Composition of Substances and Solutions
Chapter 6: Composition of Substances and Solutions
Learning Objectives
6.3 Molarity - Describe the fundamental properties of solutions
Calculate solution concentrations using molarity
Perform dilution calculations using the dilution equation
9.1 Solutions
Definition of Solutions - Solutions are homogeneous mixtures of two or more substances, meaning they have a uniform composition and appearance throughout. They form when there is sufficient attractive forces (intermolecular forces) between the solute and solvent molecules to overcome the forces holding the solute particles together and the forces between solvent molecules. Solutions are typically composed of two main components:
Solvent: The component present in the largest amount, which dissolves the solute. It determines the physical state of the solution.
Solute: The component present in a smaller amount, which is dissolved by the solvent. A solution can contain one or more solutes.
Learning Goal: Identify the solute and solvent in a solution; describe the process of solution formation, emphasizing the intermolecular interactions involved.
Solutes
Characteristics of Solutes - Solutes exhibit several key characteristics within a solution:
They may be liquids, gases, or solids, and their presence affects the overall properties of the solution.
They are spread evenly throughout the solvent at the molecular or ionic level, making the mixture homogeneous.
Solutes mix with solvents, resulting in the solute sharing the same physical state as the solvent (e.g., a solid solute becomes dispersed in a liquid solvent, forming a liquid solution).
Solutes cannot be separated from the solvent by physical means such as filtration because their particles are too small.
Solutes can often be separated by phase changes, such as evaporation of the solvent, leaving the solute behind.
Solutes are generally not visible to the naked eye within the solution, though they may impart color to the solution (e.g., copper(II) sulfate solutions are blue).
Example: - A solution of copper(II) sulfate (CuSO₄) in water forms as individual particles (ions) of solute dissolve and become evenly dispersed among the solvent (water) molecules. The blue color is characteristic of the hydrated copper(II) ions.
Types of Solutes and Solvents
Solutes and solvents can exist in any of the three common states of matter: solids, liquids, or gases. For example, air is a gaseous solution (oxygen solute in nitrogen solvent), brass is a solid solution (zinc solute in copper solvent), and saltwater is a liquid solution (salt solute in water solvent).
Water as a Solvent
Water is known as the universal solvent and is one of the most common solvents in nature, particularly for biological and chemical processes.
It is a highly polar molecule due to the significant electronegativity difference between oxygen and hydrogen, creating polar O—H bonds and an overall bent molecular geometry.
Water molecules form extensive hydrogen bonds with each other and with other polar or ionic substances. These hydrogen bonds are crucial for its solvent properties and its role in many biological compounds and systems.
Solutions with Ionic Solutes
When an ionic compound, such as sodium chloride (NaCl) crystals, dissolves in water, the process is called hydration.
During hydration, the polar water molecules surround each ion. The partially negative oxygen atoms of water are attracted to the positive cations (like Na⁺), while the partially positive hydrogen atoms are attracted to the negative anions (like Cl⁻). This ion-dipole interaction pulls the ions away from the crystal lattice and into solution, creating a solution with separate, hydrated ions.
Learning Check: Identify the Solute
A: 2 g of sugar and 100 mL of water - Solute: Sugar (present in lesser amount)
B: 60.0 mL of ethyl alcohol and 30.0 mL of methyl alcohol - Solute: Methyl alcohol (present in lesser amount)
C: 55.0 mL of water and 1.50 g of NaCl - Solute: NaCl (present in lesser amount)
D: Air: 200 mL of O₂ and 800 mL of N₂ - Solute: O₂ (present in lesser amount, nitrogen is the solvent)
Learning Check: Solubility in Water
Question: Which of the following solutes will dissolve in water? Why? (Recall the "like dissolves like" rule: polar solvents dissolve polar/ionic solutes, nonpolar solvents dissolve nonpolar solutes)
A. Na₂SO₄: Will dissolve, because it is an ionic compound and water is a polar solvent. The strong ion-dipole interactions overcome the lattice energy.
B. Gasoline (nonpolar): Will not dissolve, because gasoline is primarily nonpolar hydrocarbons, and water is polar. "Like dissolves like" principle applies.
C. I₂: Will not dissolve, because iodine is a nonpolar molecule, and water is polar. The forces between I₂ molecules and water molecules are not sufficient for dissolution.
D. HCl: Will dissolve, because hydrogen chloride is a polar molecule that also ionizes in water to form H⁺ and Cl⁻ ions, making it highly soluble in the polar solvent water.
9.2 Electrolytes and Nonelectrolytes
Role of Electrolytes - Electrolytes are substances that produce ions when dissolved in solution, allowing the solution to conduct electricity. In the body, electrolytes play an absolutely critical role in maintaining proper cellular function, nerve impulse transmission, muscle contraction, hydration, and pH balance. Essential electrolytes include sodium (Na⁺), potassium (K⁺), chloride (Cl⁻), bicarbonate (HCO₃⁻), and magnesium (Mg²⁺), which are routinely measured in blood tests to assess physiological status.
Learning Goal: Identify solutes as electrolytes (strong or weak) or nonelectrolytes based on their ability to produce ions in solution.
Strong Electrolytes
Definition: Strong electrolytes are substances that dissociate or ionize 100% (completely) in water, producing a high concentration of positive and negative ions. This complete dissociation means that virtually all the solute particles exist as ions in solution.
Characteristic: Solutions containing strong electrolytes are excellent conductors of electric current, capable of lighting a light bulb brightly. Examples include strong acids (e.g., HCl, H₂SO₄), strong bases (e.g., NaOH, KOH), and most soluble ionic salts (e.g., NaCl, K₃PO₄).
Dissociation equation representation: Often shown with a single arrow
→indicating a complete reaction, e.g., NaCl(s)
ightarrow Na^+(aq) + Cl^-(aq)
Weak Electrolytes
Definition: A weak electrolyte dissociates or ionizes only slightly (partially) in water, resulting in a solution that contains relatively few ions and a significant amount of undissociated molecules. This partial ionization means an equilibrium exists between the un-ionized compound and its ions.
Characteristic: Solutions of weak electrolytes conduct electric current, but only weakly, often causing a light bulb to glow dimly. Examples include weak acids (e.g., acetic acid, CH₃COOH) and weak bases (e.g., ammonia, NH₃).
Dissociation equation representation: Represented by a double arrow
⇌to indicate an equilibrium between the dissolved molecules and their ions, e.g., CH3COOH(aq) ightleftharpoons H^+(aq) + CH3COO^-(aq)
Nonelectrolytes
Definition: Nonelectrolytes dissolve in water as intact molecules without producing any ions.
Characteristic: Because they do not produce ions, solutions of nonelectrolytes do not conduct electric current. They dissolve by forming intermolecular forces with water, but their molecular structure remains unchanged.
Example: CH₃OH (methanol), C₁₂H₂₂O₁₁ (sucrose), and C₂H₅OH (ethanol) are common nonelectrolytes. When they dissolve in water, their molecules are dispersed throughout the solution but do not ionize.
Learning Check: Complete Reactions for Strong Electrolytes in Water
Reaction 1: CaCl₂(s) → Solution: Ca^{2+}(aq) + 2Cl^-(aq)
Reaction 2: K₃PO₄(s) → Solution: 3K^+(aq) + PO_4^{3-}(aq)
Learning Check: Dissociation of Strong Electrolyte
Which of the following reactions represent the dissociation of a strong electrolyte in water? (A strong electrolyte shows complete ionization.)
A. CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq) (Weak acid, partial dissociation)
B. C₆H₁₂O₆(s) → C₆H₁₂O₆(aq) (Nonelectrolyte, dissolves without ionization)
C. Na₂SO₄(s) → 2Na⁺(aq) + SO₄²⁻(aq) (Ionic salt, complete dissociation)
Learning Check: Write Equations for the Formation of Solutions
(Use the appropriate arrow: → for complete dissociation/dissolving without ionization, ⇌ for partial dissociation)
Dissociation of K₂CrO₄(s): K₂CrO₄(s) → 2K^+(aq) + CrO_4^{2-}(aq) (Strong electrolyte, complete dissociation)
Partial Dissociation of H₃PO₄(aq): H₃PO₄(s) ⇌ H^+(aq) + H2PO4^-(aq) (Weak acid, partial dissociation)
Dissolving Sugar: C₁₂H₂₂O₁₁(s) → C₁₂H₂₂O₁₁(aq) (Nonelectrolyte, dissolves without ionization)
Learning Objectives
6.4 Other Units for Solution Concentrations - Define various concentration units beyond molarity: mass percentage, volume percentage, mass-volume percentage, parts-per-million (ppm), and parts-per-billion (ppb).
Perform computations relating a solution’s concentration to its components’ volumes and/or masses, utilizing these different concentration units.
9.4 Solution Concentrations
Example: To prepare a 1-L solution of potassium iodide (KI), one would add 5.0 g of KI to a 1-L volumetric flask and then add enough distilled water until the total volume reaches exactly 1 L (to the calibration mark), ensuring proper mixing.
Learning Goal: Calculate the concentration of a solute in a solution using various units; use these concentration values as conversion factors to determine the amount of solute or the total amount of solution.
Concentration
Units of Concentration: The choice of concentration unit depends on the application, the physical states of the solute and solvent, and the desired precision.
Mass percent (m/m)
Volume percent (v/v)
Mass/volume percent (m/v)
Molarity (moles solute/liters solution)
Other units like parts-per-million (ppm) and parts-per-billion (ppb) are used for very dilute solutions.
Mass Percent
Definition: Mass percent (m/m) expresses the concentration as the mass of solute in 100 g of solution. It is commonly used when both solute and solvent are solids or when precise mass measurements are available.
\text{Mass percent (m/m)} = \frac{\text{mass of solute (g)}}{\text{mass of solute (g)} + \text{mass of solvent (g)}} \times 100\%
\text{Mass percent (m/m)} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 100\%Example Calculation: - When 8.00 g of KCl is dissolved in 42.00 g of water to form a 50.00 g KCl solution, the mass percent concentration is calculated as: \frac{\text{8.00 g KCl}}{\text{50.00 g solution}} \times 100\% = 16.0\% \text{ (m/m)} . This means there are 16.0 g of KCl for every 100 g of the solution.
Calculating Mass Percent with KCl
To calculate mass percent, the required quantities are:
Grams of solute (KCl): 8.00 g
Grams of solvent (water): 42.00 g
Grams of solution (KCl + water): 8.00 g + 42.00 g = 50.00 g
Learning Check: Calculate Mass Percent
Given: 15.0 g of Na₂CO₃ (solute) and 235 g of H₂O (solvent).
Options: - A. 15.0% (m/m) Na₂CO₃
B. 6.38% (m/m) Na₂CO₃
C. 6.00% (m/m) Na₂CO₃
Solution: Calculate the Mass Percent of Na₂CO₃
STEP 1: State the given/needed quantities.
Mass of Na₂CO₃ = 15.0 g
Mass of H₂O = 235 g
Mass of solution = 15.0 g + 235 g = 250 g
STEP 2: Write the concentration expression for mass percent.
\text{Mass percent (m/m)} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%STEP 3: Substitute quantities and calculate.
\text{Mass percent (m/m)} = \frac{\text{15.0 g Na}2\text{CO}3}{\text{250 g solution}} \times 100\% = 6.00\%Answer: C, 6.00% Na₂CO₃. This indicates that 6.00 grams of Na₂CO₃ are present in every 100 grams of the solution.
Volume Percent
Definition: Volume percent (v/v) expresses the concentration as the percent volume (mL) of a liquid solute relative to the total volume (mL) of solution. It is commonly used when mixing two liquids, such as alcohol in water.
\text{Volume percent (v/v)} = \frac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}} \times 100\%Conversion factor: For example, 5.75% (v/v) ethanol means 5.75 mL of ethanol per 100 mL of solution.
Mass/Volume Percent
Definition: Mass/volume percent (m/v) expresses the concentration as the percent mass (g) of solute in a given volume (mL) of solution. This unit is frequently used in hospital and pharmacy settings for intravenous fluids and medications, where solid solutes are dissolved in liquid solvents.
\text{Mass/volume percent (m/v)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100\%Conversion factor: For example, 4.8% (m/v) HCl means 4.8 g of HCl per 100 mL of solution.
Learning Check: Write Conversion Factors
(A conversion factor expresses the relationship between the percentage and the units of solute and solution)
A: 8.50% (m/m) NaOH $\rightarrow$ \frac{\text{8.50 g NaOH}}{\text{100 g solution}} or \frac{\text{100 g solution}}{\text{8.50 g NaOH}}
B: 5.75% (v/v) ethanol $\rightarrow$ \frac{\text{5.75 mL ethanol}}{\text{100 mL solution}} or \frac{\text{100 mL solution}}{\text{5.75 mL ethanol}}
C: 4.8% (m/v) HCl $\rightarrow$ \frac{\text{4.8 g HCl}}{\text{100 mL solution}} or \frac{\text{100 mL solution}}{\text{4.8 g HCl}}
Molarity
Definition: Molarity (M) is a concentration unit defined as the number of moles of solute per liter of solution. It is one of the most common concentration units used in chemistry because it directly relates to the number of particles (moles) of solute, which is critical for stoichiometric calculations.
M = \frac{\text{moles of solute}}{\text{liters of solution}}Example: A 1.0 M solution of NaCl is defined as having one mole of NaCl dissolved in enough water to make the total volume of the solution exactly 1 liter. This means 58.44 g of NaCl (its molar mass) would be present in 1 L of solution.
Molarity Calculations Example
Problem: What is the molarity of 0.500 L NaOH solution containing 6.00 g of NaOH?
SOLUTION STEPS:
STEP 1: Identify given and needed quantities.
Given: Volume of solution = 0.500 L, Mass of NaOH = 6.00 g
Needed: Molarity (mol/L)
Conversion factor needed: Molar mass of NaOH (1 mole NaOH = 22.99 g Na + 16.00 g O + 1.008 g H = 40.00 g NaOH)
STEP 2: Write the concentration expression.
M = \frac{\text{moles of solute}}{\text{liters of solution}}STEP 3: Substitute and calculate.
First, convert grams of NaOH to moles of NaOH:
\text{6.00 g NaOH} \times \frac{\text{1 mole NaOH}}{\text{40.00 g NaOH}} = \text{0.150 moles NaOH}Now, calculate molarity:
M = \frac{\text{0.150 moles NaOH}}{\text{0.500 L solution}} = \text{0.300 M NaOH}
Learning Check: Molarity Calculation Example
Problem: What is the molarity of 0.225 L of KNO₃ solution containing 34.8 g of KNO₃?
Molar mass of KNO₃ = 39.10 + 14.01 + (3 * 16.00) = 101.11 g/mol
Options: - A. 0.344 M
B. 1.53 M
C. 15.5 M
Solution:
Moles of KNO₃ = \text{34.8 g} \times \frac{\text{1 mol}}{\text{101.11 g}} = \text{0.34418 mol KNO}_3
Molarity = \frac{\text{0.34418 mol}}{\text{0.225 L}} = \text{1.5297 M}
Answer: B, 1.53 M KNO₃ (rounded to three significant figures).
Learning Check: Grams Calculation for NaOH Solution
Problem: How many grams of NaOH are needed to prepare 75.0 g of a 14.0% (m/m) NaOH solution?
Options: - A. 10.5 g of NaOH
B. 75.0 g of NaOH
C. 536 g of NaOH
Solution Steps:
STEP 1: Identify given quantities.
Mass of solution = 75.0 g
Mass percent (m/m) = 14.0%
STEP 2: Plan for calculation.
Use the mass percent as a conversion factor to go from grams of solution to grams of solute.
STEP 3: Write equalities and conversion factors.
14.0% (m/m) NaOH means 14.0 g NaOH / 100 g solution
STEP 4: Set up and calculate mass.
\text{75.0 g solution} \times \frac{\text{14.0 g NaOH}}{\text{100 g solution}} = \text{10.5 g NaOH}Answer: A, 10.5 g NaOH.
Learning Check: Calculating Volume of Ethanol Solution
Problem: How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 mL of ethanol?
Options: - A. 2.56 mL
B. 12.9 mL
C. 39.1 mL
Solution:
Use the volume percent as a conversion factor.
5.75% (v/v) means 5.75 mL ethanol / 100 mL solution.
\text{2.25 mL ethanol} \times \frac{\text{100 mL solution}}{\text{5.75 mL ethanol}} = \text{39.13 mL solution}Answer: C, 39.1 mL of solution (rounded to three significant figures).
Learning Check: Grams of AlCl₃ Calculation for Solution
Problem: How many grams of AlCl₃ are needed to prepare 125 mL of a 0.150 M solution?
Molar mass of AlCl₃ = 26.98 + (3 * 35.45) = 133.33 g/mol
Options: - A. 20.0 g of AlCl₃
B. 16.7 g of AlCl₃
C. 2.50 g of AlCl₃
Solution Steps:
STEP 1: Identify given quantities.
Volume of solution = 125 mL = 0.125 L
Molarity (M) = 0.150 M AlCl₃
STEP 2: Plan for calculation using molarity.
Convert liters of solution to moles of solute using molarity.
Convert moles of solute to grams of solute using molar mass.
STEP 3: Write conversion factors.
Molarity: 0.150 mol AlCl₃ / 1 L solution
Molar mass: 1 mol AlCl₃ / 133.33 g AlCl₃
STEP 4: Set up and calculate mass.
\text{0.125 L solution} \times \frac{\text{0.150 mol AlCl}3}{\text{1 L solution}} \times \frac{\text{133.33 g AlCl}3}{\text{1 mol AlCl}3} = \text{2.4999 g AlCl}3Answer: C, 2.50 g AlCl₃ (rounded to three significant figures).
53 Molarity Concept Explanation
Molarity (M): Molarity precisely quantifies the concentration of a substance, defined as the number of moles of a solute dissolved in each liter of the total solution volume. It's a temperature-dependent concentration unit because volume can change with temperature.
Solution Preparation: To accurately prepare a solution of a known molarity, a precise amount (mass) of solute is weighed using an analytical balance. This solute is then quantitatively transferred to a volumetric flask, which is designed to hold a precise volume. A small amount of solvent is added to dissolve the solute, and once dissolved, more solvent is carefully added up to the flask's calibration mark to achieve the desired exact volume. This method ensures high accuracy in concentration.
55 Molarity Formula
Molarity serves as a powerful conversion factor, allowing interconversion between moles of solute and liters of solution:
M = \frac{\text{moles of solute}}{\text{liters of solution}}