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Unit 4 Solutions

Unit 4 Solutions. 4.3.1 Solubility curves 4.3.2 solubility tables 4.3.3 solubility constant 4.3.4 precipitation 4.3.5 common ion effect. 4.3.1 Solubility Curves. Solubility curves tell us what mass of solute will dissolve in 100g (or 100mL) of water over a range of temperatures.

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Unit 4 Solutions

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  1. Unit 4 Solutions 4.3.1 Solubility curves 4.3.2 solubility tables 4.3.3 solubility constant 4.3.4 precipitation 4.3.5 common ion effect

  2. 4.3.1 Solubility Curves Solubility curvestell us what mass of solute will dissolve in 100g (or 100mL) of water over a range of temperatures. For most substances, solubility increases as temperature increases. Can you find any exceptions on the chart?

  3. Here's an example of reading the chart. Find the curve for KClO3. At 30°C approximately 10g of KClO3 will dissolve in 100g of water. If the temperature is increased to 80°C, approximately 40g of the substance will dissolve in 100g of water.

  4. Here are some for you to try. What mass of solute will dissolve in 100mL of water at the following temperatures. Determine which of the three substances is most soluble in water at 15°C. 1. KNO3at 70°C   2. NaCl at 100°C   3. NH4Cl at 90°C  

  5. Solution: Solubility 1. KNO3at 70°C 140g / 100mL 2. NaCl at 100°C 40g / 100mL 3. NH4Cl at 90°C 72g / 100mL 4. NaCl is the most soluble at 15°C

  6. The curves indicate the concentration of a saturated solution– (the maximum amount of solute that will dissolve at that specific temperature). The molar concentration of the substance can be calculated, example: Determine the molarity of a saturated NaCl solution at 25°C.

  7. Solution We can see from the curve that about 38 g of NaCl dissolves in 100mL at 25°C. We want to convert 38g /100mL-1to mol·L-1. First let's convert mL to L: Next we need to convert from grams to moles, we will need to use the molar mass of NaCl, which is 58.5 g·mol-1

  8. Solubility Chart Values below thecurve represent unsaturated solutions – i.e more solute could be dissolved at that temperature. Values above a curve represent supersaturated solutions, i.e a solution which holds more solute that can normally dissolve in that volume.

  9. Some examples: What term - saturated, unsaturated, or supersaturated - best describes: a solution that contains 70g of NaNO3 per 100 mL H2O at 30°C a solution that contains 60g of dissolved KCl per 100 mL H2O at 80°C

  10. Solution The NaNO3 solution is unsaturated. At 30°C a saturated solution would be able to dissolve approximately 95 g of NaNO3 . Since there are only 70g in the solution, 25 more grams of NaNO3 could dissolve. The KCl solution is supersaturated. At 80°C a saturated KCl solution contains 50 g KCl per 100 mL H2O. This solution is holding 10 g of excess KCl.

  11. 3.2 Solubility Tables & the Solubility of Ionic Compounds Solubility tables allow us to predict whether a certain ionic compound will dissolve in water at 25°C. Reading the table is not difficult. It tells which positive cationsare soluble when combined with negative anions

  12. Soluble or not??? NaCl, CH3COO- - BaSO4 Cu(NO3)2 Pb(CH3COO)2 PbCl2

  13. Saturated Solutions & Equilibrium It is important to remember that saturated systems are equilibrium systems. For example, the equation illustrating a saturated solution of silver sulfate would be: Ag2SO4 (s) 2Ag+(aq) + SO42-(aq) remember that equilibrium means "equal rates" not "equal amounts". in a saturated solution at equilibrium you will not be able to detect any changes in concentration because the solid dissolves into ions at the same rate that the ions form back into a solid (crystallization). Equilibrium does NOT mean that the concentrations of reactants and products are equal.

  14. Practice problems 4.3.2

  15. 4.3.3 The Solubility Product Constant, Ksp Recall the equilibrium constant expression for any general reaction: aA + bBcD + dD Keq= [C]c × [D]d [A]a × [B]b

  16. We can apply this mathematical relationship to solutions. We will refer to our equilibrium constant as Ksp, where "sp" stands for "solubility product" For our silver sulfate saturated solution: Ag2SO4 (s) 2Ag+(aq) + SO42-(aq) we can write our solubility product constant expression as Ksp= [Ag+]2[SO42-] [Ag2SO4]

  17. Remember that the concentrations of solids and liquids are NOT included in the expression because their concentrationsremain constant. Therefore, we will write our solubility product constant expression for our saturated silver sulfate solution as: Ksp= [Ag+]2[SO42-] Please note that it is VERY IMPORTANT to include the ion charges when writing this equation

  18. Try this example: Write the expression for the solubility product constant, Ksp, for Ca3(PO4)2. Step 1: balance equation. Ca3(PO4)2 (s) 3 Ca2+(aq) + 2 PO43-(aq) Step 2: Write the expression for Ksp: Ksp= [Ca2+]3[PO43-]2

  19. Solubility Product Tablesgive Ksp values for various ionic compounds (keep it handy). Because temperature affects solubility, values are given for specific temperatures (usually 25°C).

  20. Ksp provides us with useful information: A low value of Ksp indicates a substance that has a low solubility (or insoluble); for ionic compounds this means that there will be few ions in solution. Iron(II) sulfide, FeS, is an example of a low Ksp : Ksp= 4 ×10-19 In a saturated solution of FeS there would be few Fe2+ or S2- ions.

  21. A large value of Ksp indicates a soluble substance; for ionic compounds there will be many ions in solution. An example lead(II) chloride, PbCl2Ksp = 1.8 ×10-4 A saturated solution of PbCl2 would have a high concentration of Pb2+ and Cl- ions.

  22. 1.Calculating Ksp when concentration of a saturated solution is known The concentration of a saturated solution of BaSO4 is 3.90 × 10-5 M. Calculate Ksp for barium sulfate at 25°C  

  23. Solution Write a balanced equation: BaSO4 (s) Ba2+(aq) + SO42- (aq) Next, write the Ksp expression:     Ksp= [Ba2+][SO42-] Find the concentration of the individual ions for our equation. since 1 mole of BaSO4 produces 1 mole of Ba2+ and also 1 mole of SO42-, then . . . [BaSO4] = 3.90 × 10-5M, [Ba2+] = 3.90 × 10-5M, [SO42-] = 3.90 × 10-5M Substitute values into the Ksp expression and solve Ksp= [Ba2+][SO42-] = (3.90 × 10-5)(3.90 × 10-5) Ksp = 1.52 × 10-9 Answer

  24. Example The concentration of lead ions in a saturated solution of PbI2 at 25°C is 1.3 × 10-3 M. What is its Ksp?  

  25. Solution 1. Write a balanced equation: PbI2 (s) Pb2+(aq) + 2 I-(aq) 2. Write the Ksp expression:     Ksp= [Pb2+][I-]2 3. Determine the concentration of the ions. [PbI2] = 1.30 × 10-3M; [Pb2+] = 1.30 × 10-3M; [I-] = 2 × 1.30 × 10-3 = 2.60 × 10-3 4. Solve the Kspexpression: Ksp= [Pb2+][I-]2 = (1.30 × 10-3)(2.60 × 10-3)2 = (1.30 × 10-3)(6.76× 10-6) Ksp= 8.79 × 10-9M

  26. Calculating ion concentrations when Ksp is known. Ksp for MgCO3 at 25°C is 2.0 × 10-8. What are the ion concentrations in a saturated solution at this temperature?  

  27. Solution 1. Balance equation: MgCO3(s) Mg2+(aq) + CO32-(aq) 2. Write the Ksp expression:  Ksp= [Mg2+][CO32-] 3. Let the unknown ion concentrations equal x. The balanced equation tells us that both Mg2+ and CO32- will have the same concentration! 4. Substitute values into the equation and solve for the unknown:     Ksp= [Mg2+][CO32-] 2.0 × 10-8 = (x) (x) = x2 X2 = 2.0 × 10-8 X = √2.0 × 10-8 = 1.4× 10-4M So, x = [Mg2+] = 1.4× 10-4M and  x = [CO32-] = 1.4× 10-4M

  28. 4.3.4 Precipitation Reactions Sometimes when two aqueous solutions are mixed together a solid is produced. This solid is called a precipitate, and the reaction is known as a precipitation reaction. We can use the Solubility Table to predict whether a precipitate will form.

  29. Example ZnBr2 (aq) + 2 AgNO3 (aq) → Zn(NO3)2 + 2 AgBr But if we look at our solubility table, we see that Ag+ and Br- forms an insoluble compound and form a solid. On the other hand, we see in our solubility table that Zn2+ and NO3- form a soluble compound. Thus these two ions will remain in solution We now need to return to our double displacement reaction and add the physical states to the two products: ZnBr2 (aq) + 2 AgNO3 (aq) → Zn(NO3)2 (aq) + 2 AgBr(s)

  30. Let’s write our solutions out in long form, remember - anything that is (aq) is really present as separate ions. This, however, ONLY applies to aqueous states. Thus we can rewrite our equation as: Zn2+(aq) + 2 Br-(aq) + 2 Ag+(aq) + 2 NO3-(aq) →    Zn2+(aq) + 2 NO3-(aq)+ 2 AgBr(s) Notice that some ions are the same on both sides of the equation, indicating they did not change during the reaction. Ions that are present in a reaction but do not participate are called spectator ions.

  31. We can remove spectator ions from equations to highlight what changes occurred during a reaction. Reactions that the spectator ions have been removed are called net ionic equations. Therefore, for the overall equation: Zn2+(aq) + 2 Br-(aq) + 2 Ag+(aq) + 2 NO3-(aq) → Zn2+(aq) + 2 NO3-(aq)+ 2 AgBr(s) The net ionic equation is: 2 Ag+(aq) + 2 Br-(aq) → 2 AgBr(s)

  32. Keep in mind that a reaction does not always take place. Let's look at what happens when we mix solutions of Mg(CH3COO)2 and (NH4)2SO4. Write a balanced equation and predict if the products. Then predict the physical states of the products (reactants are aq.) Mg(CH3COO)2 (aq) + (NH4)2SO4 (aq)→ MgSO4 + 2 NH4CH3COO We find that both MgSO4 and NH4CH3COO are soluble compounds, meaning they do notform precipitates. Thus, there is NO REACTION occurring, all participants simply remain in solution!

  33. Lab 4.3.4

  34. Selective Precipitation Knowledge of precipitation reactions also allows us to use selective precipitation to separate mixtures of ions. For example . . . We have a solution that contains the following cations: Ag+, Cu2+, and Mg2+; each of these cationsis associated with nitrate ions, NO3-. Compounds containing nitrate are ALWAYS soluble. So the nitrate ions will remain spectator ions, so we will not need to consider them We wish to isolate each ion by causing each one to precipitate out of solution. Once a solid forms, we can filter the solid precipitate out, leaving the other ions in solution. We are given the following solutions to use: Na2S, NaCl, and NaOH.

  35. Sodium ions, Na+, always form soluble compounds. So Na+ will remain as a spectator ion . So we are given solutions of the following anions: S2-, Cl-, and OH- We add these one at a time to our cation mixture. We need to determine the order so that we have ONE AND ONLY ONE precipitate occur each addition, removing one of the cations from the solution. Then continue until all cations are isolated. Prepare a chart, with the cations we need to separate along one axis and anions along the other axis:

  36. Use a table of solubility to determine which combinations produce soluble compounds and which ones form a precipitate (ppt- precipitate).

  37. We need to determine the correct order for adding the negative ions Only by adding Cl- first can we separate out one of the ions, Ag+. Our first precipitate would be AgCl(s).Note that both S2- and OH- also form a precipitate with Ag+, but it has been removed using Cl-- Now add S2-to again form a single precipitate - CuS(s), thus removing Cu2+ This will leave only Mg2+ still in solution. Sometimes this last cation is allowed to remain in solution,if add OH-we get our final precipitate, Mg(OH)2 (s) To summarize: First add NaCl to remove Ag+ , forming the precipitate AgCl(s) Second, add Na2S to remove Cu2+, forming CuS(s) Last add NaOH to remove Mg2+, forming Mg(OH)2 (s)

  38. Practice problems 4.3.4 Assignment 4.3.4

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