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Let’s study solutions

Let’s study solutions. Solutions homogeneous mixtures of two or more substances solvent & one or more solutes Solutes spread evenly throughout cannot separate filtration Separated evaporation. n ot visible unless colored. Most common solvent “water”. polar molecule.

hedley-osborn
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Let’s study solutions

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  1. Let’s study solutions Solutions • homogeneous mixtures of two or more substances • solvent & one or more solutes Solutes • spread evenly throughout • cannot separate • filtration • Separated • evaporation. • not visible • unless colored

  2. Most common solvent “water” • polar molecule. • forms hydrogen bonds • hydrogen atom in one molecule - oxygen atom of different water molecule Na+ and Cl- ions • surface of a NaCl crystal attracted • polar water molecules. • hydrated solution • many H2O molecules surrounds

  3. Like dissolves like Two substances form solution • an attraction between • particles of solute and solvent. • polar solvent dissolves polar solutes • water and sugar • water ionic solutes (NaCl) • non-polar solvent hexane (C6H14) • dissolves nonpolar solutes • (oil or grease) Substances in water are: • strong electrolytes produce ions, conduct an electric current • NaCl(s) + H2O Na+(aq) + Cl− (aq) • weak electrolytes produce few ions • HF(g) + H2O(l) H3O+(aq) + F- (aq) • non-electrolytes do not produce ions Water and polar solute

  4. Solubility • maximum amount of solute • dissolves in specific amount of solvent • expressed grams of solute in 100 grams of solvent water g of solute 100 g water Unsaturated solutions • less than maximum amount of solute • dissolve more solute Saturated solutions • contain maximum amount of solute than dissolves • undissolved solute at bottom of container Dissolved solute

  5. Effect of Temperature on Solubility Solubility • Depends on temperature. • Of most solids increases as temperature increases. • Of gases decreases as temperature increases.

  6. How to calculate percent concentration? Solubility amount of solute amount of solution • mass or volume of solute in a solution expressed in “100” percent = amt.solute x 100 amt. solute + amt. solvent • a. mass percent (m/m) = g of solute x 100 100 g of solution b. volume % (v/v) = mL of solute x 100 100 mL of solution c. mass/volume % (m/v) = g of solute x 100 100 mL of solution

  7. Example calculating percentages! calculation of mass percent • grams of solute (g KCl) and • grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% 50.00 g KCl solution

  8. Using percent concentration as conversion factors! How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2 g solution g NaCl STEP 3 Write the 10.0% (m/m) as conversion factors. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4 Set up using the factor that cancels g solution. 225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution

  9. Learning Check A solution is prepared by mixing 15.0 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. A) 15.0% (m/m) Na2CO3 B) 6.38% (m/m) Na2CO3 C) 6.00% (m/m) Na2CO3

  10. Solution C) 6.00% (m/m) Na2CO3 STEP 1mass solute = 15.0 g Na2CO3 mass solution = 15.0 g + 235 g = 250. g STEP 2 Use g solute/ g solution ratio STEP 3mass %(m/m) = g solute x 100 g solution STEP 4 Set up problem mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00% Na2CO3 250. g solution

  11. What is Molarity(M)? Molarity (M) • concentration term for solutions • gives moles of solute in 1 L solution • moles of solute liter of solution M = m v • or m = M x v or v = m M 1.00 M NaCl solution prepared • weigh 58.5 g NaCl (1.00 mole) and • add water to make 1.00 liter of solution

  12. Steps for calculation of molarity What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1Given 6.00 g NaOH in 0.500 L solution Need molarity (mole/L) STEP 2Plan g NaOH mole NaOHmolarity STEP 3Conversion factors 1 mole NaOH = 40.0 g 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Calculate molarity. 6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1 L

  13. Learning Check What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3? A) 0.557 M B) 1.44 M C) 1.71 M

  14. Solution C) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3 84.0 g NaHCO3 0.557 mole NaHCO3 = 1.71 M NaHCO3 0.325 L

  15. Learning Check What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3? A) 0.344 M B) 1.53 M C) 15.5 M

  16. Solution B) 1.53 M 34.8 g KNO3 x 1 mole KNO3 = 0.344 mole KNO3 101.1 g KNO3 M = mole = 0.344 mole KNO3 = 1.53 M L 0.225 L In one setup: 34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L

  17. Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. TABLE 7.8

  18. Calculations Using Molarity How many grams of KCl are needed to prepare 125mL of a 0.720 M KCl solution? STEP 1Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2Plan L KClmoles KCl g KCl

  19. Learning Check How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? A) 20.0 g AlCl3 B) 16.7g AlCl3 C) 2.50 g AlCl3

  20. Solution C) 2.50 g AlCl3 0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3 1 L 1 mole

  21. Learning Check How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3? A) 12.0 mL B) 83.3 mL C) 190. mL

  22. Solution 24.0 g HNO3 x 1 mole HNO3 x 1000 mL = 63.0 g HNO3 2.00 mole HNO3 Molarity factor inverted = 190. mL HNO3

  23. Dilution In a dilution • water is added. • volume increases. • concentration decreases.

  24. Comparing Initial and Diluted Solutions In the initial and diluted solution, • the moles of solute are the same. • the concentrations and volumes are related by the following equations: For percent concentration: C1V1 = C2V2 initial diluted For molarity: M1V1 = M2V2 initial diluted

  25. Dilution Calculations with Percent What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? Prepare a table: C1= 14.0% (m/v) V1 = 25.0 mL C2= 2.00% (m/v) V2 = ? Solve dilution equation for unknown and enter values: C1V1 = C2V2 V2 =V1C1 = (25.0 mL)(14.0%) = 175 mL C2 2.00%

  26. Learning Check What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

  27. Solution What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table: C1= 9.00 %(m/v) V1 = 10.0 mL C2= ? V2 = 60.0 mL Solve dilution equation for unknown and enter values: C1V1 = C2V2 C2 =C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v) V2 60.0 mL

  28. Dilution Calculations with Molarity What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L? Prepare a table: M1= 0.600 M V1 = 0.180 L M2= ? V2 = 0.540 L Solve dilution equation for unknown and enter values: M1V1 = M2V2 M2 =M1V1 = (0.600 M)(0.180 L) = 0.200 M V2 0.540 L

  29. Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? A) 27.0 mL B) 60.0 mL C) 90.0 mL

  30. Solution What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table: M1= 1.80 M V1 = 15.0 mL M2= 0.300MV2 = ? Solve dilution equation for V2 and enter values: M1V1 = M2V2 V2 =M1V1 = (1.80 M)(15.0 mL) = 90.0 mL M2 0.300 M

  31. How colloids and suspension different than solutions? Examples • Fog • Whipped cream • Milk • Cheese • Blood plasma • Pearls Solutions • contain small particles (ions or molecules). • are transparent. • do not separate. • cannot be filtered. • do not scatter light. Colloids • have medium size particles. • cannot be filtered. • can be separated by semipermeable membranes. • scatter light (Tyndall effect). Suspensions • have very large particles. • settle out. • can be filtered. • must be stirred to stay suspended, blood platelets, muddy water

  32. What is osmosis? • water (solvent) flows from lower solute concentration into higher solute concentration. • level of solution with higher concentration rises • concentrations of two solutions become equal with time Osmotic pressure • produced by solute particles • dissolved in solution. • equal to pressure that • prevent flow of additional water • into more concentrated solution. • greater as number of dissolved particles in solution increases. isotonic solution • same osmotic pressure hypotonic solution • has a lower osmotic pressure hypertonic solution • has a higher osmotic pressure

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