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Solutions and Solubility

Solutions and Solubility. Definitions. A solution is a homogeneous mixture A solute is dissolved in a solvent . solute is the substance being dissolved solvent is the liquid in which the solute is dissolved an aqueous solution has water as solvent

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Solutions and Solubility

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  1. Solutions and Solubility

  2. Definitions • A solution is a homogeneous mixture • A solute is dissolved in a solvent. • solute is the substance being dissolved • solvent is the liquid in which the solute is dissolved • an aqueous solution has water as solvent • A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve. • A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve, but crystallizes at the same rate so that there “appears” to be nothing happening.

  3. Substances that make up the solution can be: • Solid- salt dissolved in water • Liquid- HCl dissolved in water • Gas- Carbon Dioxide dissolved in water- pop • Not every liquid dissolves every solid ex. NaCl dissolves in water but not gasoline, parrafin wax dissolves in gasoline but not water. • Liquids are similar- liquids that dissolve other liquids are miscible- ethanol dissolves in water. • - liquids that do not dissolve each other are immiscible- oil and water.

  4. A common phrase that is used to determine which substances dissolve which is “like dissolves like”. Meaning polar substances dissolve polar substances and nonpolar dissolve non polar. • Examples of solutions not in liquid phase: • Air- made of N2, O2 CO2 and others • Metal Alloys- brass- made up of zinc and copper

  5. Reactions involving Precipitates • When two liquids (with dissolved substances) mix and a solid forms it is called a precipitate.

  6. Remember- Net Ionic Equations • When a solution of Pb(NO3)2 is mixed with a solution of KI the result is a precipitate of PbI and a solution of KNO3. • Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) • Remember- dissociation equations • Net Ionic Equation

  7. Solution and Concentration • 4 ways of expressing concentration • Molarity(M):moles solute / Liter solution • Mass percent:(mass solute / mass of solution) * 100 • Molality* (m)- moles solute / Kg solvent • Mole Fraction(A) - moles solute / total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

  8. Making Molar Solutions From Liquids(More accurately, from stock solutions)

  9. Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled) 4.48 - 4.50 4.86 - 4.87 5.00

  10. % Concentration • % (w/w) = • % (w/v) = • % (v/v) =

  11. The Dilution formula E.g. if we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? M1 = 3 mol/L, V1 = 1 L, V2 = 6 L M1V1 = M2V2, M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M Why does the formula work? Because we are equating mol to mol: V1 = 1 L M1 = 3 M V2 = 6 L M2 = 0.5 M M1V1 = 3 mol M2V2 = 3 mol

  12. Practice problems Q – What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2, M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L Q – 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M HCl solution. What is the final concentration of HCl? (hint: first calculate total number of moles and total number of L) # mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L) = 3 mol + 1 mol = 4 mol # L = 1 L + 0.5 L = 1.5 L # mol/L = 4 mol / 1.5 L = 2.67 mol/L Do 1 – 8 . Try 6 two ways

  13. 1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? 2. You have 200 mLof 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? 3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M? 4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl? 5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl? 6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water? 7. Water is added to 4 L of 6 M antifreeze until it is 1.5M. Whatisthe totalvolumeofthenew solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,whatistheconcentrationoftheremaining HF?

  14. Dilution problems (1-6, 6 two ways) 1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = 0.03125 L = 31.25 mL 2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L) M2 = 1.2 M 3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M) V2 = 0.4 L or 400 mL

  15. Dilution problems (4 - 6) 4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 L # mol/L = 2.6 mol / 1 L = 2.6 mol/L 5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol + 0.4 mol = 1.9 mol # mol/L = 1.9 mol / 5 L = 0.38 mol/L 6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/L Or, using M1V1 = M2V2, M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

  16. Dilution problems (7, 8) 7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M) V2 = 16 L 8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.

  17. %m = 3.5 g CoCl2 100g H2O = 3.5% (m/m) % Concentration: % Mass Example 3.5 g of CoCl2 is dissolved in 100mL solution. Assuming the density of the solution is 1.0 g/mL, what is concentration of the solution in % mass?

  18. Concentration: Molarity Example If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4? As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4 Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L . Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M 0.250 L solution

  19. Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ? Concentration 0.500 M Sucrose 250 mL of 0.348 M sucrose Dilution When a solution is diluted, solvent is added to lower its concentration. The amount of solute remains constant before and after the dilution: moles BEFORE = moles AFTER C1V1 = C2V2 A bottle of 0.500 M standard sucrose stock solution is in the lab. Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution.

  20. Factors Affecting Solubility 1. Nature of Solute / Solvent. - Like dissolves like (IMF) 2. Temperature - i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3. Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already close together, extra pressure will not increase solubility. ii) gas - Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent.

  21. Solubilities of Solids vs Temperature Solubilities of several ionic solid as a function of temperature. MOST salts have greater solubility in hot water. A few salts have negative heat of solution, (exothermic process) and they become less soluble with increasing temperature.

  22. Temperature & the Solubility of GasesThe solubility of gases DECREASES at higher temperatures

  23. Soft Drinks • Soft drinks contain “carbonated water” – water with dissolved carbon dioxide gas. • The drinks are bottled with a CO2 pressure greater than 1 atm. • When the bottle is opened, the pressure of CO2 decreases and the solubility of CO2 also decreases • Therefore, bubbles of CO2 escape from solution.

  24. Colligative Properties Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure Colligative Properties are properties of a liquid that change when a solute is added. The magnitude of the change depends on the number of solute particles in the solution, NOT on the identity of the solute particles.

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