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The End of Equilibrium!

The End of Equilibrium!

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The End of Equilibrium!

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  1. The End of Equilibrium! (well, for us!) TEXT 692019 AND YOUR QUESTION TO 37607

  2. Ksp What is the solubility of FeCO3? Solubility = MAXIMUM amount of a compound that can dissolve in water. This is actually an equilibrium. TEXT 692019 AND YOUR QUESTION TO 37607

  3. Equilibrium problems involve 3 parts: • Balanced equation • “K-equation” • ICE chart What is the balanced equation for dissolving something? TEXT 692019 AND YOUR QUESTION TO 37607

  4. FeCO3 (s) Fe2+(aq)+ CO32-(aq) What is the “K-equation”? K = [Fe2+][CO32-] The “K” is the PRODUCT of the SOLUBLE ions. Hence, this reaction is called a “solubility product”. Ksp = [Fe2+][CO32-] Ksp(FeCO3) = 3.07x10-11 TEXT 692019 AND YOUR QUESTION TO 37607

  5. What is the solubility of FeCO3? Ksp(FeCO3) = 3.07x10-11 FeCO3 (s) Fe2+(aq)+ CO32-(aq) I S 0 0 C -x +x +x E 0.000000000001 x x Ksp = 3.07x10-11 = [Fe2+][CO32-] = x*x x = SQRT(3.07x10-11) = 5.54x10-6 M TEXT 692019 AND YOUR QUESTION TO 37607

  6. Clicker question What is the solubility of Ba3(PO4)2 at 298K? Ksp(Ba3(PO4)2) = 6x10-39 A. 8x10-20 M B. 2x10-8 M C. 3x10-9 M D. 9x10-9 M E. 3x10-20 M TEXT 692019 AND YOUR QUESTION TO 37607

  7. Ba3(PO4)2 (s) 3 Ba2+ (aq) + 2 PO43-(aq) I S 0 0 C -x +3x +2x E - 3x 2x Ksp = 6x10-39 = (3x)3(2x)2 = 27x3*4x2 5.56x10-41 = x5 x = 8.89x10-9 M = 9x10-9 M TEXT 692019 AND YOUR QUESTION TO 37607

  8. More common units for solubility… …are g/L. If you wanted g/L TEXT 692019 AND YOUR QUESTION TO 37607

  9. Precipitation Reaction The reverse reaction: Solubility Product: Ba3(PO4)2 (s) 3 Ba2+ (aq) + 2 PO43-(aq) Precipitation: 3 Ba2+ (aq) + 2 PO43-(aq)  Ba3(PO4)2 (s) It’s just K “upside down” TEXT 692019 AND YOUR QUESTION TO 37607

  10. How do you know if something precipitates? Ba3(PO4)2 (s) 3 Ba2+ (aq) + 2 PO43-(aq) Ksp = 6x10-39 What is the Ksp? It’s the limit on the amount of ions in solution. Ksp = [Ba2+]3[PO43-]2 Remember our old friend “Q”? TEXT 692019 AND YOUR QUESTION TO 37607

  11. What’s Q? Q is just the concentrations of products and reactants when you are NOT at equilibrium. Ksp = [Ba2+]3[PO43-]2 = 6x10-39 Q = [Ba2+]3[PO43-]2 = any other number TEXT 692019 AND YOUR QUESTION TO 37607

  12. Q is less than K means… • You are NOT at equilibrium. • You could dissolve more solid: the products (dissolved ions) are too small. TEXT 692019 AND YOUR QUESTION TO 37607

  13. Q is more than K means… • You are NOT at equilibrium. • You have TOO MANY products (dissolved ions). They can’t stay dissolved, they need to precipitate out! TEXT 692019 AND YOUR QUESTION TO 37607

  14. A little precipitation question: 500 mL of 0.100 M Fe(NO3)3 is mixed with 250 mL of 0.100 M KOH. What, if anything, precipitates from the solution? What mass of precipitate is formed? TEXT 692019 AND YOUR QUESTION TO 37607

  15. What COULD form…? KOH(s)  K+ (aq) + OH-(aq) Fe(NO3)3(s) Fe3+ (aq) + 3 NO3- (aq) A beaker of KOH and Fe(NO3)3 has neither KOH nor Fe(NO3)3, it’s all ions! TEXT 692019 AND YOUR QUESTION TO 37607

  16. The 1st Rule of Chemistry… Opposites attract! Positive ions like negative ions. Negative ions like positive ions. Postive ions hate positive ions. Negative ions hate negative ions. Fe3+ NO3- K+ OH- TEXT 692019 AND YOUR QUESTION TO 37607

  17. Only possible products are… KOH or KNO3 Fe(OH)3 or Fe(NO3)3 We know that KOH and Fe(NO3)3 don’t form…that’s what we started with. What about KNO3 and Fe(OH)3? Fe3+ NO3- K+ OH- TEXT 692019 AND YOUR QUESTION TO 37607

  18. How many “Ks” are there in the beaker? • None of the below • 3 • 2 • 4 • 5 TEXT 692019 AND YOUR QUESTION TO 37607

  19. What about KNO3 and Fe(OH)3? They are both possible products of the reaction. Could they both form? Which one forms first? Do they form together? How would you know? Ksp TEXT 692019 AND YOUR QUESTION TO 37607

  20. When you have 2 possible reactions… BIGGEST K wins! Or, in this case, SMALLEST Ksp Kprecipitation = 1/Ksp Small Ksp means big Kprecipitation. TEXT 692019 AND YOUR QUESTION TO 37607

  21. Precipitation is just the reverse of dissolution. KNO3 (s) K+ (aq) + NO3-(aq) Ksp (KNO3) = HUGE (K+ salts are very soluble and nitrates are very soluble) Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) Ksp (Fe(OH)3)=2.79x10-39 TEXT 692019 AND YOUR QUESTION TO 37607

  22. So the only reaction to consider is… Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) Ksp (Fe(OH)3)=2.79x10-39 All equilibrium problems have 3 parts…yada yada yada… TEXT 692019 AND YOUR QUESTION TO 37607

  23. Ksp (Fe(OH)3)=2.79x10-39 Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I C E Ksp = 2.79x10-39 = [Fe3+][OH-]3 What do we know? TEXT 692019 AND YOUR QUESTION TO 37607

  24. Don’t forget the dilution 500 mL of 0.100 M Fe(NO3)3 is mixed with 250 mL of 0.100 M KOH. So… Dilution is the solution! 0.100 M x 0.500 L = 0.05 mol/0.750 L = 0.0667 M 0.100 M x 0.250 L = 0.025 mol/0.750 L = 0.0333 M TEXT 692019 AND YOUR QUESTION TO 37607

  25. Ksp (Fe(OH)3)=2.79x10-39 Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I 00.067 0.033 C +x -x -3x E x0.067-x 0.033-3x Ksp = 2.79x10-39 = [0.067-x][0.033-3x]3 This is an algebraic mess BUT…K is really small. TEXT 692019 AND YOUR QUESTION TO 37607

  26. K is really small… …which means… Fe(OH)3 is not very soluble. So, x is going to be huge! We can use that to our advantage. We can mathematically precipitate out ALL of the Fe(OH)3 and then redissolve it! TEXT 692019 AND YOUR QUESTION TO 37607

  27. What is product limiting? Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I - 0.067 0.033 C - -x -3x E - 0.067-x 0.033-3x 0.067-x = 0 X = 0.067 0.033 – 3x = 0 X=0.011 The hydroxide runs out first! TEXT 692019 AND YOUR QUESTION TO 37607

  28. What is product limiting? Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I - 0.067 0.033 C - -0.011 -3(0.011) E - 0.056 0 0.067-x = 0 X = 0.067 0.033 – 3x = 0 X=0.011 The hydroxide runs out first! TEXT 692019 AND YOUR QUESTION TO 37607

  29. Double your ICE, double your pleasure! Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I - 0.067 0.033 C +0.011 -0.011 -3*(0.011) I 0.011 0.056 0 C -x +x +3x E 0.011-x 0.056+x 3x Ksp = 2.79x10-39 = [0.056+x][3x]3 Look how much simpler that is. Even better, let’s try and solve it the easy way! TEXT 692019 AND YOUR QUESTION TO 37607

  30. Double your ICE, double your pleasure! Ksp = 2.79x10-39 = [0.056+x][3x]3 Assume x<<0.056! 2.79x10-39 = [0.056][3x]3 = 0.056*27x3 1.8452x10-39 = x3 1.23x10-13 = x! Pretty good assumption. TEXT 692019 AND YOUR QUESTION TO 37607

  31. Double your ICE, double your pleasure! Fe(OH)3 (s)  Fe3+ (aq) + 3 OH-(aq) I - 0.067 0.033 C +0.011 -0.011 -3*(0.011) I 0.011 0.056 0 C - 1.23x10-13 + 1.23x10-13 +3(1.23x10-13) E 0.011 0.056 3.68x10-13 0.011 M Fe(OH)3 precipitate 0.011 M Fe(OH)3*0.75 L * 106.9 g/mol = 0.9 g Fe(OH)3 TEXT 692019 AND YOUR QUESTION TO 37607

  32. Neat trick, huh? Actually, that is another little trick in your ICE arsenal… We know what to do when x is small. Now, if we suspect x is large, we can try this little trick. In fact, you could always forcibly do a reaction to change the initial condition. After all, in the end the equilibrium will decide where it finishes. TEXT 692019 AND YOUR QUESTION TO 37607

  33. Write a balanced equation for the dissolution of FeCO3(s) in water. A. FeCO3 (s) ↔ Fe3+ (aq) + CO33- (aq) B. FeCO3(s) + H2O (l) ↔ Fe2+(aq) + CO32-(aq) C. FeCO3 (s) ↔ Fe3+(aq) + CO32- (aq) D. FeCO3 (s) ↔ Fe2+ (aq) + CO32- (aq) E. FeCO3 (s) ↔ Fe (aq) + CO3(aq) TEXT 692019 AND YOUR QUESTION TO 37607

  34. Write a balanced equation for the dissolution of MgCO3(s) in water. A. MgCO3 (s) ↔ Mg3+ (aq) + CO33- (aq) B. MgCO3(s) + H2O (l) ↔ Mg2+(aq) + CO32-(aq) C. MgCO3(s) ↔ Mg2+ (aq) + CO32- (aq) D. MgCO3(s) → Mg2+ (aq) + CO32- (aq) E. MgCO3(s) ↔ Mg (aq) + CO3(aq) TEXT 692019 AND YOUR QUESTION TO 37607

  35. Write a balanced equation for the dissolution of Mg(NO3)2 (s) in water. A. Mg(NO3)2 (s) ↔ Mg2+ (aq) + NO32- (aq) B. Mg(NO3)2(s) ↔ Mg+(aq) + NO3-(aq) C. Mg(NO3)2(s) ↔ Mg2+ (aq) + 2 NO3-(aq) D. Mg(NO3)2(s) ↔ Mg+(aq) + 2 NO3-(aq) E. Mg(NO3)2 (s) ↔ Mg2+(aq) + 2 NO32- (aq) TEXT 692019 AND YOUR QUESTION TO 37607

  36. Write a balanced equation for the dissolution of Fe(NO3)2 (s) in water. A. Fe(NO3)2 (s) ↔ Fe2+ (aq) + NO32- (aq) B. Fe(NO3)2(s) ↔ Fe+(aq) + NO3-(aq) C. Fe(NO3)2(s) ↔ Fe2+ (aq) + 2 NO32-(aq) D. Fe(NO3)2(s) ↔ Fe+(aq) + 2 NO3-(aq) E. Fe(NO3)2 (s) ↔ Fe2+(aq) + 2 NO3-(aq) TEXT 692019 AND YOUR QUESTION TO 37607

  37. Write a balanced equation for the dissolution of K2CO3(s) in water. A. K2CO3 (s) ↔ 2 K+ (aq) + CO32- (aq) B. K2CO3(s) ↔ K2+(aq) + CO32-(aq) C. K2CO3(s) ↔ 2K+(aq) + CO3-(aq) D. K2CO3(s) ↔ K+(aq) + CO32-(aq) E. K2CO3(s) ↔ K(aq) + CO3(aq) TEXT 692019 AND YOUR QUESTION TO 37607

  38. If I dissolve Mg(NO3)2, Fe(NO3)2, and K2CO3 in water, what species are present in my beaker? • Mg, NO3, Fe, NO3, K, CO3, H2O • Mg2+, NO3-, Fe3+ , NO3-, K2+ , CO32-, H2O • Mg2+, Fe3+ , NO3-, K+, CO32-, H2O • Mg2+, Fe2+, NO3-, K+ , CO32-, H2O • Mg2+, Fe3+ , CO32-, H2O TEXT 692019 AND YOUR QUESTION TO 37607

  39. Given the following Ksp values, which salt is LEAST soluble in water? A. Ksp[FeCO3] = 3.07x10-11 B. Ksp[MgCO3] = 6.82x10-6 C. Ksp[Mg(NO3)2] = 3.14x104 D. Ksp [Fe(NO3)2] = 6.28x103 E. Ksp [K2CO3] = 1.26x105 TEXT 692019 AND YOUR QUESTION TO 37607

  40. Be a little careful about stoichiometry… Imagine the following chloride salts: Ksp (XCl2) = 12 Ksp (YCl) = 9 Which salt is MORE SOLUBLE? • XCl2 • YCl • I need more information TEXT 692019 AND YOUR QUESTION TO 37607

  41. TEXT 692019 AND YOUR QUESTION TO 37607

  42. A. I did Problem 8 B. I did not get to Problem 8 yet C. I don’t know how to do Problem 8 TEXT 692019 AND YOUR QUESTION TO 37607

  43. What precipitates? • K2CO3 • Mg(NO3)2 • Fe(NO3)2 • FeCO3 • MgCO3 TEXT 692019 AND YOUR QUESTION TO 37607

  44. Consider a solution that is 2.3×10-2 M in Fe(NO3)2 and 1.5×10-2 M in Mg(NO3)2. What must the concentration of potassium carbonate be to get the iron ions to start to precipitate? • None – the iron ions won’t precipitate. • 11.62 M • 5.5x10-6 M • 1.3x10-9 M • 0.0286 M TEXT 692019 AND YOUR QUESTION TO 37607

  45. The question is when Q>Ksp. But before it can exceed Ksp, Q must equal Ksp, so As soon as my carbonate is 1 tiny bit above that, precipitation starts. TEXT 692019 AND YOUR QUESTION TO 37607

  46. Consider a solution that is 2.3×10-2 M in Fe(NO3)2 and 1.5×10-2 M in Mg(NO3)2. What must the concentration of potassium carbonate be to get the magnesium ions to start to precipitate? • 0.028 M • 2.1x106 M • 4.5x10-4 M • 0.0286 M • 723 M TEXT 692019 AND YOUR QUESTION TO 37607

  47. As soon as my carbonate is 1 tiny bit above that, precipitation starts. TEXT 692019 AND YOUR QUESTION TO 37607

  48. What minimum concentration of K2CO3 is required to cause the precipitation of the cation that precipitates first? • 2.1x106 M • 4.5x10-4M • 5.5x10-6 M • 1.3x10-9 M • 0.0286 M TEXT 692019 AND YOUR QUESTION TO 37607

  49. Puzzle 1. How much of the first cation is still in solution when the second cation begins to precipitate? • 4.55x10-4 M • 6.75x10-8 M • 1.5x10-2 M • 1.3x10-2 M • 2.3x10-2 M TEXT 692019 AND YOUR QUESTION TO 37607

  50. TEXT 692019 AND YOUR QUESTION TO 37607