1. Chapter 15 & 16: Applications of Aqueous Equilibrium

2. Common Ion Effect  The addition of an ion already present(common) in a system causes equilibrium to shift away from the common ion. http://www.precisiongraphics.com/portfolio/animation/page/4/

3. The Common Ion Effect  For example, the addition of NaF to a solution of HF. The HF is dissolved and at equilibrium. Adding a common ion, in this case, F -, to the solution, shifts the equilibrium and makes the HF less soluble.  This can be explained by the use of LeChatelier's Principle.

4. The addition of a common ion to a solution of a weak acid makes the solution less acidic.  HC 2 H 3 O 2  H + + C 2 H 3 O 2 - If we add NaC 2 H 3 O 2, equilibrium shifts to undissociated HC 2 H 3 O 2, raising pH.  The new pH can be calculated by putting the concentration of the anion into the K a equation and solving for the new [H + ].

5. Ex 1. A solution contain 1.0 M HF and 1.5 M KF. What is the pH of the solution?  If the pH of 1.0 M HF is originally 1.57, what did the addition of the salt do to the pH and why? The pH goes up because there is a shift in equilibrium to the left as F - is added. The amount of H + at equilibrium is decreased.

6. 15.2 Buffered Solutions  Buffered solution- A solution that resists changes in pH when hydroxide ions or protons are added. A buffer solution usually consists of a solution of a weak acid and its salt or a weak base and its salt.

7. Ex. HC 2 H 3 O 2 /C 2 H 3 O 2 - buffer system  Addition of strong acid: H + + C 2 H 3 O 2 -  HC 2 H 3 O 2  Addition of strong base: OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 - NH 3 /NH 4 + buffer system Addition of strong acid: H + + NH 3  NH 4 + Addition of strong base: OH - + NH 4 +  NH 3 + H 2 O

8. So, how does a buffer work?  As you can see above, when hydrogen ions are added to a buffer system, they react with the basic portion of the buffer. This drives the production of the weak acid portion of the system. If hydroxide ions are added to the buffer system, they react with the acidic portion of the buffer, producing the conjugate base.

9. Buffer Capacity  Buffer capacity- The amount of acid or base that can be absorbed by a buffer system without a significant change in pH.  In order to have a large buffer capacity, a solution should have large concentrations of both buffer components.  A buffer should maintain a system at a desired pH. To find what acids are best for a certain buffer, look at their K a values. A buffer with a pH of 4.0 can be obtained from an acid with a pK a of 4.0.

10. Hints for Solving Buffer Problems: 1.Write an equation for the dissociation of the acid or base into your RICE table. 2.Write the concentration of the acid and conjugate base or base and conjugate acid. You will have two values in the “initial” section of your table. 3.Solve for x just as you always would, but subtracted and added x values will probably be negligible in you expression. 4.Solve and check your answer to see that it makes sense.

11. Ex 2. A solution is 0.120 M in acetic acid and 0.0900 M in sodium acetate. Calculate the [H + ] at equilibrium. The K a of acetic acid is 1.8 x 10 -5.

12. Once a buffered solution is set up, the ions are at equilibrium. However, addition of common ions can shift that equilibrium.  Here are some hints for working with this kind of problem: 1.Determine major species involved initially. 2.If a base is added, react it with the acid part of the buffer. If an acid is added, react it with the basic part of the buffer. Show all stoichiometry for this process in a before, change, after (BCA) table. 3.Find the molarity of the major species after the reaction, and use them in a RICE table. 4.Set up equilibrium expression (K a or K b ) and solve. 5.Check the logic of your answer.

13. Ex 3. Calculate the pH of the buffer system in example 2 when 100.0 mL of 0.100 M HCl is added to 455 mL of solution.

14. The Henderson-Hasselbach Equation  One way to calculate the pH of a buffer system in which the ration [HA]/[A - ] is known is with the Henderson-Hasselbach equation.  For a particular buffering system, all solutions that have the same ratio of [A - ]/[HA] have the same pH.  Optimum buffering occurs when [HA] = [A - ] and the pK a of the weak acid used should be as close as possible to the desired pH of the buffer system.

15. The Henderson-Hasselbach Equation  This equation comes from the equilibrium expression.  which can be rearranged to  This means that if you know the concentration of the acid and conjugate base, you can find the hydrogen ion concentration. This can be used to find pH. The simplified version of this equation is shown above.  pH = pK a + log [base] pH = pK a + log [A - ]  [acid] [HA]

16.  The Henderson-Hasselbach (HH) equation needs to be used cautiously. It is sometimes used as a quick, easy equation to plug numbers into. A K a or K b problem requires a greater understanding of the factors involved and can always be used instead of the HH equation. The Henderson-Hasselbach Equation

17. Ex. 4 A buffer is prepared by adding 0.25 M NH 3 (K b = 1.8 x 10 -5 ) and 0.40 M NH 4 Cl.  A. Calculate the pH of this solution.

18. B. calculate the pH that results when 0.10 mole of gaseous HCl is added to 1.0 L of the buffered solution.

19. c. Use the Henderson-Hasselbach equation to calculate part B of this problem.  pH = pK a + log [base] [acid]  Since this is a base reaction, use K b, the base/acid, and then you will subtract to find pH.  pOH = (-log 1.8 x 10 -5 ) + ([0.50]/[0.15])  pH =8.73

20. Acid-Base Titrations

21. 15.4 Titrations and pH Curves  In an acid/base titration, the titrant (commonly a base) is added to a solution of unknown concentration until the substance being analyzed is just consumed (stoichiometric point or equivalence point).  Titrant-solution of known concentration (in buret)  pH or titration curve -plot of pH as a function of the amount of titrant added.

22. http://www.youtube.com/watch?v=yirkozUyG74 Titration Curves

23. There are three main types of acid/base titrations. We will examine each one and discuss how to calculate the concentration of species during the titration.

24. 1.Strong Acid-Strong Base  Simple reaction H + + OH -  H 2 O  The pH is easy to calculate because all reactions go to completion.  At the equivalence point, the solution is neutral.

25. Ex. 5 12.5 mL of 0.200 M HNO 3 is titrated with 0.100 M NaOH.  Calculate the pH at the start of the titration. Write this point on the graph.  No NaOH has been added, so only HNO 3, a strong acid, is present.  [H + ] = [HNO 3 ]  [H + ] = 0.200M  pH = -log (0.200)  pH = 0.699

26. Calculate the pH of the solution after 2.5 mL of NaOH has been added. Add this point to the graph (0.0025 L NaOH)(0.100M) = 2.5x10 -4 mol OH - (0.0125 L HNO 3 )(0.200M) = 2.5x10 -3 mol H + H + + OH -  H 2 O 2.5x10 -3 + 2.5x10 -4  --- -2.5x10 -4 -2.5x10 -4  --- 2.25x10 -3 0  ---  Once the reaction has occurred, only H + ions remain.  [H + ] = 2.25x10 -3 mol/0.0150 L = 0.150 M  pH = -log (0.150)  pH = 0.824

27. Calculate the pH of the solution after 5.0 mL of NaOH has been added. Add this point to the graph. (0.0050 L NaOH)(0.100M) = 5.0x10 -4 mol OH - (0.0125 L HNO 3 )(0.200M) = 2.5x10 -3 mol H + H + + OH -  H 2 O 2.5x10 -3 + 5.0x10 -4  --- -5.0x10 -4 -5.0x10 -4  --- 2.00x10 -3 0  ---  Once the reaction has occurred, only H + ions remain.  [H + ] = 2.00x10 -3 mol/0.0175 L = 0.114 M  pH = -log (0.114)  pH = 0.942

28. Calculate the pH of the solution after 25.0 mL of NaOH has been added. Add this point to the graph. (0.0250 L NaOH)(0.100M) = 2.5x10 -3 mol OH - (0.0125 L HNO 3 )(0.200M) = 2.5x10 -3 mol H + H + + OH -  H 2 O 2.5x10 -3 + 2.5x10 -3  --- - 2.5x10 -3 - 2.5x10 -3  --- 0 0  ---  Only water remains. This is the equivalence point!  pH = 7.00

29. Calculate the pH of the solution after 35.0 mL of NaOH has been added. Add this point to the graph. (0.0350 L NaOH)(0.100M) = 3.5x10 -3 mol OH - (0.0125 L HNO 3 )(0.200M) = 2.5x10 -3 mol H + H + + OH -  H 2 O 2.50x10 -3 + 3.50x10 -3  --- - 2.50x10 -3 -2.50x10 -3  --- 0 0.00100  ---  Once the reaction has occurred, only OH - ions remain.  [OH - ] = 0.00100 mol/0.0475 L = 0.0211 M  pOH = -log (0.0211)  pH = 12.323

30. Calculate the pH of the solution after 50.0 mL of NaOH has been added. Add this point to the graph. (0.0500 L NaOH)(0.100M) = 5.00x10 -3 mol OH - (0.0125 L HNO 3 )(0.200M) = 2.5x10 -3 mol H + H + + OH -  H 2 O 2.50x10 -3 + 5.00x10 -3  --- - 2.50x10 -3 -2.50x10 -3  --- 0 0.00250  ---  Once the reaction has occurred, only OH - ions remain.  [OH - ] = 0.00250 mol/0.0625 L = 0.0400 M  pOH = -log (0.0400)  pH = 12.602

31. 2.Weak Acid-Strong Base  The reaction of a strong base with a weak acid is assumed to go to completion.  Before the equivalence point, the concentration of weak acid remaining and the conjugate base formed are determined.  At halfway to the equivalence point, pH = pK a.  At the equivalence point, a basic salt is present and the pH will be greater than 7. After the equivalence point, the strong base will be the dominant species and a simple pH calculation can be made after the stoichiometry is done.

32. Ex. 6 A 25.0 mL sample of 0.10 M acetic acid is titrated with 0.10 M NaOH.  Calculate the pH at the start of the titration. Write this point on the graph.  No NaOH has been added, so only HC 2 H 3 O 2, a weak acid, is present.  So, do a RICE table. HC 2 H 3 O 2  H + + C 2 H 3 O 2 - 0.10  0 + 0 -x  +x +x 0.10 - x  x x

33. Calculate the pH of the solution after 5.0 mL of NaOH has been added. Add this point to the graph. HC 2 H 3 O 2  H + + C 2 H 3 O 2 -

34. Calculate the pH of the solution after 12.5 mL of NaOH has been added. Add this point to the graph

35. Calculate the pH of the solution after 20.0 mL of NaOH has been added. Add this point to the graph. HC 2 H 3 O 2  H + + C 2 H 3 O 2 -

36. Calculate the pH of the solution after 25.0 mL of NaOH has been added. Add this point to the graph. C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH -

37. Calculate the pH of the solution after 40.0 mL of NaOH has been added. Add this point to the graph.

38. 2.Weak Base-Strong Acid  Before the equivalence point, a weak base equilibria exists. Calculate the stoichiometry and then the weak base equilibria.  At the equivalence point, an acidic salt is present and the pH is below 7. After the equivalence point, the strong acid is the dominant species. Use the [H + ] to find the pH.

39. Ex. Calculate the pH when 50.0 mL of 0.050 M NH 3 is titrated with 0.10 M HCl. (K b of NH 3 = 1.8 x 10 -5 )  Calculate the pH at the start of the titration. Write this point on the graph.  Only a weak base is present, so, do a RICE table with K b. NH 3 + H 2 O  NH 4 + + OH - 0.050 ---  0 0 -x ---  +x +x 0.050 - x ---  x x

40. Calculate the pH of the solution after 7.5 mL of HCl has been added. Add this point to the graph. NH 3 + H 2 O  NH 4 + + OH -

41. Calculate the pH of the solution after 12.5 mL of HCl has been added. Add this point to the graph.

42. Calculate the pH of the solution after 25.0 mL of HCl has been added. Add this point to the graph. NH 4 +  NH 3 + H +

43. Calculate the pH of the solution after 35.0 mL of HCl has been added. Add this point to the graph.

44. Calculate the pH of the solution after 50.0 mL of HCl has been added. Add this point to the graph.

45. Acid-Base Indicators  End point - the point in a titration where the indicator changes color  Indicators are usually weak acids, HIn. They have one color in their acidic (HIn) form and another color in their basic (In - ) form.  A very common indicator, phenolphthalein, is colorless in its HIn form and pink in its In- form. It changes color in the range of pH 8-10.

46. Choosing an Indicator  When choosing an indicator, we want the indicator end point and the titration equivalence point to be as close as possible.  Since strong acid-strong base titrations have a large vertical area, color changes will be sharp and a wide range of indicators can be used. For titrations involving weak acids or weak bases, we must be more careful in our choice of indicator.

47. So Which Indicator Do I Use?  Usually 1/10 of the initial form of the indicator must be changed to the other form before a new color is apparent.  The following equations can be used to determine the pH at which an indicator will change color:  For titration of an acid:pH = pK a + log 1/10 = pK a - 1  For titration of a base:pH = pK a + log 10/1 = pK a + 1  The useful range of an indicator is usually its pK a ±1.  When choosing an indicator, determine the pH at the equivalence point of the titration and then choose an indicator with a pK a close to that.

48. Indicators

49. The useful range of an indicator is usually its pK a ±1. When choosing an indicator, determine the pH at the equivalence point of the titration and then choose an indicator with a pK a close to that. You should know the ranges of color change for bromothymol blue, phenolphthalein and litmus.

50. The pH Curve for the Titration of 100.0 mL of 0.10 M HCI with 0.10 M NaOH The pH Curve for the Titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH

51. Chapter 16: Solubility Equilibrium

52. 16.1 Solubility Equilibria and the Solubility Product  Saturated solutions of salts are another type of chemical equilibria.  For a saturated solution of AgCl, the equation would be: AgCl(s)  Ag + (aq) + Cl - (aq)  The solubility product expression would be: K sp = [Ag + ][Cl - ]  The AgCl(s) is left out since solids are left out of equilibrium expressions (constant concentrations).  For Ag 2 CO 3, Ag 2 CO 3  2Ag + + CO 3 2- K sp = [Ag + ] 2 [CO 3 2- ]  The K sp of AgCl is 1.6 x 10 -10.  This means that if the product of [Ag + ][Cl - ] < 1.6 x 10 -10, the solution is unsaturated and no solid would be present.  If the product = 1.6 x 10 -10, the product is exactly saturated and no solid would be present.  If the product > 1.6 x 10 -10, the solution is saturated and a solid (precipitate) would form.

53. The Ion Product Constant, Q  The product of the ions (raised to the power of their coefficients) is called the ion product constant or Q.  If K sp > Q, no ppt forms.  If K sp < Q, ppt forms.

54. Ex. The molar solubility of silver sulfate is 1.5 x 10 -2 mol/L. Calculate the solubility product of the salt. Remember that molar solubility is “x”! Since 1.5 x 10 -2 mol/L of Ag 2 SO 4 dissolve, 1.5 x 10 -2 mol/L of SO 4 2- form and 2(1.5 x 10 -2 mol/L) of Ag + form. K sp = [Ag + ] 2 [SO 4 2- ] x = 1.5 x 10 -2 Ag 2 SO 4  2Ag + + SO 4 2- ---- 0 0 -x +2x +x ---- 3.0 x 10 -2 1.5 x 10 - K sp = (3.0 x 10 -2 ) 2 (1.5 x 10 -2 ) = 1.4 x 10 -5

55. Ex. Calculate the molar solubility of calcium phosphate. The K sp of calcium phosphate is 1.2 x 10 -26. Remember that molar solubility is “x”! K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 Ca 3 (PO 4 ) 2  3Ca 2+ + 2PO 4 3- ---- 0 0 -x +3x +2x ---- 3x 2x 1.2 x 10 -26 = (108)x 5 = 2.6 x 10 -6 K sp = [3x] 3 [2x] 2 = (108)x 5

56. Ex. What is the molar solubility of lead(II) iodide in a 0.050 M solution of sodium iodide? K sp = [Pb 2+ ][I - ] 2 PbI 2  Pb 2+ + 2I - ---- 0 0.050 -x +x +2x ---- x 0.050+2x 1.4 x 10 -8 = (0.0025)x = 5.6 x 10 -6 1.4x10 -8 = [x][0.050+2x] 2 Don’t forget to put in the initial concentration of the common ion! 1.4x10 -8 = [x][0.050] 2 This shows the decreased solubility of a salt in the presence of a common ion.

57. Ex. Exactly 200 mL of 0.040 M BaCl 2 are added to exactly 600 mL of 0.080 M K 2 SO 4. Will a precipitate form? Q = (1.0 x 10 -2 M Ba 2+ )(6.0 x 10 -2 M SO 4 2- ) = 6.0 x 10 -4 Q > K sp 6.0 x 10 -4 > 1.1 x 10 -10 First, write a balanced equation to figure out what ppt will form? BaCl 2 + K 2 SO 4  BaSO 4 + 2KCl Barium sulfate is the likely precipitate. Find the molarity of each of the ions that will potentially form the ppt. (0.200 L)(0.040M) BaCl 2 = 8.0 x 10 -3 mol Ba 2+ /0.800L total volume = 1.0 x 10 -2 M Ba 2+ (0.600 L)(0.080M) K 2 SO 4 = 4.8 x 10 -2 mol SO 4 2- /0.800L total volume = 6.0 x 10 -2 M SO 4 2- K sp = [Ba 2+ ][SO 4 2- ] = 1.1 x 10 -10 (look this up in table) A precipitate of BaSO 4 forms.