Solutions Aqueous dissolved soluble dissociated homogeneous etc Recall

Solutions! Aqueous, dissolved, soluble, dissociated, homogeneous, etc…

Recall that matter is classified and categorized: MATTER Pure Substances Elements Compounds Mixtures Heterogeneous mixtures Homogeneous mixtures

Solutions • A solution is a homogenous mixture of two or more substances in a single physical state. • A solution is made up of a solute and a solvent. • Solute – the substance that is dissolved (usually appears to change state – for example, sugar in water doesn’t appear to stay solid). • Solvent – the substance that does the dissolving; present in greater amounts.

Types of Solutions 1. Solids – two solids evenly mixed (ex: alloys, like gold jewelry – molten (liquid) gold and copper mixed). 2. Gaseous – two or more gases mixed (ex: air) 3. Liquid – solvent and solution are liquids; solute can be solid, liquid or gas (ex: vinegar – acetic acid in water, soft drinks – CO 2 and sugar with H 2 O).

OUR FOCUS: Aqueous - A special type of liquid solution, with water as the solvent.

The formation of Solutions The dissolving process for ionic compounds: • Ionic substances (salts) are composed of ions with positive and negative charges. • Cations are attracted to the negative ends of nearby water molecules; Anions are attracted to the positive ends of nearby water molecules (this is called an ion-molecule interaction). • If the attraction is strong enough, the ion will be pulled away from the surface of the crystal. • The ion in solution will be surrounded by a shell of water molecules.

The formation of Solutions The dissolving process for ionic compounds: https: //www. youtube. com/watch? v=xdedxfhcp Wo

An equation can be written to describe this process: Ca. Cl 2 +2 Ca (aq) + 2 Cl (aq) When Ca. Cl 2 dissolves in water the compound will dissociate into Ca 2+ ions and Cl- ions.

The formation of Solutions The dissolving process for covalent compounds: • Covalent compounds are molecules. As with ionic compounds dissolving, there attractions between the solute and solvent particles. • Attractions may be classified as dipole-dipole forces, hydrogen bonds, or van der Waals forces. • If the attraction is strong enough, an outer molecule will be pulled away from the surface of the solid. • However, the molecules themselves stay intact – they are not separated like ions!

• Solvation – the interaction between solute and solvent particles. • Hydration – the interaction between solute and water molecules (when the solvent is water).

Highlight this (it’s important ) • Energy is required to break bonds/attractions. • Energy is released when bonds/attractions form. • If breaking attractions requires more energy than is released in forming attractions, heat will be absorbed in the overall process dissolving process is endothermic. • If breaking attractions requires less energy than is released in forming attractions, heat will be given off in the overall process dissolving process is exothermic.

Concentration of a Solution Concentration = the amount of solute in a given amount of solvent or solution. • Concentrated – relatively lots of solute • Dilute – relatively small amount of solute

We usually use the mole to help us express concentration of a solution. • There are three ways to express concentration: 1. Molarity – moles of solute per liter of total solution 2. Molality – moles of solute per kilogram of just solvent 3. Percent Solutions – ratio of volume of solute to volume or solution OR ratio of mass of solute to mass of solution

MOLARITY • Molarity (M) – the number of moles of a substance per liter of solution Moles of a substance = molarity (M) Liters of solution

Example 1 • A chemistry teacher needs to make 500. m. L of a dilute calcium chloride solution. She puts 0. 050 moles of Ca. Cl 2 in 500. m. L of solution. What is the concentration of the solution? • Calculate molarity by dividing number of moles by number of liters. (500 m. L = 0. 500 L) M = moles = 0. 050 mol = 0. 10 M liters 0. 500 L

Example 2 • A household cleaner contains 10. 0 g Na. OH in 0. 100 L of solution. What is the molarity of the cleaning solution? 1. Convert grams to moles 10. 0 g Na. OH x 1 mol = 0. 250 mol Na. OH 39. 998 g 2. Calculate molarity by dividing moles by liters Molarity = mol = 0. 250 mol = 2. 5 M L 0. 100 L

Making a solution for the original solute is a simple process: 1. Determine the mass of solute needed and measure on a balance. 2. Pour the solute into the proper size volumetric flask. 3. Add enough distilled water to make the necessary amount of solution. 4. Cover the volumetric flask and mix completely by inverting. http: //www. chem. ucla. edu/~gchemlab/soln_conc_web. htm

Example • A student needs to use 0. 50 M Pb(NO 3)2 for a lab. How many grams of Pb(NO 3)2 must be used to make 250. m. L of the solution? Briefly describe how the solution would be made. 1. Determine the number of moles needed to make the required amount of solution. M = mol L 0. 50 M Pb(NO 3)2 = x mol 0. 250 L X = 0. 125 mol Pb(NO 3)2

Example, continued 2. Convert moles to grams 0. 125 mol Pb(NO 3)2 x 331. 22 g = 41. 4025 g Pb(NO 3)2 1 mol 3. Briefly describe how to make the solution. Measure 41. 40 g of lead (II) nitrate on a balance. Put solid into a 250 m. L volumetric flask. Add distilled water until the flask is about half-full. Mix completely by covering and inverting. Add enough distilled water to make 250 m. L of solution.

http: //www. chem. ucla. edu/~gchemlab/soln_conc_web. htm Example, continued 3. Briefly describe how to make the solution. • Measure 41. 40 g of lead(II) nitrate on a balance. • Put solid into a 250 m. L volumetric flask. • Add distilled water until the flask is about halffull. • Mix completely by covering and inverting. • Add enough distilled water to make 250 m. L of solution.