Chapter 9: Solutions – Concentration, Dilution, and Solubility

Chapter 9: Solutions

9.1 Solutions

A solution is a homogeneous mixture composed of two or more substances. The major component is called the solvent, and the minor component(s) are called solute(s). Solutions are fundamental in chemistry for studying reactions, properties, and concentrations.

9.4 Concentration of Solutions

Concentration describes the amount of solute dissolved in a given quantity of solvent or solution. The most common unit is molarity (M), defined as moles of solute per liter of solution.

• Molarity (M): $M = \frac{\text{mol solute}}{\text{volume (L)}}$

• Calculating moles of solute: $\text{mol solute} = \text{molarity (M)} \times \text{volume (L)}$

• Calculating volume required: $\text{volume (L)} = \frac{\text{mol solute}}{\text{molarity (M)}}$

Example 1: How many moles are present in 75.0 mL of 0.0150 M KNO3?

• Convert volume to liters: $75.0\ \text{mL} = 0.0750\ \text{L}$

• Calculate moles: $0.0150\ \text{M} \times 0.0750\ \text{L} = 0.001125\ \text{mol} = 1.13 \times 10^{-3}\ \text{mol}$

Example 2: What volume (in mL) is required to contain 0.00500 moles of NaCl in a 0.0250 M solution?

• Calculate volume: $\frac{0.00500\ \text{mol}}{0.0250\ \text{M}} = 0.200\ \text{L} = 200\ \text{mL}$

9.5 Dilution of Solutions

Dilution is the process of reducing the concentration of a solution by adding more solvent. The amount (moles) of solute remains constant, but the volume increases, causing the molarity to decrease.

• Key Principle: $\text{moles of solute before dilution} = \text{moles of solute after dilution}$

• Mathematical relationship: $M_1 V_1 = M_2 V_2$

• $M_1$ = initial molarity, $V_1$ = initial volume

• $M_2$ = final molarity, $V_2$ = final volume

Example 1: Calculate the molarity after diluting 0.050 L of 0.10 M HCl to 1.0 L.

• $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.10\ \text{M} \times 0.050\ \text{L}}{1.0\ \text{L}} = 0.0050\ \text{M}$

Example 2: What is the molarity after diluting 0.180 L of 0.600 M HNO3 to 0.540 L?

• $M_2 = \frac{0.600\ \text{M} \times 0.180\ \text{L}}{0.540\ \text{L}} = 0.200\ \text{M}$

Example 3: What volume is required to dilute 0.240 L of 1.50 M NaOH to 0.400 M?

• $V_2 = \frac{M_1 V_1}{M_2} = \frac{1.50\ \text{M} \times 0.240\ \text{L}}{0.400\ \text{M}} = 0.900\ \text{L}$

9.3 Solubility

Solubility is the maximum amount of solute that can dissolve in a specific amount of solvent at a given temperature. It is typically expressed as grams of solute per 100 grams of solvent (usually water).

• Solubility formula: $\text{Solubility} = \frac{\text{g solute}}{100\ \text{g water}}$

• Temperature sensitivity: Most solids are more soluble at higher temperatures; gases are less soluble as temperature increases.

• Pressure effects: Pressure affects the solubility of gases in liquids (higher pressure increases gas solubility).

Types of Solutions:

• Unsaturated solution: Can dissolve more solute at the given temperature.

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

• Supersaturated solution: Contains more solute than can normally dissolve at the current temperature; unstable and may precipitate.

Example: At 40°C, the solubility of KBr is 80 g/100 g H2O. Identify the following as saturated or unsaturated:

Case	Calculation	Type
60 g KBr in 100 g H2O	$\frac{60\ g}{100\ g\ H_2O}$	Unsaturated
200 g KBr in 200 g H2O	$\frac{200\ g}{200\ g\ H_2O} = \frac{100\ g}{100\ g\ H_2O}$	Saturated
25 g KBr in 50 g H2O	$\frac{25\ g}{50\ g\ H_2O} = \frac{50\ g}{100\ g\ H_2O}$	Unsaturated

Additional info:

• Thermal pollution can affect solubility of gases in natural waters, impacting aquatic life.

• Solubility is a key concept in environmental chemistry, pharmaceuticals, and industrial processes.