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Solution Chemistry (Chp. 7)

Solution Chemistry (Chp. 7). Chemistry 2202. Topics. Definitions/properties Molar Concentration (mol/L) Dilutions % Concentration (pp. 255 – 263) Solution Process Solution Preparation Solution Stoichiometry Dissociation. T erms. molar solubility saturated unsaturated supersaturated

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Solution Chemistry (Chp. 7)

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  1. Solution Chemistry (Chp. 7) Chemistry 2202

  2. Topics • Definitions/properties • Molar Concentration (mol/L) • Dilutions • % Concentration (pp. 255 – 263) • Solution Process • Solution Preparation • Solution Stoichiometry • Dissociation

  3. Terms molar solubility saturated unsaturated supersaturated dissociation electrolyte non-electrolyte filtrate precipitate limiting reagent excess reagent actual yield theoretical yield decanting pipetting dynamic equilibrium solution solvent solute concentrated dilute aqueous miscible Immiscible alloy solubility x

  4. Define the terms in bold and italics from pp. 237 – 240. • Solids, liquids, and gases can combine to produce 9 different types of solution. Give an example of each type. • p. 242 #’s 5, 7, 9, & 10

  5. Terms alloy solubility molar solubility saturated unsaturated supersaturated dynamic equilibrium solution solvent solute concentrated dilute aqueous miscible immiscible x

  6. Factors Affecting Solubility (pp.243 – 254) • List 3 factors that affect the rate of dissolving. • How does each of the following affect solubility? • particle size • temperature • pressure

  7. Factors Affecting Solubility • What type of solvent will dissolve: • polar solutes • nonpolar solutes • ionic solutes • Why do some ionic compounds have low solubility in water? p. 254 #’s 1, 2, 4 - 6

  8. Section 7.2 (pp. 243 – 252) • State the generalizations regarding solubility and solutions (in italics) • Define terms (in bold) ion-dipole attractions electrolyte hydrated non-electrolyte

  9. Rate of Dissolving • for most solids, the rate of dissolving is greater at higher temperatures • stirring a mixture or by shaking the container increases the rate of dissolving. • decreasing the size of the particles increases the rate of dissolving.

  10. “Like Dissolves Like” • ionic solutes and polar covalent solutes both dissolve in polar solvents • non-polar solutes dissolve in non-polar solvents.

  11. Solubility • small molecules are often more soluble than larger molecules. • the solubility of most solids increases with temperature while the solubility of most liquids is not affected by temperature. • the solubility of gases decreases as temperature increases • an increase in pressure increases the solubility of a gas in a liquid.

  12. Applications 1. An opened soft drink goes ‘flat’ faster if not refrigerated. 2. Warming of pond water may not be healthy for the fish living in it. 3. After pouring 5 glasses of pop from a 2 litre container, Jonny stoppered the bottle and crushed it to prevent the remaining pop from going flat.

  13. Molar Concentration Review: - Find the molar mass of Ca(OH)2 - How many moles in 45.67 g of Ca(OH)2? - Find the mass of 0.987 mol of Ca(OH)2.

  14. Molar Concentration The terms concentrated and dilute are qualitative descriptions of solubility. A quantitative measure of solubility uses numbers to describe the concentration of a solution.

  15. Molar Concentration The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).

  16. Molar Concentration FORMULA: Molar Concentration = number of moles volume in litres C = n V

  17. eg. Calculate the molar concentration of: • 4.65 mol of NaOH is dissolved to prepare 2.83 L of solution. • 15.50 g of NaOH is dissolved to prepare 475 mL of solution. p. 268 - # 19

  18. Eg. Calculate the following: • the number of moles in 4.68 L of 0.100 mol/L KCl solution. • the mass of KCl in 268 mL of 2.50 mol/L KCl solution.

  19. the volume of 6.00 mol/L HCl(aq) that can be made using 0.500 mol of HCl. the volume of 1.60 mol/L HCl(aq) that can be made using 20.0 g of HCl. p. 268 #’s20-24

  20. Number of moles before dilution Dilution (p. 272) When a solution is diluted: • The concentration decreases • The volume increases • The number of moles remains the same ni = nf Number of moles after dilution

  21. Dilution (p. 272) ni = nf Ci Vi = Cf Vf eg. Calculate the molar concentration of a vinegar solution prepared by diluting 10.0 mL of a 17.4 mol/L solution to a final volume of 3.50 L.

  22. p. 273 #’s 25 – 27 p. 276 #’s 1, 2, 4, & 5 DON’T SHOW UP UNLESS THIS IS DONE!!

  23. Solution Preparation & Dilution • standard solution – a solution of known concentration • volumetric flask – a flat-bottomed glass vessel used to prepare a standard solution • delivery pipet – pipets that accurately measure one volume • graduated pipet – pipets that have a series of lines that can be use to measure many different volumes

  24. To prepare a standard solution: 1. calculate the mass of solute needed 2. weigh out the desired mass 3. dissolve the solute in a beaker using less than the desired volume 4. transfer the solution to a volumetric flask (rinse the beaker into the flask) 5. add water until the bottom of the meniscus is at the etched line

  25. To dilute a standard solution: • 1. Rinse the pipet several times with deionized water. • 2. Rinse the pipet twice with the standard solution. • 3. Use the pipet to transfer the required volume. • 4. Add enough water to bring the solution to its final volume.

  26. Percent Concentration • Concentration may also be given as a %. • The amount of solute is a percentage of the total volume/mass of solution. • liquids in liquids - % v/v • solids in liquids - % m/v • solids in solids - % m/m

  27. Percent Concentration p. 258 #’s 1 – 3 DSUUTID!!

  28. p. 261 #’s 5 – 9 DSUUTID!!

  29. p. 263 #’s 10 – 13 DSUUTID!!

  30. Concentration in ppm and ppb Parts per million (ppm) and parts per billion (ppb) are used for extremely small concentrations

  31. eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in: a) ppm b) ppb

  32. ppm = 0.005 g x 106 100,000 g = 0.05 ppm ppb = 0.005 g x 109 100,000 g = 50.0 ppb

  33. p. 265 #’s 15 – 17 • pp. 277, 278 #’s 11, 13, 15 – 18, 20 DON’T SHOW UP UNLESS THIS IS DONE!!

  34. Solution Stoichiometry 1. Write a balanced equation 2. Calculate moles given ORn=CV 3. Mole ratio 4. Calculate required quantity

  35. Solution Stoichiometry eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L Ca(OH)2 solution. Calculate the molar concentration of the HCl(aq) solution. p. 304: #’s 16, 17, & 18 Worksheet

  36. Sample Problems • What mass of copper metal is needed to react with 250.0 mL of 0.100 mol/L silver nitrate solution? (0.794 g Cu) Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq) Step 2 n = 0.02500 mol AgNO3 Step 3 n = 0.01250 mol Cu Step 4 m = 0.794 g Cu

  37. 2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH. (0.0150 L HCl) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Step 2 n = 0.0300 mol NaOH Step 3 n = 0.0300 mol HCl Step 4 V = 0.0150 L HCl

  38. 3. What volume of 3.00 mol/L HNO3(aq) is needed to neutralize 450.0 mL of 0.100 mol/L Sr(OH)2(aq)? (0.0300 L HNO3) 2 HNO3(aq) + Sr(OH)2(aq) → 2 H2O(l) + Sr(NO3)2(aq) Step 2 n = 0.04500 mol Sr(OH)2 Step 3 n = 0.0900 mol HNO3 Step 4 V = 0.0300 L HNO3

  39. The Solution Process (p. 299) • Dissociationoccurs when an ionic compound breaks into ions as it dissolves in water. • A dissociation equation shows what happens to an ionic compound in water. eg. NaCl(s) → Na+(aq) + Cl-(aq) K2SO4(s) → 2 K+(aq) + SO42-(aq)

  40. The Solution Process (p. 299) • Solutions of ionic compounds conduct electric current. • A solute that conducts an electric current in an aqueous solution is called an electrolyte.

  41. The Solution Process (p. 299) • Acids are also electrolytes. 6 strong acids: perchloric acid - HClO4(aq) hydrochloric acid – HCl(aq) hydroiodic acid - HI(aq) hydrobromic acid – HBr(aq) nitric acid – HNO3(aq) sulfuric acid – H2SO4(aq) • eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq) HCl(s) → H+(aq) + Cl-(aq) All other acids are weak electrolytes (poor conductors)

  42. The Solution Process (p. 299) eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq) HCl(s) → H+(aq) + Cl-(aq) • Molecular Compounds DO NOT dissociate in water. eg. C12H22O11(s) → C12H22O11(aq)

  43. The Solution Process (p. 299) • Because they DO NOT conduct electric current in solution, molecular compounds are non-electrolytes. • Low solubility compounds are weak electrolytes due to their low solubility in water. eg. AgBr BaSO4 PbI2

  44. The Solution Process (p. 299) The molar concentration of any dissolved ion is calculated using the ratio from the dissociation equation. eq. What is the molar concentration of each ion in a 5.00 mol/L MgCl2(aq) solution: 5.00 mol/L 5.00 mol/L 10.00 mol/L

  45. The Solution Process (p. 299) eg. Calculate the concentration of Al(NO3)3 in a solution with a nitrate ion concentration equal to 0.300 mol/L.

  46. The Solution Process (p. 299) eg. Calculate the molar concentration of chloride ions in a solution prepared by dissolving 10.0 g of FeCl3 to make 100.0 mL of solution

  47. The Solution Process (p. 299) eg. Calculate the mass of Na2SO4 needed to prepare 400.0 mL of solution with a sodium ion concentration equal to 0.350 mol/L.

  48. eg. What mass of calcium chloride is required to prepare 2.00 L of 0.120 mol/L Cl-(aq) solution?

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