Presentation on theme: "1 Ch. 7: Solutions Chem. 20 El Camino College. 2 Terminology The solute is dissolved in the solvent. The solute is usually in smaller amount, and the."— Presentation transcript:
2 Terminology The solute is dissolved in the solvent. The solute is usually in smaller amount, and the solvent is usually in larger amount A concentrated solution contains a relatively large amount of solute A dilute solution contains a relatively small amount of solute Solubility measures how much of a solute can dissolve in a solvent at a given temperature.
5 Terminology An unsaturated soln can hold more of a soluteunsaturated A saturated soln cannot hold more of a solute. If more solute is added (without the temp changing), it won’t dissolvesaturated A supersaturated soln is very unstable. It’s made by creating a saturated soln at high temp, and then cooling slowly. Adding a crystal will cause the excess solute to crystallize outsupersaturated Miscible refers to liquids that are soluble in each other Immiscible liquids are not soluble, like oil and water.
6 Terminology “Like dissolves like” means Polar molecules dissolve polar (and many ionic) substances Nonpolar molecules dissolve nonpolar substances Water is a very polar molecule. Which will dissolve in water: NaCl, sugar, benzene (a nonpolar molecule) Is sugar polar or nonpolar?
8 Solution Formation When water dissolves salt crystals, the + end of water is attracted to the Cl - ions, and the - end is attracted to the Na + ions The ions become hydrated, or surrounded by water molecules.
11 Electrolytes Substances can be classified as strong-, weak-, or non-electrolytes in water Strong electrolytes contain a lot of ions that conduct electricity Weak electrolytes contain only a few ions, and they conduct electricity weakly Nonelectrolytes do not contain ions. They don’t conduct electricity
14 Strong Electrolytes When a strong electrolyte is placed in water, it dissolves and forms ions Write the eqn for Ca(NO 3 ) 2 (s) dissolving in H 2 O. Ca(NO 3 ) 2 (s) Ca 2+ (aq) + 2 NO 3 - (aq) ions only Write the eqn for NaCl(s) dissolving in H 2 O. NaCl(s) Na + (aq) + Cl - (aq) ions only H2OH2O H2OH2O
15 Weak Electrolytes Some substances, such as weak acids, only form a few ions when placed in water For weak acids, H + pops off a small % of molecules Since there are only a few ions in the soln, weak electrolytes conduct electricity weakly HF(g) HF(aq) (minor: H + (aq) + F - (aq)) mostly molecules, few ions H2OH2O
16 Nonelectrolytes Many molecules do not form ions when they dissolve C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) molecules only H2OH2O
17 Classify These Based on the given information, decide whether ions only, few ions, or molecules only form in water. Are they strong, weak, or nonelectrolytes? H2OH2O LiCl(s) Li + (aq) + Cl - (aq) note: LiCl is water soluble H2OH2O CH 3 OH( l ) CH 3 OH(aq) note: CH 3 OH is a molecule that dissolves, but doesn’t form ions ions only, strong electrolyte few ions, weak electrolyte molecules only, nonelectrolyte H2OH2O HNO 2 (g) H + (aq) + NO 2 - (aq) note: HNO 2 is a weak acid
18 Factors Affecting Solubility Pressure More gas dissolves at higher pressure. Pressure has no effect on solubility of solids or liquids Temperature More solid dissolves at higher temps More gas dissolves at lower temps.
20 Solution Concentration (g solute+g solvent) mass % = g solute g solution x 100 g solute x 100 mass % = Ex. What is the % concentration of 250.0 g solution which contains 8.75 g solid? % by mass = g solute g solution x 100 = 8.75 g 250.0 g x 100 = 3.50 % Ex. What is the % conc of a soln which contains 5.33 g NaCl and 10.99 g water? g solute + g solvent g solute x 100 mass % = 5.33 g + 10.99 g 5.33 g x 100 mass % = = 32.7 %
21 Molarity Molarity is (moles solute)/(liters solution) M=mol/L Ex. Calculate the molarity of 3.5 L of solution that contains 0.50 mol hydrogen chloride. 0.50 mol 3.5 L =.14 M
22 Using Molarity as a Conversion Factor Note, if molarity is given in the problem, use it as a conversion factor with the units mol/L A solution is listed as 5.60 M. Write a conversion factor with mol on top. Write a conversion factor with mol on the bottom. 5.60 mol 1 L 5.60 mol
23 Given Molarity, Find Moles Ex. How many moles of sodium chloride are in 400. mL of a 1.45 M solution? 400. mL 1000 mL 1 L = 0.580 mol 1.45 mol 1 L Ex. How many moles of HBr are in 755 mL of a 3.50 M solution? 755. mL 1000 mL 1 L = 2.64 mol 3.50 mol 1 L
24 Given Molarity, Find Grams Ex. How many g of sodium chloride are in 50. mL of a 4.0 M solution? Ex. How many g of I 2 are in 2.00 L of a 0.75 M soln? 50. mL 1000 mL 1 L = 12 g 4.0 mol 1 L 1 mol 58.44 g 2.00 L = 381 g 0.75 mol 1 L 1 mol 253.8 g
25 Given Molarity, Find Volume Ex. How many mL of 5.60 M sucrose soln contain 2.50 mol sucrose? 2.50 mol 5.60 mol 1 L = 446 mL 1000 mL 1 L Ex. How many mL of 12.0 M HCl soln contain 0.330 mol HCl? 0.330 mol 12.0 mol 1 L = 27.5 mL 1000 mL 1 L
26 Colligative Properties When a solute is dissolved in a pure liquid, the freezing point (fp) and boiling point (bp) of the soln will change.pure liquid soln FP depression means that a solute will lower the fp of the soln compared to the pure solvent BP elevation means that a solute will raise the bp of a soln compared to the pure solvent
27 Colligative Properties Colligative Properties are only determined by the number of solute particles in a quantity of solvent Molality m=(mol solute)/(kg solvent) Freezing Point Depression T f = K f m T f is the change in temperature Subtract T f from the freezing pt (for H 2 O, the freezing pt is 0 o C) K f is the molal freezing pt depression constant For H 2 O, K f = 1.86 o C/m
28 Colligative Properties Boiling Point Elevation T b = K b m T b is the change in temperature Add T b to the boiling pt (for H 2 O, the bp is 100 o C) K b is the molal boiling pt elevation constant For H 2 O, K b = 0.52 o C/m
29 Ex. Determine the T f of 3.0 kg of water when 5.0 mol antifreeze is added. T f =K f m = 1.86 o C m 5.0 mol solute 3.0 kg H 2 O = 3.1 o C fp = 0 - T f = - 3.1 o C Calculate the freezing pt of the above soln.
30 Ex. Determine T b of 3.0 kg of water when 5.0 mol antifreeze is added. T b =K b m = 0.52 o C m 5.0 mol solute 3.0 kg H 2 O = 0.87 o C bp = 100 + T b = 100.87 o C Determine the bp of the above soln.
31 Ex. Determine the freezing pt of 1.53 kg of water when 6.2 mol formic acid (HCO 2 H) is added. T f =K f m = 1.86 o C m 6.2 mol HCO 2 H 1.53 kg H 2 O = 7.5 o C fp = -7.5 o C