-polar substances but not in water.

  • Effect of Pressure:

    • Pressure impacts gas solubility in liquids.

    • Henry’s Law states that gas solubility in a liquid increases with increasing pressure of the gas.

    • Example: Opening a soda can decreases pressure, resulting in bubbles as CO2 escapes.

  • Effect of Temperature:

    • Gas solubility decreases with increased temperature.

    • Solid solubility typically increases with temperature (with exceptions).

Concentration of Solutions

  • Concentration describes the amount of solute in a solution.

  • Molarity (M):

    • Defined as moles of solute per liter of solution.

    • Formula:

    • M = moles of solute / volume of solution in liters

    • Units are mol·L⁻¹ or M.

Practice: Molarity Calculation

  • Example 1: Molarity of HCl in 26.8 mL solution:

    • M = 4.28 x 10⁻² moles / (26.8 mL = 0.0268 L) = 1.60 M

  • Example 2: Molarity of KBr in 1.75 L solution:

    • KBr mass = 25.5 g; Moles of KBr = 25.5 g / 119.00 g/mol = 0.214 moles.

    • M = 0.214 moles / 1.75 L = 0.122 M

Preparing Solutions of Known Concentrations

  1. Combining a Weighed Solute with Solvent:

    • Example: Prepare 2.00 L of 1.50 M KBr.

    • Required mass = moles = M × V = 1.50 M × 2.00 L = 3.00 moles.

Mass = 3.00 mol × 119.00 g/mol = 357 g KBr

  1. Calculating Mass for Specific Volume:

  • Example: What mass of KBr (in grams) do you need to make 250.0 mL of a 1.50 M KBr solution?

  • Moles of KBr = M × V = 1.50 mol/L × 0.250 L = 0.375 mol KBr.

  • Mass of KBr = 0.375 mol × 119.00 g/mol = 44.6 g KBr.

  • Practice Exercise: What mass (in grams) of Mg(NO3)2 do you need to make 145 mL of a 1.50 M aqueous solution of Mg(NO3)2?

    • Required moles = 1.50 mol/L × 0.145 L = 0.2175 mol.

    • Molar mass of Mg(NO3)2 = 24.31 g/mol (Mg) + (2 × 14.01 g/mol (N)) + (6 × 16.00 g/mol (O)) = 148.31 g/mol.

    • Mass = 0.2175 mol × 148.31 g/mol = 32.2 g Mg(NO3)2

  1. Diluting a More Concentrated Solution:

    • Example: Dilute 2.00 M KBr to 0.100 M in 500 mL.

    • Moles required = 0.100 M × 0.500 L = 0.0500 mol.

    • Volume of concentrated solution = n / M = 0.0500 mol / 2.00 M = 25.0 mL

Practice Exercise

  • Dilution Problem: If you dilute 25.0 mL of 1.50 M HCl to 500.0 mL, find the final concentration.

    • Answer: 0.0750 M

To solve the last dilution problem, we used the dilution formula, which states that the product of the concentration and volume before dilution equals the product of the concentration and volume after dilution. The formula is:

C₁V₁ = C₂V₂

Where:

  • C₁ is the initial concentration (1.50 M),

  • V₁ is the initial volume (25.0 mL),

  • C₂ is the final concentration, and

  • V₂ is the final volume (500.0 mL).

Since we are trying to find the final concentration (C₂), we can rearrange the formula:

C₂ = (C₁V₁) / V₂

Plugging in the values:

C₂ = (1.50 M * 25.0 mL) / 500.0 mL = 0.0750 M

Thus, the final concentration after dilution is 0.0750 M.