Solutions and Molarity

1. Solutions and Molarity Describe the types of solutions. Define the vocabulary words. List and explain the factors that influence solubility and the rate at which a solute dissolves in a solvent. Explain and calculate molarity.

2. Types of Solutions • Gas in a gas – Gases mix freely and will always form a solution unless they react (Example: air) • Solid in a solid – Alloy: a homogeneous mixture of metals (Example: brass is a mixture of copper and zinc) • Liquid in a liquid Miscible – when liquids can be mixed together to form a solution (Example: ethylene glycol and water form antifreeze) Immiscible – when liquids cannot be mixed (Example: oil and water) “Like dissolves like” – polar dissolves polar Polar and nonpolar (vinegar and oil) are immiscible

3. Solid in a liquid Energy is required to separate particles of a solid (Endothermic) Energy is released when solute particles and solvent particles are attached (Exothermic) Goes back and forth: Dynamic equilibrium Differences are responsible for different solubilities

4. Gases in liquids – when gas is attracted to solvent particles---release energy Free---move toward entropy Examples: CO2 in soda and O2 in seawater • Entropy (S) – disorder or randomness Systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy)

5. Solubility Curves

6. Refer to Solubility Curves • At 10 °C, how many grams of KI will dissolve • At 50 °C, how many grams of KNO3 will dissolve • At 30 °C, how many grams of NH4Cl will dissolve • At 75 °C, how many grams of KCl will dissolve

7. Factors Influencing the Rate at which a Solute Dissolves in a Solvent • 1. Agitation – stirring; a surface phenomenon • 2. Particle size • 3. Temperature – the only factor that affects both the rate of solution and the solubility

8. Temperature Effects on Solubility • In general, an increase in temperature increases the solubility of solids in liquids • In general, an increase in temperature decreases the solubility of gases in liquids (gases escape)

9. Pressure Effects • Pressure has little effect on the solubility of solids in liquids • Pressure increases the solubility of gases in liquids (nail being hammered into wood) • Henry’s Law – At a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid

10. Concentration of Solutions • Dilute – contains relatively little solute • Concentrated – contains relatively large amount of solute

11. Vocabulary Words • Solubility - the amount of a substance that dissolves in a given quantity of solvent at specified conditions of temperature and pressure to produce a saturated solution • Solute – dissolving particles • Solvent – the dissolving medium in a solution (Example: water) • Solvation – a process that occurs when an ionic solute dissolves; in solution, the ions are surrounded by solvent particles

12. Saturated – a solution that holds as much solid as it normally can at a given temperature (If more solid is added, it will not dissolve) • Unsaturated – a solution which has not yet reached the limit of solubility at a given temperature • Supersaturated – rare solution that contains MORE dissolved solute than it can normally hold at a given temperature (Crystallization)

13. Objectives • Be able to calculate the concentration of a solution (molarity). • Be able to determine the amount of solute (moles) in a given amount of solution. • Be able to determine the amount of solution (volume) that contains a given amount of solute.

14. Molarity • The number of moles of solute dissolved per liter (1000mL) of solution • Water – 1 mL = 1 g • M= moles of solute ÷ liters of solution • Moles of solute = liters of solution x molarity • Problem: What is the molarity of the solution obtained by dissolving 90g of glucose (C6H12O6) in 1000 grams of water?

15. Answer • 1 mole of glucose = 180 g • 90g ÷ 180 g = .5 mole • 1000 g = 1000 mL • .5M per 1000 mL water • Problems: How many grams are needed to make a molar solution of a. 1M glucose b. 2M glucose c. .5M glucose

16. Answer • 1 mole of glucose = 180 grams • a. 1M = 180 grams • b. 2M = 2 x 180 g = 360 grams • c. .5M = .5 x 180 g = 90 grams

17. To Determine Molarity • 1. List your givens and unknowns. • 2. Calculate how many moles. • 3. Calculate moles per liter. Problem: You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

18. Answer • 90g/58.44g = 1.54 mol NaCl • 1.54/3.5 = 0.440 M NaCl • Do Practice Problem 1 on page 421.

19. To Determine Moles • 1. List the givens and unknowns. • 2. Multiply the concentration (molarity) by the volume. • Problem: You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

20. Answer • .8 x .5 = 0.4 mol HCl • Do practice problem 2 on page 421.

21. To Determine the Volume of Solution • 1. List the givens and unknowns. • 2. Calculate the moles of solute. • 3. Divide the moles of solute by the concentration (molarity). • Problem: To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate in solution as a reactant. All you have on hand is 5 L of a 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction?

22. Answer • 23.4/194.2 = 0.120 mol • .12/6 = 0.020 L • Do practice problem 3 on page 421.

23. Preparing Molar Solutions • The solute will take up some of the available space in the volumetric flask. • Steps 1. The solute should be added to some of the solvent and dissolved. 2. Then solvent is added to the 1L mark on the volumetric flask. • If these steps are not followed, the total volume of the mixture is likely to exceed the desired volume. • A volumetric pipet measures volumes even more accurately.

24. Making Dilutions • Dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change. • Moles of solute = molarity (M) x liters of solution (V) • Moles of solute = M1 x V1 = M2 x V2 • Problem: How many milliliters of a stock solution of 2.00M MgSO4 would you need to prepare 100.0 mL of 0.400 M MgSO4?

25. Answer • 0.400M x 100.0 mL ÷ 2.00M = 20.0 mL • Thus, 20.0 mL of the initial solution must be diluted by adding enough water to raise the volume to 100.0 mL • OR 0.400M is 1/5th of 2.00M 1/5th of 100mL is 20 mL

26. Percent Solutions • If both the solute and solvent are liquids, a convenient way to make a solution is to measure volumes. Example: 20 mL of pure alcohol is diluted with water to a total volume of 100 mL – The concentration of alcohol is 20% (v/v) • A commonly used relationship for solutions of solids dissolved in liquids is percent (mass/volume).

27. Problems • 1. What is the percent by volume of ethanol (C2H6O), or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? • 2. How many grams of glucose (C6H12O6) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution?

28. Answers • 1. 85/250 x 100% = 34% ethanol (v/v) • 2. In a 2.8% solution, each 100 mL of solution contains 2.8 grams of glucose 100mL is 1/10th of a liter, so you need 28 grams per liter 2 L x 28 g = 56 grams of glucose