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Solutions

Chemistry I2 – Chapters 7 & 8. Solutions. Some Definitions. A solution is a homogeneous mixture of 2 or more substances . One constituent is usually regarded as the SOLVENT and the others as SOLUTES . Parts of a Solution.

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Solutions

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  1. Chemistry I2 – Chapters 7 & 8 Solutions

  2. Some Definitions A solution is a homogeneous mixture of 2 or more substances . One constituent is usually regarded as the SOLVENTand the others as SOLUTES.

  3. Parts of a Solution • 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

  4. Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

  5. Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: • Warm the solvent so that it will dissolve more, then cool the solution • Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

  6. Miscibility • Miscible: compounds that dissolve readily in each other in any proportion. Ex: water and alcohol. • Immiscible: Liquids that do not readily dissolve in each other Ex: oil and water

  7. Solubility • Solubility is relative, most substances will dissolve by extremely small amounts. • Soluble: the solute will dissolve in a particular solvent greater than 1 g / 100 mL • Insoluble: the solute will dissolve only 0.1 g / 100 mL, or not dissolve in the solvent

  8. K+(aq) + MnO4-(aq) IONIC COMPOUNDSCompounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO4 in water

  9. Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions _________________________ They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  10. Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol

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

  12. 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

  13. PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O] = 0.0841 M

  14. USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? 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 = M•V

  15. Practice For homework: • Pg 242 # 1-6, 9 • Pg 268 # 19 & 20 (Read pg 266-268)

  16. Solutions (continued) Day 2 Mrs. Kay

  17. Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1) 12 g 2) 48 g 3) 300 g

  18. Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this!

  19. mol solute m of solution = kilograms solvent Two Other Concentration Units MOLALITY, m % by mass grams solute grams solution % by mass =

  20. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol.

  21. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by mass). Calculate molality Calculate weight %

  22. Learning Check A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% Na2CO3 2) 6.4% Na2CO3 3) 6.0% Na2CO3

  23. Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

  24. Try this molality problem • 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 molal salt water

  25. Preparing Solutions • Weigh out a solid solute and dissolve in a given quantity of solvent. • Dilute a concentrated solution to give one that is less concentrated.

  26. Practice for homework: • Read pg 255-263 • Do Pg 261 #5-8 (%m/m) • Do Pg 268 # 19-20 (molarity)

  27. Changing Concentration: • Our calculations for this unit are based on solutions being carefully made with known concentrations. • Two ways to have a known concentration: • Dissolving a measured mass of pure solute in a certain volume of solution. (We practiced this as molarity questions) • Diluting a solution of known concentration to the new concentration you’re trying to achieve.

  28. Diluting a Standard solution: • Use the formula: C1v1 = C2v2 You will be given 3 of the 4 variables and need to solve for the unknown. C1 = the original concentrated solution v1= the amount of original concentrated solution used. C2= the new diluted concentration of solution v2= the new volume of diluted solution

  29. Practice: • I have 2.0 L of 0.10 M of sulfuric acid. This usually sold as an 18 M concentrated solution. How much of the concentrated solution do I need to make my new solution? YOU KNOW: C1=18 M; C2=0.10 M; v2= 2.0 L So, v1 = C2v2 C1 v1 = (0.10 M)(2.0 L) (18 M) v1 = 0.011 L or 11 mL

  30. More Practice: • Page 273 # 25-27 • Finish the Handout called “Molarity Review ” # 5-9

  31. CHEM 11ASolution Stoichiometry You can solve for amounts of product made or reactants needed even though they are stated as mol/L rather than g or moles. You need to use the basics that you learned in stoichiometry from grade 11 chemistry and apply it to what you’ve learned about solutions Let’s work through an example

  32. How many grams of Ca(NO3)2 can be prepared by reacting 75 mL of 2.6 mol/L HNO3 with an excess of Ca(OH)2? 2 HNO3(l) + Ca(OH)2(s) Ca(NO3)2(aq) + 2H2O(l) • What are you given? V HNO3= 0.075L and M=2.6 mol/L • Find the moles of HNO3 used: 2.6 mol/L x 0.075 L = 0.195 mol • Use stoichiometry to find the ratio of acid moles to calcium nitrate. 0.195 mol (acid) x 1 mol (calcium nitrate) = 0.0975 mol 2 mol (acid)

  33. 2 HNO3 + Ca(OH)2 Ca(NO3)2 + 2H2O 4. 0.0975 mol calcium nitrate needs to be converted to mass. 0.0975 mol Ca(NO3)2 X 164.04 g = 15.99 g mol Ca(NO3)2 So, we will make 15.99 g of calcium nitrate during this reaction.

  34. Practice: • Handout # 56-60

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