1. Applications of Aqueous Equilibria Chapter 15

2. Common Ion Effect Calculations Calculate the pH and the percent dissociation of a.200M HC 2 H 3 O 2 (K a =1.8x10-5) Buffer Buffer Calculate the pH and the percent dissociation of a.200M HC 2 H 3 O 2 in the presence of.500 M NaC 2 H 3 O 2

3. Common Ion Effect Calculations Continued ICE ICE [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] [H + ] [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] [H + ] Initial (mol/L).200 0 0 Change (mol/L) - x + x + x Equil. (mol/L).200 - x x x

4. Common Ion Effect Calculations Continued [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] [H + ] [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] [H + ] Initial (mol/L). 200.500 0 Change (mol/L) - x + x + x Equil. (mol/L).200 - x.500+x x

5. A Buffered Solution... resists change in its pH when either H + or OH  are added. 1.0 L of 0.50 M H 3 CCOOH + 0.50 M H 3 CCOONa pH = 4.74 Adding 0.010 mol solid NaOH raises the pH of the solution to 4.76, a very minor change.

6. Preparation of Buffered Solutions Buffered solution can be made from: 1.a weak acid and its salt (e.g. HC 2 H 3 O 2 & NaC 2 H 3 O 2 ). 1.a weak acid and its salt (e.g. HC 2 H 3 O 2 & NaC 2 H 3 O 2 ). 2. a weak base and its salt (e.g. NH 3 & NH 4 Cl). Other examples of buffered pairs are: H 2 CO 3 & NaHCO 3 H 3 PO 4 & NaH 2 PO 4

7. Buffer Calculations A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution. Ka= (1.8 x 10 -5 )

8. Henderson-Hasselbalch Equation A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution. Ka= (1.8 x 10 -5 ) pH = pK a + log([A - ]/[HA]) pH= -log (1.8 x 10 -5 ) + log ([.50]/[.50]) pH = 4.74 + 0 = 4.74

9. Henderson-Hasselbalch Equation -Useful for calculating pH when the [A  ]/[HA] ratios are known.

10. Calculate the pH of a solution that contains.250M formic acid HCOOH (Ka=1.8x10-4) And.100M sodium formate HCOONa

11. Buffer practice Calculate the pH of a solution that contains.500M formic acid HCOOH (Ka=1.8x10-4) And.200M sodium formate HCOONa

12. Textbook774 23 a,b,D 24 a,b,c,d

13. Titration (pH) Curve A plot of pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

14. Titration curve for a strong base added to a strong acid -- the equivalence point has a pH of 7.

15. Titration curve for the addition of a strong base to a weak acid-- pH is above 7.00.

16. Titration curve for the addition of a strong acid to a weak base -- the pH at equivalence is below 7.00.

17. Strong Acid - Strong Base Titration Before equivalence point, [H + ] is determined by dividing number of moles of H + remaining by total volume of solution in L.Before equivalence point, [H + ] is determined by dividing number of moles of H + remaining by total volume of solution in L. At equivalence point, pH is 7.00.At equivalence point, pH is 7.00. After equivalence point [OH - ] is calculated by dividing number of moles of excess OH - by total volume of solution in L.After equivalence point [OH - ] is calculated by dividing number of moles of excess OH - by total volume of solution in L.

18. Titration curve for a strong base added to a strong acid -- the equivalence point has a pH of 7.

19. The equivalence point is defined by the stoichiometry, not the pH.

20. Determining the End Point in a Titration Two methods are used: pH meter acid-base indicator

21. Acid-Base Indicator... marks the end point of a titration by changing color. The color change will be sharp, occurring with the addition of a single drop of titrant. The equivalence point is not necessarily the same as the end point. Indicators give a visible color change will occur at a pH where: pH = pK a  1 pH = pK a  1

22. The useful pH ranges of several common indicators -- the useful range is usually pK a  1. Why do some indicators have two pH ranges?

23. On the left is the pH curve for the titration of a strong acid and a strong base. On the right is the curve for a weak acid and a strong base.

24. Solubility Product For solids dissolving to form aqueous solutions. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = solubility product constant and K sp = [Bi 3+ ] 2 [S 2  ] 3 K sp = [Bi 3+ ] 2 [S 2  ] 3 Why is Bi 2 S 3(s) not included in the solubilty product expression?

25. Ba(OH) 2 (s) Ag 2 CrO 4 )(s) Ca 3 (PO 4 ) 2 (s)

26. Solubility Product Relative solubilities can be predicted by comparing K sp values only for salts that produce the same total number of ions.Relative solubilities can be predicted by comparing K sp values only for salts that produce the same total number of ions. AgI (s) K sp = 1.5 x 10 -16 CuI (s) K sp = 5.0 x 10 -12 CaSO 4(s) K sp = 6.1 x 10 -5 CaSO 4(s) > CuI (s) > AgI (s)

27. Solubility Product CuS (s) K sp = 8.5 x 10 -45 Ag 2 S (s) K sp = 1.6 x 10 -49 Bi 2 S 3(s) K sp = 1.1 x 10 -73 Bi 2 S 3(s) > Ag 2 S (s) > CuS (s) Why does this order from most to least soluble appear to be out of order?

28. Solubility Product “Solubility” = s = concentration of Bi 2 S 3 that dissolves. The [Bi 3+ ] is 2s and the [S 2  ] is 3s. Note:K sp is constant (at a given temperature) s is variable (especially with a common ion present) ion present) Solubility product is an equilibrium constant and has only one value for a given solid at a given temperature. Solubility is an equilibrium position.

29. Solubility Product Calculations Cupric iodate has a measured solubility of 3.3 x 10 -3 mol/L. What is its solubility product? Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) 3.3 x 10 -3 M ---> 3.3 x 10 -3 M + 6.6 x 10 -3 M K sp = [Cu 2+ ][IO 3 - ] 2 K sp = [3.3 x 10 -3 ][6.6 x 10 -3 ] 2 K sp = 1.4 x 10 -7

30. Solubility from Ksp NiCO 3 Ksp 1.4x10-7 Ba 3 (PO 4 ) 2 Ksp=6x10-39 PhBr 2 Ksp=4.6x10-6

31. AgI (s) K sp = 1.5 x 10 -16 CuBr (s) K sp = 5.0 x 10 -12 MgSO 4(s) K sp = 6.1 x 10 -5

32. AgCl Ksp 1.5x10 -10 Ag 2 CrO4 Ksp=9.0x10 -12 Ag3PO4 Ksp 1.8 x10 -18

33. Common Ion Effect CaF 2(s) Ca 2+ (aq) + 2F - (aq) What will be the effect on this equilibrium if solid sodium fluoride is added? Explain. Equilibrium will shift to the left, due to Le Chatelier’s Principle. Solubility product must stay constant, so the amount of Ca 2+ & F - must decrease by forming solid CaF 2.

34. Solubility Product Calculations Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) K sp = [Cu 2+ ][IO 3 - ] 2 If solid cupric iodate is dissolved in HOH; double & square the iodate concentration. If solid cupric iodate is dissolved in HOH; double & square the iodate concentration. If mixing two solutions, one containing Cu 2+ and the other IO 3 -, then use the concentration of iodate and only square it.

35. Ion Product, Q sp If 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 is added to 300.0 mL of 2.00 x 10 -2 M KIO 3, will Ce(IO 3 ) 3 precipitate? [Ce 3+ ] = (4.00 x 10 -3 M)X(750.0 mL) (750.0 mL + 300.0 mL) (750.0 mL + 300.0 mL) [Ce 3+ ] = 2.86 x 10 -3 M [IO 3 - ] = (2.00 x 10 -2 M) x (300.0 mL) (750.0 mL + 300.0 mL) (750.0 mL + 300.0 mL) [IO 3 - ] = 5.71 x 10 -3 M

36. Ion Product, Q sp Continued Q sp = [Ce 3+ ] 0 [IO 3 - ] o 3 Q sp = [2.86 x 10 -3 ][5.72 x 10 -3 ] 3 Q sp = 5.32 x 10 -10 Q sp > K sp  Ce(IO 3 ) 3 will precipitate. K sp = 1.9 x 10 -10

37. A 200.ml solution of 1.3x10 -3 M AgNO 3 is mixed with 100ml of a 4.5 x10 -5 M Na 2 S will precipitation occur? Ksp=1.6x10 -49

38. pH & Solubility If a solid precipitate has an anion X - that is an effective base (HX is a weak acid), then the salt MX will show increased solubility in an acidic solution. Salts containing OH -, S 2-, CO 3 2-, C 2 O 4 2-, & CrO 4 2- are all soluble in acidic solution. Limestone caves are made up of insoluble CaCO 3, but dissolve in acidic rain water (H 2 CO 3 ).

39. Solubility Product Calculations If a 0.010 M solution of sodium iodate is mixed with a 0.0010 M cupric nitrate, will a precipitate form? 2 NaIO 3(aq) + Cu(NO 3 ) 2(aq) ---> Cu(IO 3 ) 2(s) + 2 NaNO 3(aq) Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) Q sp = [Cu 2+ ][IO 3 - ] 2 Q sp = [1.0 x 10 -3 ][1.0 x 10 -2 ] 2 Q sp = 1.0 x 10 -7 Q sp < K sp  no precipitate forms.