1. C h a p t e rC h a p t e r C h a p t e rC h a p t e r 16 Applications of Aqueous Equilibria Chemistry 4th Edition McMurry/Fay Chemistry 4th Edition McMurry/Fay

2. Slide 2 Buffer Solutions01 A Buffer Solution : is a solution of a weak acid and a weak base (usually conjugate pair); both components must be present. A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers are very important to biological systems!

3. Slide 3 Buffer Solutions02 Base is neutralized by the weak acid. Acid is neutralized by the weak base.

4. Slide 4 Buffer Solutions03 Buffer solutions must contain relatively high concentrations of weak acid and weak base components to provide a high “buffering capacity”. The acid and base components must not neutralize each other. The simplest buffer is prepared from equal concentrations of an acid and its conjugate base.

5. Slide 5 Buffer Solutions04 a) Calculate the pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution? a) pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COO – Na + is: 4.74 (see earlier slide)

6. Slide 6 Buffer Solutions04 b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution? CH 3 CO 2 H (aq) + H 2 O (aq) CH 3 CO 2 – (aq) + H 3 O + (aq) I 1.01.0 0 add 0.10 mol HCl + 0.10 – 0.10 C -x -x +x E 1.1 – x 0.9 + x x K a = 1.8 x 10 -5 = = (0.9+x)(x) (0.9)x (1.1 – x) (1.1) x = ( 1.1 / 0.9 ) 1.8 x 10 -5 x = 2.2 x 10 -5 pH = 4.66 reacts w/

7. Slide 7 Water – No Buffer What is the pH after the addition of 0.10 mole of HCl to 1.0 L of pure water? HCl – strong acid, completely ionized. H + concentration will be 0.10 molar. pH will be –log(0.10) = 1.0 Adding acid to water: Δ pH = 7.0 - 1.0 = 6.0 pH units Adding acid to buffer: Δ pH = 4.74 - 4.66 = 0.09 pH units!! The power of buffers!

8. Slide 8 Titration: a procedure for determining the concentration of a solution using another solution of known concentration. Titrations involving strong acids or strong bases are straightforward, and give clear endpoints. Titration of a weak acid and a weak base may be difficult and give endpoints that are less well defined. Acid–Base Titrations01

9. Slide 9 The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.) For a strong acid/strong base titration, the equivalence point should be at pH 7. Acid–Base Titrations02 H + (aq) + OH – (aq) → H 2 O (l) neutral

10. Slide 10 The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.) Titration of a weak acid with a strong base gives an equivalence point with pH > 7. Acid–Base Titrations02 HA (aq) + OH – (aq) → H 2 O (l) + A – (aq) basic

11. Slide 11 Acid–Base Titration Examples03 Titration curve for strong acid–strong base: Note the very sharp endpoint (vertical line) seen with strong acid – strong base titrations. The pH is changing very rapidly in this region. add one drop of base: get a BIG change in pH

12. Slide 12 Titration of 0.10 M HCl03 Volume of Added NaOH 1. Starting pH (no NaOH added) 1.00 2. 20.0 mL (total) of 0.10 M NaOH. 1.48 3. 30.0 mL (total) of 0.10 M NaOH. 1.85 4. 39.0 mL (total) of 0.10 M NaOH. 2.90 5. 39.9 mL (total) of 0.10 M NaOH. 3.90 6. 40.0 mL (total) of 0.10 M NaOH. 7.00 7. 40.1 mL (total) of 0.10 M NaOH.10.10 8. 41.0 mL (total) of 0.10 M NaOH.11.08 9. 50.0 mL (total) of 0.10 M NaOH12.05 pH

13. Slide 13 Acid–Base Titrations04 pH at the equivalence point will always be >7 w/ weak acid/strong base Titration curve for weak acid–strong base: The endpoint (vertical line) is less sharp with weak acid – strong base titrations. strong acid weak acid strong acid equivalence point weak acid equivalence point

14. Slide 14 Acid–Base Titration Curves weak acid very weak acid With a very weak acid, the endpoint may be difficult to detect.

15. Slide 15 Acid–Base Titrations09 Strong Acid–Weak Base: The (conjugate) acid hydrolyzes to form weak base and H 3 O +. At equivalence point only the (conjugate) acid is present. pH at equivalence point will always be <7.

16. Slide 16 Aqueous Solubility Rules for Ionic Compounds A compound is probably soluble if it contains the cations: a. Li +, Na +, K +, Rb + (Group 1A on periodic table) b. NH 4 + A compound is probably soluble if it contains the anions: a. NO 3– (nitrate), CH 3 CO 2 – (acetate, also written C 2 H 3 O 2 – ) b. Cl –, Br –, I – (halides) except Ag +, Hg 2 2+, Pb 2+ halides c. SO 4 2– (sulfate) except Ca 2+, Sr 2+, Ba 2+, and Pb 2+ sulfates Other ionic compounds are probably insoluble. Solubility Equilibria01 Old way to analyze solubility. Answer is either “yes” or “no”.

17. Slide 17 Solubility Equilibria02 measure New method to measure solubility: Consider solution formation an equilibrium process: MCl 2 (s)  M 2+ (aq) + 2 Cl – (aq) Give equilibrium expression K c for this equation: K C = [M 2+ ][Cl – ] 2 K sp This type of equilibrium constant K c that measures solubility is called: K sp

18. Slide 18 Solubility Equilibria03 Solubility Product: is the product of the molar concentrations of the ions and provides a measure of a compound’s solubility. MX 2 (s)  M 2+ (aq) + 2 X – (aq) K sp = [M 2+ ][X – ] 2 “Solubility Product Constant”

19. Slide 19 Al(OH) 3 1.8 x 10 –33 BaCO 3 8.1 x 10 –9 BaF 2 1.7 x 10 –6 BaSO 4 1.1 x 10 –10 Bi 2 S 3 1.6 x 10 –72 CdS8.0 x 10 –28 CaCO 3 8.7 x 10 –9 CaF 2 4.0 x 10 –11 Ca(OH) 2 8.0 x 10 –6 Ca 3 (PO 4 ) 2 1.2 x 10 –26 Cr(OH) 3 3.0 x 10 –29 CoS4.0 x 10 –21 CuBr4.2 x 10 –8 CuI5.1 x 10 –12 Cu(OH) 2 2.2 x 10 –20 CuS6.0 x 10 –37 Fe(OH) 2 1.6 x 10 –14 Fe(OH) 3 1.1 x 10 –36 FeS6.0 x 10 –19 PbCO 3 3.3 x 10 –14 PbCl 2 2.4 x 10 –4 PbCrO 4 2.0 x 10 –14 PbF 2 4.1 x 10 –8 PbI 2 1.4 x 10 –8 PbS3.4 x 10 –28 MgCO 3 4.0 x 10 –5 Mg(OH) 2 1.2 x 10 –11 MnS3.0 x 10 –14 Hg 2 Cl 2 3.5 x 10 –18 HgS 4.0 x 10 –54 NiS 1.4 x 10 –24 AgBr 7.7 x 10 –13 Ag 2 CO 3 8.1 x 10 –12 AgCl 1.6 x 10 –10 Ag 2 SO 4 1.4 x 10 –5 Ag 2 S 6.0 x 10 –51 SrCO 3 1.6 x 10 –9 SrSO 4 3.8 x 10 –7 SnS 1.0 x 10 –26 Zn(OH) 2 1.8 x 10 –14 ZnS 3.0 x 10 –23 Solubility Equilibria - K sp Values04

20. Slide 20 Solubility Equilibria05 The solubility of calcium sulfate (CaSO 4 ) is found experimentally to be 0.67 g/L. Calculate the value of K sp for calcium sulfate. The solubility of lead chromate (PbCrO 4 ) is 4.5 x 10 –5 g/L. Calculate the solubility product of this compound. Calculate the solubility of copper(II) hydroxide, Cu(OH) 2, in g/L.

21. Slide 21 Equilibrium Constants - Review The reaction quotient (Q c ) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction. Q c K c System forms more reactants (left)

22. Slide 22 Equilibrium Constants - Q c Predicting the direction of a reaction. Q c < K c Qc > KcQc > KcQc > KcQc > Kc

23. Slide 23 Solubility Equilibria06 We use the reaction quotient ( Q c ) to determine if a chemical reaction is at equilibrium: compare Q c and K c K sp values are also a type of equilibrium constant, but are valid for saturated solutions only. We can use “ion product” (IP) to determine whether a precipitate will form: compare IP and K sp

24. Slide 24 Solubility Equilibria06 Ion Product (IP): solubility equivalent of reaction quotient ( Q c ). It is used to determine whether a precipitate will form. IP < K sp IP = K sp IP > K sp Unsaturated (more solute can dissolve) Saturated solution Supersaturated; precipitate forms.

25. Slide 25 Solubility Equilibria07 A BaCl 2 solution (200 mL of 0.0040 M) is added to 600 mL of 0.0080 M K 2 SO 4. Will precipitate form? (K sp for BaSO 4 is 1.1 x 10 -10 ) 0.200 L x 0.0040 mol/L =.00080 moles Ba 2+ [Ba 2+ ] = 0.00080 mol  0.800 L = 0.0010 M 0.600 L x 0.0080 mol/L =.00480 moles SO 4 2– [SO 4 2– ] = 0.00480 mol  0.800 L = 0.0060 M [ Ba 2+ ] [ SO 4 2– ]

26. Slide 26 Solubility Equilibria07 IP= [ Ba 2+ ] 1 x [ SO 4 2– ] 1 = (0.0010) x (0.0060) = 6.00 x 10 -6 A BaCl 2 solution (200 mL of 0.0040 M) is added to 600 mL of 0.0080 M K 2 SO 4. Will precipitate form? (K sp for BaSO 4 is 1.1 x 10 -10 ) IP> K sp, so ppt forms

27. Slide 27 Solubility Equilibria07 Exactly 200 mL of 0.0040 M BaCl 2 are added to exactly 600 mL of 0.0080 M K 2 SO 4. Will a precipitate form? If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will precipitation occur?

28. Slide 28 The solubility product (K sp ) is an equilibrium constant; precipitation will occur when the ion product (IP) exceeds the K sp for a compound. If AgNO 3 is added to saturated AgCl, the increase in [Ag + ] will cause AgCl to precipitate. IP = [Ag + ] 0 [Cl – ] 0 > K sp The Common-Ion Effect and Solubility

29. Slide 29 The Common-Ion Effect and Solubility MgF 2 becomes less soluble as F - conc. increses

30. Slide 30 The Common-Ion Effect and Solubility CaCO 3 is more soluble at low pH.

31. Slide 31 Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10 –3 M silver chloride solution. Calculate the solubility of AgBr (in g/L) in: (a) pure water (b) 0.0010 M NaBr The Common-Ion Effect and Solubility