Characteristics of solutions Solution – homogeneous mixture Solution – homogeneous mixture a) parts of a solution i) solute – substance being dissolved ii) solvent – substance doing dissolving both can be either solid, liquid, or gas

Solubility Soluble – substance can dissolve in a solvent Soluble – substance can dissolve in a solvent ex: salt in water Insoluble – substance cannot dissolve in a solvent Insoluble – substance cannot dissolve in a solvent ex: sand in water

Solvation In Aqueous Solutions Solvation – process of surrounding solute particles with solvent particles Solvation – process of surrounding solute particles with solvent particles Why are some substances soluble in a solvent and some others are not? must be compatibility between solute and solvent

Dissolution of sodium Chloride

“like dissolves like” Defn – rule used to determine if substance will dissolve in another Defn – rule used to determine if substance will dissolve in another - based on attractive forces between solute and solvent

“like dissolves like” polar solvents – dissolve polar molecular compounds and ionic compounds polar solvents – dissolve polar molecular compounds and ionic compounds ex: salt and water, alcohol and vinegar nonpolar solvents – dissolve nonpolar compounds only nonpolar solvents – dissolve nonpolar compounds only ex: oil and gasoline

Factors Affecting Rate of Solvation How can you dissolve something faster??? How can you dissolve something faster??? a) increase temp of solvent this accelerates particles creating more particle collisions

Factors Affecting Rate of Solvation b) agitate the solution more particle collisions between solute and solvent more particle collisions between solute and solvent c) Increase surface area of solute breaking into smaller pieces allows more solute to be in contact w/ solvent

Solubility Defn – max amt of solute that can dissolve in a solvent at a specific temp Defn – max amt of solute that can dissolve in a solvent at a specific temp how much solute can be put into solvent?

Unsaturated Solution Defn – less than max amt of solute dissolved Defn – less than max amt of solute dissolved if I put sugar into water and all sugar is dissolved, solution is unsaturated

Saturated Solution Defn – contains max amt of solute dissolved Defn – contains max amt of solute dissolved if I put sugar into water and not dissolves (you can see the sugar), the solution is saturated

Supersaturated Solution Defn – contains more solute than saturated solution at the same conditions Defn – contains more solute than saturated solution at the same conditions a saturated solution made at high temp cools slowly. Slow cooling allows excess solute to remain dissolved in solution at lower temperature very unstable

Solubility Curve (generic) Saturated- Line represents max amount solute that will dissolve at a given temperature Temperature Solubility (g solute/ 100 g H 2 O) Unsaturated (below line) Supersaturated (above line)

How does temp affect solubility? The higher the temp, higher the solubility (for most cases)

Solution Concentration Concentration – how much solute dissolved in amount of solvent Concentration – how much solute dissolved in amount of solvent what is difference between concentrated and diluted?

Concentrated vs. Dilute

Concentration 3 different units of concentration 3 different units of concentration a)percent by mass b) molarity (M) c) molality (m)

Percent by mass Formula FormulaOR

Percent by mass Ex prob: If 3.6 g NaCl is dissolved in 100 g H 2 O, what is the percent by mass? Ex prob: If 3.6 g NaCl is dissolved in 100 g H 2 O, what is the percent by mass? solute What is the solute? NaCl What is the solvent? H2OH2OH2OH2O

Percent by mass Mass of solute (NaCl) = Mass of solute (NaCl) = 3.6 g Mass solvent (H 2 O) = Mass solvent (H 2 O) = 100 g Mass solution = Mass solution = = g

Percent by mass Percent by mass = 3.6 g x x g = 3.5 % NaCl

Molarity Defn - # of moles per liter of solution Defn - # of moles per liter of solution Formula mol solute Formula mol solute L solution L solution unitmol = M (capital M) unitmol = M (capital M) L

Molarity Ex prob #1 Molarity Ex prob #1 A solution has a volume of 250 mL and has 0.70 mol NaCl. What is the molarity? A solution has a volume of 250 mL and has 0.70 mol NaCl. What is the molarity? 2.8 mol/L or 2.8 M L = 0.70 mol

Molarity ex prob #2 What is the molarity of a solution made of 47.3 g NaOH in 500 mL water? What is the molarity of a solution made of 47.3 g NaOH in 500 mL water? step 1: convert grams to moles 47.3 g NaOH 40 g NaOH 1 mol NaOH = mol NaOH

Molarity ex prob #2 Step 2: divide moles by volume (L) 2.37 mol/L NaOH or 2.37 M NaOH L = mol

Molarity ex prob #3 How many moles of solute are present in 1.5 L of 2.4 M NaCl? How many grams? How many moles of solute are present in 1.5 L of 2.4 M NaCl? How many grams? # moles = volume x molarity 1.5 L L 2.4 mol NaCl x = 3.6 mol NaCl

Molarity ex prob #3 moles to grams moles to grams 3.6 mol NaCl 1 mol NaCl 58.5 g NaCl = g NaCl

Diluting Solutions Defn – add more solvent to original solution Defn – add more solvent to original solution Formula Formula M 1 V 1 = M 2 V 2 M 1 is more concentrated than M 2

Diluting Solutions What volume of a 2.0 M stock solution is needed to make 0.50 L of a M solution? What volume of a 2.0 M stock solution is needed to make 0.50 L of a M solution? M1=M1= V1=V1= M2=M2= V2=V2= 2.0 M0.300 M ? 0.50 L (2.0 M) V 1 = (0.300 M)(0.50 L) V 1 = L

Diluting Solutions If you dilute 20.0 mL of a 3.5 M solution to make mL of solution, what is the molarity of the dilute solution? If you dilute 20.0 mL of a 3.5 M solution to make mL of solution, what is the molarity of the dilute solution? M1=M1= V1=V1= M2=M2= V2=V2= 3.5 M? 20.0 mL 100 mL (3.5 M)(20.0 mL) = M 2 (100.0 mL) M 2 = 0.7 M

Molality Defn - # moles of solute in one kg solvent Defn - # moles of solute in one kg solvent Formulamol solute Formulamol solute kg solvent Unitsmol = m (lower case m) Unitsmol = m (lower case m) kg kg

Molality ex problem What is the molality of a solution with 8.4 g NaCl in 255 g of water? What is the molality of a solution with 8.4 g NaCl in 255 g of water? Step 1: convert grams to moles 8.4 g NaCl 58.5 g NaCl 1 mol NaCl = 0.14 mol NaCl

Molality ex problem Step 2: divide by mass (kg) 0.55 mol/kg NaCl or 0.55 m NaCl kg = 0.14 mol NaCl

Colligative Properties of Solutions Solutes affect the physical properties of their solvents Solutes affect the physical properties of their solvents Colligative properties (defn) – properties that depend only on the number of solute particles present, not their identity Colligative properties (defn) – properties that depend only on the number of solute particles present, not their identity Ex: boiling point, freezing point Ex: boiling point, freezing point

Electrolytes Defn – substances that break up (ionize) in water to produce ions; can conduct electricity Defn – substances that break up (ionize) in water to produce ions; can conduct electricity - consist of acids, bases, ionic compounds Ex: NaCl  Na 1+ + Cl 1- H 2 SO 4  2 H + + SO 4 2-

Nonelectrolytes Defn – do not break up (ionize) in water, they stay the same; doesn’t conduct electricity Defn – do not break up (ionize) in water, they stay the same; doesn’t conduct electricity - usually molecular/covalent compounds Ex:sugar C 6 H 12 O 6  C 6 H 12 O 6 ethanolC 2 H 5 OH  C 2 H 5 OH

Determining # of solute particles For ionic cmpds/acids For ionic cmpds/acids  sum # moles of ions ex: NaCl  1 Na Cl 1- = 2 particles CaBr 2  1 Ca Br 1- = 3 particles

Determining # of solute particles For covalent compounds For covalent compounds  # is always 1 ex: sugar C 6 H 12 O 6  1 particle ethanolC 2 H 5 OH  1 particle

Freezing Point Depression Defn (ΔT f ) – difference in temp between solution’s fp and pure solvent’s fp Defn (ΔT f ) – difference in temp between solution’s fp and pure solvent’s fp FormulaΔT f = K f x m x i FormulaΔT f = K f x m x i # particles molality molal fp constant

Ex problem What is the freezing pt of water if 12.3 g NaCl is added to 200 g water? What is the freezing pt of water if 12.3 g NaCl is added to 200 g water? (K f = 1.86 °C/m, fp = 0°C) 1 mol NaCl 58.5 g NaCl 12.3 g NaCl = 0.21 mol NaCl m = 0.21 mol NaCl kg water = 1.05 m

Ex problem ΔT f = K f x m x i ΔT f = K f x m x i ΔT f = (1.86 °C/m)(1.05 m)(2) = 3.91 °C New f.p. = 0°- 3.91°= -3.91°C

Boiling Point Elevation Defn (ΔT b ) – difference in temp between solution’s bp and pure solvent’s bp Defn (ΔT b ) – difference in temp between solution’s bp and pure solvent’s bp FormulaΔT b = K b x m x i FormulaΔT b = K b x m x i # particles molality molal bp constant

Ex problem What is the new boiling point of acetone if 13.7 g C 6 H 12 O 6 is dissolved in 200 g acetone? (K b = 1.71°C/m, bp = 56°C) What is the new boiling point of acetone if 13.7 g C 6 H 12 O 6 is dissolved in 200 g acetone? (K b = 1.71°C/m, bp = 56°C) 1 mol C 6 H 12 O g C 6 H 12 O g C 6 H 12 O 6 = mol m = mol C 6 H 12 O kg acetone = m

Ex problem ΔT b = K b x m x i ΔT b = K b x m x i ΔT b = (1.71 °C/m)(0.381 m)(1) = 0.65 °C New bp = 56°+ 0.65°= 56.65°C