🧪 Solutions & Concentration Chemistry 🧪

Solutions and concentration form the foundation of quantitative chemistry. Understanding how to express and calculate different concentration units enables chemists to predict reactions, prepare accurate solutions, and analyze chemical systems effectively.

🎯 Percentage Composition in Solutions and Concentration

Percentage composition expresses solutions and concentration as the mass of solute per 100 parts of solution. This fundamental concept appears in three main forms:

Weight/Weight Percentage (w/w)

% (w/w) = (Mass of solute / Mass of solution) × 100

Example Problem:

Calculate the w/w percentage of a solution containing 25g NaCl in 200g water.

Solution:

Mass of solution = 25g + 200g = 225g

% (w/w) = (25/225) × 100 = 11.11%

Weight/Volume Percentage (w/v)

% (w/v) = (Mass of solute in g / Volume of solution in mL) × 100

Volume/Volume Percentage (v/v)

% (v/v) = (Volume of solute / Volume of solution) × 100

⚗️ Mole Fraction in Solutions and Concentration

Mole fraction represents the ratio of moles of one component to total moles in solutions and concentration calculations.

χ_A = n_A / (n_A + n_B + n_C + …)
where χ_A + χ_B + χ_C + … = 1

Numerical Problem:

Find the mole fraction of ethanol in a solution containing 2 moles ethanol and 8 moles water.

Solution:

Total moles = 2 + 8 = 10 moles

χ_ethanol = 2/10 = 0.2

χ_water = 8/10 = 0.8

🔬 Molarity: The Most Common Solutions and Concentration Unit

Molarity defines solutions and concentration as moles of solute per liter of solution. It’s temperature-dependent and widely used in laboratory work.

Molarity (M) = Moles of solute / Volume of solution (L)

Advanced Numerical Problem:

Calculate the molarity of H₂SO₄ solution prepared by dissolving 49g H₂SO₄ in water to make 500mL solution.

Solution:

Molar mass of H₂SO₄ = 98 g/mol

Moles = 49g ÷ 98 g/mol = 0.5 mol

Volume = 500mL = 0.5L

Molarity = 0.5 mol ÷ 0.5L = 1.0 M

🌡️ Molality: Temperature-Independent Solutions and Concentration

Molality expresses solutions and concentration as moles of solute per kilogram of solvent. Unlike molarity, molality remains constant with temperature changes.

Molality (m) = Moles of solute / Mass of solvent (kg)

Challenging Problem:

Calculate molality when 58.5g NaCl dissolves in 2kg water.

Solution:

Molar mass of NaCl = 58.5 g/mol

Moles of NaCl = 58.5g ÷ 58.5 g/mol = 1.0 mol

Molality = 1.0 mol ÷ 2 kg = 0.5 m

⚖️ Normality: Equivalent-Based Solutions and Concentration

Normality measures solutions and concentration based on equivalent weight, crucial for acid-base and redox reactions.

Normality (N) = Number of equivalents / Volume of solution (L)
Equivalents = Mass / Equivalent weight

Complex Numerical Problem:

Find normality of 0.1M H₂SO₄ solution.

Solution:

H₂SO₄ is diprotic (n = 2)

Normality = Molarity × n

N = 0.1M × 2 = 0.2N

📊 PPM and PPB: Trace Solutions and Concentration

Parts per million (ppm) and parts per billion (ppb) express extremely dilute solutions and concentration levels.

ppm = (Mass of solute / Mass of solution) × 10⁶
ppb = (Mass of solute / Mass of solution) × 10⁹

Environmental Chemistry Problem:

Calculate ppm of lead if 0.05mg Pb exists in 1L water (density = 1g/mL).

Solution:

Mass of solution = 1L × 1000g/L = 1000g

Mass of Pb = 0.05mg = 0.00005g

ppm = (0.00005g / 1000g) × 10⁶ = 0.05 ppm

🔗 Relationships Between Solutions and Concentration Units

Understanding interconversions between different solutions and concentration units proves essential for advanced chemistry applications.

Molarity to Normality Relationship

Normality = Molarity × n-factor

Molarity to Molality Relationship

Molality = (Molarity × 1000) / (1000 × density – Molarity × Molar mass)

PPM to Molarity Conversion

Molarity = (ppm × density × 10) / Molar mass

🧮 Advanced Numerical Problems: Solutions and Concentration

Multi-Step Problem:

A solution contains 20% HCl by weight with density 1.1 g/mL. Calculate its molarity, molality, and normality.

Solution:

Step 1 – Molarity:

Mass of HCl in 1L = 1000mL × 1.1 g/mL × 0.20 = 220g

Moles of HCl = 220g ÷ 36.5 g/mol = 6.03 mol

Molarity = 6.03 M

Step 2 – Molality:

Mass of water = 1100g – 220g = 880g = 0.88 kg

Molality = 6.03 mol ÷ 0.88 kg = 6.85 m

Step 3 – Normality:

HCl is monoprotic (n = 1)

Normality = 6.03 N

Dilution Problem:

How much water must be added to 200mL of 0.5M NaOH to make it 0.2M?

Solution:

Using M₁V₁ = M₂V₂

0.5M × 200mL = 0.2M × V₂

V₂ = 500mL

Water to add = 500mL – 200mL = 300mL

📚 External References for Solutions and Concentration

For additional learning on solutions and concentration, explore these authoritative sources: