1. Continuation of Acid-Base Chemistry

2. CALULATE THE pH OF A STRONG ACID Compute the pH and equilibrium concentrations of all species in a 2 x 10 -4 M solution of HCl. 1) Species: H +, Cl -, HCl 0, OH - 2) Mass action laws: 3) Mass balance: [HCl 0 ] + [Cl - ] = 2 x 10 -4 M 4) Charge balance: [H + ] = [Cl - ] + [OH - ]

3. Assumptions: HCl is a very strong acid so [H + ] >> [OH - ] and [Cl - ] >> [HCl 0 ] Now the only source of H + and Cl - are the dissociation of HCl, so [H + ] = [Cl - ] (this is also apparent from the charge balance) Thus, pH = - log (2 x 10 -4 ) = 3.70, and [Cl - ] = 2 x 10 - 4 M. [OH - ] = K w /[H + ] = 10 -14 /2 x 10 -4 = 5 x 10 -11 M

4. MONOPROTIC ACIDS What are the pH and concentrations of all species in a 0.1 mol L -1 HF solution? 1)Write out important species: 2) Write out all independent reactions and their equilibrium constants 3) Write out mass-balance expressions. 4) Write out the charge-balance expression. 5) Make reasonable assumptions.

5. MONOPROTIC ACIDS What are the pH and concentrations of all species in a 0.1 mol L -1 HF solution? 1) Write out important species: H +, OH -, HF 0, F -. 2) Write out all independent reactions and their equilibrium constants: HF 0  H + + F - H 2 O(l)  H + + OH -

6. 3) Write out mass-balance expressions: 0.1 mol L -1 =  F = [F - ] + [HF 0 ] 4) Write out the charge-balance expression: [H + ] = [F - ] + [OH - ] 5) Make reasonable assumptions: HF is an acid, so [H + ] >> [OH - ] the charge-balance becomes [H + ]  [F - ] = X and the mass-balance becomes: [HF 0 ] = 0.1 - X

7. 6) Solve quadratic equation: a = -1; b = -10 -3.2 = -6.31x10 -4 ; c = 10 -4.2 = 6.31x10 -5 X 1 = -0.00825; X 2 = 0.00765 [H + ] = [F - ] = 7.65x10 -3 mol L -1 ; pH = -log [H + ] = 2.12 [HF 0 ] = 0.1 - 0.00765 = 0.0924 mol L -1

8. 7) Check assumption: 1.318x10 -12 << 7.65x10 -3, so [OH - ] << [H + ]. What if we assumed [HF 0 ] >> [F - ], i.e., [HF 0 ]  0.1? This might be valid because HF is a weak acid. 10 -3.2 = X 2 /0.1 X 2 = 10 -4.2 X = 10 -2.1 = 0.00794 [H + ] = [F - ] = 7.94x10 -3 mol L -1 ; pH = 2.10 [HF 0 ] = 0.1 - 0.00794 = 0.092 mol L -1 The above answer is only 8% different from 0.1. It seems in any case where K A < 10 -3.2, the above assumption should be good!

9. POLYPROTIC ACID What is the pH and concentration of all species in a 0.1 mol L -1 solution of H 3 PO 4 ? 1) Species: H +, OH -, H 3 PO 4 0, H 2 PO 4 -, HPO 4 2-, PO 4 3- 2) Mass action expressions: H 3 PO 4 0  H 2 PO 4 - + H + H 2 PO 4 -  HPO 4 2- + H + HPO 4 2-  PO 4 3- + H +

10. H 2 O(l)  H + + OH - 3) Mass-balance 0.1 mol L -1 = [H 3 PO 4 0 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ] 4) Charge-balance [H + ] = [H 2 PO 4 - ] + 2[HPO 4 2- ] + 3[PO 4 3- ] + [OH - ] 5) Assumptions a) Because H 3 PO 4 0 is an acid [H + ] >> [OH - ] b) Because H 2 PO 4 - and HPO 4 2- are very weak acids and H 3 PO 4 0 is only moderately weak: [H 3 PO 4 0 ] > [H 2 PO 4 - ] >> [HPO 4 2- ] >> [PO 4 3- ] so, 0.1 = [H 3 PO 4 0 ] + [H 2 PO 4 - ] and [H + ] = [H 2 PO 4 - ] = X.

11. 10 -3.1 - 10 -2.1 X - X 2 = 0 X 1 = 0.0245; X 2 = -0.0324 [H + ] = [H 2 PO 4 - ] = 0.0245 mol L -1 ; pH = 1.61 [H 3 PO 4 0 ] = 0.1 - 0.0245 = 0.0755 mol L -1 So [HPO 4 2- ] = 10 -7.0 mol L -1

12. [PO 4 3- ] = 10 -17.79 = 1.62x10 -18 mol L -1 [OH - ] = 10 -14 /10 -1.61 = 10 -12.39 = 4.07x10 -13 mol L -1

13. GRAPHICAL APPROACH TO EQUILIBRIUM CALCULATIONS Consider the monoprotic acid HA: C T = 10 -3 = [HA 0 ] + [A - ]; so [A - ] = C T - [HA 0 ] K A [HA 0 ] = [H + ][A - ] K A [HA 0 ] = [H + ](C T - [HA 0 ]) K A [HA 0 ] = [H + ]C T - [H + ][HA 0 ] K A [HA 0 ] + [H + ][HA 0 ] = [H + ]C T

14. C T K A - K A [A - ] = [H + ][A - ] C T K A = [A - ]([H + ] + K A ) 1) At pH > K A so [H + ] + K A  [H + ] [HA 0 ] = C T ([H + ]/[H + ]) = C T log [HA 0 ] = log C T [A - ] = C T K A /[H + ] log [A - ] = log C T - pK A + pH C T - [A - ] = [HA]

15. 2) pH = pK A ; [H + ] = K A so [H + ] + K A = 2[H + ] [HA 0 ] = C T [H + ]/(2[H + ]) = C T /2 log [HA 0 ] = log C T - log 2 = log C T - 0.301 [A - ] = C T [H + ]/(2[H + ]) = C T /2 log [A - ] = log C T - log 2 = log C T - 0.301 3) pH > pK A ; [H + ] << K A so K A + [H + ]  K A [HA 0 ] = C T [H + ]/K A log [HA 0 ] = log C T + pK A - pH [A - ] = C T K A /K A = C T log [A - ] = C T

16. Speciation diagram for HA with pK A = 5.5 and C T = 10 -3

17. SPECIATION DIAGRAM FOR A DIPROTIC SYSTEM Consider H 2 S with pK 1 = 7.0, pK 2 = 13.0 S T = 10 -3 M = [H 2 S 0 ] + [HS - ] + [S 2- ]

18. 1) pH K 1 > K 2 log [H 2 S 0 ] = log S T log [HS - ] = log (S T K 1 ) + pH log [S 2- ] = log (S T K 2 K 1 ) + 2pH

19. 2) pH = pK 1 K 2 log [H 2 S 0 ] = log S T - 0.301 log [HS - ] = log S T - 0.301 log [S 2- ] = log (S T K 2 /2) + pH

20. 3) pK 1 [H + ] > K 2 log [H 2 S 0 ] = log (S T /K 1 ) - pH log [HS - ] = log S T log [S 2- ] = log (S T K 2 ) + pH

21. 4) pK 1 [H + ] = K 2 log [H 2 S 0 ] = log (S T /2K 1 ) - pH log [HS - ] = log S T - 0.301 log [S 2- ] = log S T - 0.301

22. 5) pK 1 K 2 > [H + ] log [H 2 S 0 ] = log (S T /K 1 K 2 ) - 2pH log [HS - ] = log (S T /K 2 ) - pH log [S 2- ] = log S T

23. Bjerrum plot showing the activities of reduced sulfur species as a function of pH for a value of total reduced sulfur of 10 -3 mol L -1.

24. Easier approach: 10 -7 = [HS - ] [H + ] / H 2 S [HS - ] / [H 2 S] = 10 -7 / [H + ] 10 -12.9 = [S 2- ] [H + ] / HS - [S 2- ] / [HS] = 10 -12.9 / [H + ] The lower the pH, the higher [H + ], the smaller 10 -7 / [H + ], the smaller [HS - ] / [H 2 S] and the higher [H 2 S] The higher the pH, the lower [H + ], the higher 10 -12.9 / [H + ], the higher [S 2- ] / [HS - ] and the higher [S 2- ]

25. This type of diagram is called a Bjerrum diagram,after its originator. Although the previous examples were calculated in detail, it is possible to sketch the relationships based only on the Ks of the dissociation reactions. These will show the pH ranges of the dominant species; i.e. the species that is controlling the chemistry, except for a small region in the vicinity of the K value. This is illustrated in the following slide.

27. THE CO 2 -H 2 O SYSTEM - I Carbonic acid is a weak acid of great importance in natural waters. The first step in its formation is the dissolution of CO 2 (g) in water according to: CO 2 (g)  CO 2 (aq) At equilibrium we have: Once in solution, CO 2 (aq) reacts with water to form carbonic acid: CO 2 (aq) + H 2 O(l)  H 2 CO 3 0

28. THE CO 2 -H 2 O SYSTEM - II In practice, CO 2 (aq) and H 2 CO 3 0 are combined and this combination is denoted as H 2 CO 3 *. It’s formation is dictated by the reaction: CO 2 (g) + H 2 O(l)  H 2 CO 3 * For which the equilibrium constant at 25°C is: Most of the dissolved CO 2 is actually present as CO 2 (aq); only a small amount is actually present as true carbonic acid H 2 CO 3 0.

29. BJERRUM PLOTS These are used for closed systems with a specified total carbonate concentration. They plot the log of the concentrations of various species in the system as a function of pH. The species in the CO 2 -H 2 O system: H 2 CO 3 *, HCO 3 -, CO 3 2-, H +, and OH -. At each pK value, conjugate acid-base pairs have equal concentrations. At pH < pK 1, H 2 CO 3 * is predominant, and accounts for nearly 100% of total carbonate. At pK 1 < pH < pK 2, HCO 3 - is predominant, and accounts for nearly 100% of total carbonate. At pH > pK 2, CO 3 2- is predominant.

30. Bjerrum plot showing the activities of inorganic carbon species as a function of pH for a value of total inorganic carbon of 10 -3 mol L -1. In most natural waters, bicarbonate is the dominant carbonate species!

31. A Bjerrum plot shows the relative importance of the various species in an acid- base system under closed conditions (i.e., the total concentration of all species is constant). For example, for the CO 2 -H 2 O system, a Bjerrum plot shows the concentrations of H 2 CO 3 *, HCO 3 -, CO 3 2-, H +, and OH -, under the condition that the sum of the concentrations of H 2 CO 3 *, HCO 3 - and CO 3 2- is constant. The Bjerrum plot is constructed based partially on the concepts discussed in slide 6. That is: 1) At each pK value, conjugate acid-base pairs have equal concentrations; 2) At pH pK 2, CO 3 2- is predominant. The Bjerrum plot is also constructed assuming that activity coefficients can be neglected. When pH < pK 1, and H 2 CO 3 * is predominant, the concentrations/activities of the other carbonate species can be derived by rearranging the mass-action expressions for the dissociation reactions, and the mass-balance constraint that the sum of the concentrations of H 2 CO 3 *, HCO 3 - and CO 3 2- is constant. For example, rearranging the equation given in the notes to slide 6 yields: log a HCO 3 - = pH - pK 1 + log a H 2 CO 3 * At pH < pK 1, the concentration of H 2 CO 3 * is approximately equal to the total concentration of all carbonate species, and is hence, approximately constant. Thus, the equation shows that, at pH < pK 1, the concentration of bicarbonate increases one log unit for each unit increase in pH. Similar equations can be derived for all the carbonate species in each of the pH ranges of the diagram. For more details, consult Faure (1998) Principles and Applications of Geochemistry, Prentice-Hall (Chapter 9, pp. 123-124).