1. 1.5 Solutions

2. Assessment Objectives 1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm–3 and mol dm–3). 1.5.2 Solve problems involving concentration, amount of solute and volume of solution.

3. A solution is a homogeneous mixture of 2 or more substances. It is made of a solvent and a solute.

4. SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution SoluteSolventExample solidsolid Metal alloys solidliquid Kool aid liquidliquid Alcoholic drinks gasliquidPepsi gasgasair

5. Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Solution: Homogeneous mixtures of two or more substances. Most common is solid, liquid or gas dissolved in a liquid (usually water). These are called aqueous solutions. Can have other solutions such as solid-solid (alloy) or gas-gas (air). Solute: The dissolved particles. Usually the substance in the least amount. Solvent: The dissolving medium. Usually the substance in the greater amount.

6. Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Properties of Solutions Solubility Solubility is the amount of solute that dissolves in a given amount of solvent at a given temperature to produce a saturated solution. Units: grams solute/100 g solvent NaCl: solubility of 36.2 g/ 100 g water at 25 o C Any amount less than that is an unsaturated solution. A solution that contains more solute than it should theoretically is supersaturated.

7. Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

8. Concentration If something is concentrated it has more moles/grams of solute per amount of solvent. Ex: 12 M HCl If something is dilute it has less moles/grams of solute per amount of solvent Ex: 0.10 M HCl

9. PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

10. Step 1: Change mL to L. 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles Step 3: Convert moles to grams. (0.0125 mol)(90.00 g/mol) = 1.13 g moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

11. IB Note Square brackets [0.60] means 0.60 M or 0.60 mol/L or 0.60 mol dm -3 This is what IB uses! Note: Can Also use g dm -3 (g/dm 3 ) Is the number of grams of solute dissolved per dm 3 of solution

12. Questions Calculate the concentration of the solution formed when 4.00 moles of glucose are dissolved in 5.00 dm -3 If 2.00g of sodium hydroxide is dissolved in 200 cm 3 of water, determine the concentration of the resulting solution. What mass of hydrated copper (II) sulphate crystals is present in 17.3 cm 3 of a 0.279 mol dm -3 solution of copper(II)sulphate.

13. Questions Calculate the volume to which 20cm -3 of 7.63 mol dm -3 hydrochloric acid must be diluted to produce a solution with a concentration of exactly 5.00 mol dm -3

14. Solve problems involving concentration, amount of solute and volume of solution. Find the concentration in g dm -3 and mol dm -3 of a solution containing 2.00 g sodium hydroxide in 125 cm 3 of solution. g dm -3 125 cm 3 = 0.125 dm 3 2.00 g/0.125 dm 3 = 16.0 g dm -3 mol dm -3 2.00 g NaOH = 2.00 g/ 40.01 g mol -1 =.0500 mol.0500 mol/0.125 dm 3 = 0.400 mol dm -3

15. Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in 23.65 cm 3 of 0.100 mol dm -3 HCl(aq)?

16. Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in 23.65 cm 3 of 0.100 mol dm -3 HCl(aq)? Molarity = mol dm -3 so Molarity x dm 3 = mol mol = 0.100 mol dm -3 x.02365 dm -3 =.002365 mol HCl grams =.002365 mol x 36.46 g mol -1 =.0862 g HCl

17. Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ?

18. Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ? Molarity = mol dm -3 so dm -3 = mol/Molarity mol = 28.6 g / 158.04 g mol -1 =.181 mol dm -3 =.181 mol/1.25 mol dm -3 =.145 dm 3

19. Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)?

20. Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)? Find total moles.200 dm -3 x 3.00 mol dm -3 =.600 mol HCl.300 dm -3 x 1.50 mol dm -3 =.450 mol HCl Total moles = 1.050 mol Find total volume:.200 dm 3 +.300 dm 3 =.500 dm 3 Concentration: 1.050 mol/.500 dm 3 = 2.10 mol dm -3

21. Terms to Know Element Compound Empirical formula Molecular formula Structural formula Percent composition Percentage yield Molar mass Mole Avogadro’s constant Chemical equation Molar ratio Ionic equation Solid Liquid Gas Ideal gas equation Molar volume of a gas Solute Solvent Solution Concentration