solution

COMPONENTS OF A SOLUTION SOLUTE – substance being dissolved; present in lower proportion SOLVENT – the dissolving medium; present in greater proportion

TYPES OF SOLUTIONS SOLUTE SOLVENT EXAMPLE GAS O2 in N2 (air) LIQUID CO2 in soda SOLID H2 in palladium H2O(g) in air Alcohol in H2O Hg in Ag (dental amalgam) Sugar solution Steel (C in Fe)

SOLVATION – dissolving of solids accompanied by heat of solution ENDOTHERMIC – dissolution process absorbs heat from the surrounding EXOTHERMIC – dissolution process releases heat into the surrounding

FACTORS AFFECTING SOLUBILITY 1. Nature of solute and solvent – the solute-solvent interaction must be greater than the solute-solute interactions and solvent –solvent interactions (or like dissolves like – polar solutes combine with polar solvent; non-polar solutes combine with non-polar solvents

2. Pressure – The greater the pressure above a liquid, the greater is the solubility of the gas in the liquid. 3. Temperature – The lower the temperature of a liquid, the greater is the solubility of the gas in the liquid.

RELATIVE (or comparative) WAYS OF EXPRESSING CONCENTRATION UNSATURATED or DILUTE – contains a relatively small amount of solute SATURATED – contains the maximum amount of solute in that a given quantity of solvent can dissolve in a given temperature SUPERSATURATED – contains more solute than saturated solutions - prepared by adding more solute to a saturated solution at a higher temperature

QUANTITATIVE WAYS OF EXPRESSING CONCENTRATION PERCENTAGE COMPOSITION PARTS PER MILLION MOLE FRACTION MOLARITY MOLALITY

PERCENTAGE CONCENTRATION % by mass of solute = mass of solute X 100% mass of solution % by mass of solvent mass of solvent X 100% % by volume of solute volume of solute X 100% volume of solution % by volume of solvent volume of solvent X 100%

(Ex.) What is the % by mass of sugar prepared by dissolving 18 g of sugar in 54 g of water? Solution: Mass of solute (sugar) = 18 g Total mass of solution = mass of sugar + mass of H2O = 18 g + 54 g = 72 g % sugar (by mass) = mass of sugar X 100% mass of solution = 18 g X 100% = 25 % 72 g

PARTS PER MILLION (ppm) ppm of solute = mass of solute X 1 000 000 mass of solution ppm of solvent mass of solvent X 1 000 000

MOLARITY (M) – number of moles (n) of solute per liter of solution (V)

Ex. A solution is prepared by dissolving 12 g of NaOH in water to make 80 mL of solution. What is the molarity of the solution? Solution: Molar Mass (MM) of NaOH = (23 + 16 + 1) g = 40 g/mole Number of moles (n) = Given Weight (GW) Molar Mass (MM) = 12 g = 0.3 mole 40 g/mole Molarity (M) = 0.3 mole = 3.75 mole = 3.75 M 0.08 L L

MOLALITY (m) – number of moles (n) of solute kilogram of solvent (kg)

Ex. What is the molality of a solution if 196 g of H2SO4 is dissolved in 500 grams of water? Solution: Molar Mass of H2SO4 = 2(1)g + 1(32)g + 4(16)g = 98 g/mole Number of moles (n) = Given Weight (GW) Molar Mass (MM) = 196 g = 2 moles 98 g/mole m = 2 moles = 4 m 0.5 kg

MOLE FRACTION (mf) – number of moles of solute OR solvent divided by the total number of moles in the solution

Solution: Molar Mass of H2SO4 = 2(1)g + 1(32)g + 4(16)g = 98 g/mole Ex. What is the mole fraction of the solute if 196 g of H2SO4 is dissolved in 500 grams of water? Solution: Molar Mass of H2SO4 = 2(1)g + 1(32)g + 4(16)g = 98 g/mole Number of moles (n) = Given Weight (GW) Molar Mass (MM) H2SO4 = 196 g = 2 moles H2SO4 98 g/mole H2O = 500 g = 27.8 moles H2O 18 g/mole mf H2SO4 = 2 moles H2SO4 = 0.067 2 moles H2SO4 + 27.8 moles H2O