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Equilibrium Chemistry

Equilibrium Chemistry. CE 541. Important to: Determine the relationship between constituents in water Understand the effect of alterations of water on the different chemical species present Limitations of Equilibrium Calculations Dynamic changes (in wastewater and surface water)

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Equilibrium Chemistry

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  1. Equilibrium Chemistry CE 541

  2. Important to: • Determine the relationship between constituents in water • Understand the effect of alterations of water on the different chemical species present • Limitations of Equilibrium Calculations • Dynamic changes (in wastewater and surface water) • Due to exposure to sun • Due to exposure to pollution (organic and inorganic) • Rapid reactions (reaction between acids and bases) • Very slow reactions (oxidation-reduction in natural waters) • Precipitation reactions • Lack of information on accurate equilibrium constants for many of the reactions taking place in natural waters

  3. Ion Activity Coefficients “The activity of ion or molecule can be found by multiplying its molar concentration by an activity coefficient, ” {A} =  [A] {A} = activity [A] = concentration  = activity coefficient For practical reasons and rough calculations [A] is used in place of {A}

  4. Some investigators found that “the activity coefficients for ions in an electrolyte were related to the concentration of charged particles in the solution”. They introduced the ionic strength as an empirical measure of the interactions among all the ions in a solution. •  = ionic strength • Ci = molar concentration of the ith ion • Zi = charge of the ith ion Langelier estimated  as (TDS  2.5  10-5)

  5. Other investigators found that, for dilute solution, there is a relationship between  and  as follows: This relationship is used for dilute solutions with ionic strength < 0.1 (in Environmental Engineering, most waters of interest are more dilute than this, except seawater) This relationship is used for solutions with  up to 0.5 M.

  6. Conclusion There is no good relationship that provides a satisfactory estimate of  for  > 0.5 M. In this course,  will be assumed to be equal to 1 unless otherwise mentioned. Study Example 1 page 108.

  7. Problem 4.2 Calculate the activity coefficient and activity of each ion in a solution containing 75 mg/l Na+, 25 mg/l Ca2+, 10 mg/l Mg2+, 125 mg/l Cl-, 50 mg/l HCO3-, and 48 mg/l SO42-.

  8. Solutions to Equilibrium Problems Le Chatelier’s Principle states that “A chemical system will respond to change with processes which tend to reduce the effect of the change” For any chemical reaction • Principle of conservation of mass must be obeyed • Electroneutrality must be maintained “All positively charged species in solution must be balanced by equivalent numbers of negatively charged species” • Proton condition “Species with an excess of protons must be balanced by the species with a deficiency in protons” • All reactions involved must proceed towards a state of equilibrium.

  9. Steps to Solve Equilibrium Problem Involving Aqueous Phase only • Define the equilibrium problem • what chemical reactions are taking place • what is reacting with what • List all constituents of the system • all systems involving water include • H2O • H+ • OH- • Include all ions, elements and neutral species present initially

  10. Steps to Solve Equilibrium Problem Involving Aqueous Phase only • For each element present initially • list all forms or species which are likely to contain the element and which are likely to present after equilibrium is attained • Identify concentrations of all species for each element so that mass and charge balances can be made • List all appropriate equilibrium relationships between species of concern

  11. Steps to Solve Equilibrium Problem Involving Aqueous Phase only • List associated equilibrium constants • List all mass and charge balance relationships for the system • List proton conditions • Steps 5 to 8 will produce a number of equations equal to the unknown species. • Solve the equations simultaneously

  12. Steps to Solve Equilibrium Problem Involving Aqueous Phase only • If gaseous or solid phase are involved, then equations expressing mass and charge balances between and within each phase must be included. Study Example 2

  13. Problem 4.4 A solution is prepared by diluting 10-2 mol of ammonia to 1 liter with distilled water. Calculate the equilibrium concentration for each chemical species in the water.

  14. Acids and Bases • strong acids and bases ionize completely in dilute solutions (water) • weak acids and bases ionize partially in dilute solutions (water) • acids increase H+ concentration • Bases increase OH- concentration

  15. [H+][OH-] = Kw [ ] = activity or approximately molar concentration [H+] is expressed in terms of pH. pH has an effect on: • Equilibrium between most of the chemical species • Effectiveness of coagulation • Potential of water to be corrosive • Suitability of water to microorganisms • Other quality characteristics of the water • Thus it is very important to understand the factors that have an effect on pH of water.

  16. The pH and p(x) Concept In pure water (no other materials): • Activity = Molar concentration • [H+] = [OH-] • Kw = 10-14 @ 25 C • [H+][OH-] = 10-14 • [H+] = [OH-] = 10-7 • pH = 7 (is the neutral pH)

  17. pH Meters • scale 0 to 14 • pH less than 7 indicates acidic condition; [H+] > [OH-] • pH more than 7 indicates basic condition; [H+] < [OH-] • electrodes measure hydrogen-ion activity not molar concentration

  18. The concept of expressing [H+] activity can be used with other ions. So • x = concentration of a given chemical species or equilibrium constant. Then

  19. Since,[H+][OH-] = Kw = 10-14 @ 25 C Then,-log[H+] – log[OH-] = -log Kw pH + pOH = pKw pKw = 14 @ 25 C for weak acids and bases: • pKA is the negative log of the ionization constant for weak acids • pKB is the negative log of the ionization constant for weak bases Tables 4-1 and 4-2 show KA, pKA, KB, and pKB for weak acids and bases.

  20. For weak acid and its conjugate base or weak base and its conjugate acid: pKA + pKB = 14 @ 25 C or KAKB = 10-14 = Kw Example Boric acid haspKA = 9.24 Borate haspKB = 4.76 pKA + pKB = 14

  21. Solving Acid – Base Equilibrium Problems Assumptions • equilibrium occurs very fast (thus neglect kinetic considerations) • strong acids and bases are completely ionized in water (except when the added concentration is ≈ 10-7)

  22. Tools to be used in Solving Problems • equilibrium relationships • mass balance • charge balance • proton condition • How to Solve Problems • identify unknowns • generate equations = unknowns • solve equations simultaneously • use graphical solutions • use computers for complex problems Study Examples 3, 4, 5, and 6

  23. Problem 4.5 Calculate the equilibrium pH of a solution containing (a) 10-3 M H2SO4; (b) 10-8 M H2SO4

  24. Logarithmic Concentration Diagram Log C – pH diagram represents mass balance of each constituent at every pH value. To construct the diagram, develop equations of C as a function of pH, Kw, KA, CT

  25. Logarithmic Concentration Diagram for Monoprotic Acids and Bases Diagram of Monoprotic Acid 0.02 M acetic acid solution. Use: the above acetic acid equations can be used for all monoprotic acids. Monoprotic acid is an acid which yields one proton when ionized.

  26. Line of [H+] is obtained from: log [H+] = -pH Line of [OH-] is obtained from: log [OH-] = pH – pKw Concentration of Acetic Acid and Acetate can be obtained from:

  27. Intersection of [HAc] and [Ac-] lines is called the System Point and is located at: pH = pKA To left of the system point, [H+] > KA, so Both lines pass through the system point

  28. To the right of the system point, [H+] < KA, so Both lines pass through the system point

  29. Just below the system point and from[HAc] + [Ac-] = CT [HAc] = [Ac-] = (1/2)CT log [(1/2)CT] = log CT + log (1/2) = log CT – 0.3 So, curves of horizontal and diagonal lines intersect at a point 0.3 below the system point.

  30. General Procedure for Construction of logC-pH Diagrams for Monoprotic Acids and Bases • Draw horizontal line representing log CT • Locate the system point at pH = pKA • Draw 45 lines sloping to left and right of the system point • Locate a point of 0.3 logarithmic units below the system point • Connect horizontal and 45 lines with short curves passing through the points • [H+] and [OH-] lines are drawn as 45 lines which intersect at pH = 7 and log C = -7 • Change in concentration will only shift the [HAc] and [Ac-] curves up or down

  31. Diagram of Monoprotic Base • Similar procedure • System point at pH = pKw – pKB 0.01 M NH3 Solution For ammonia, pKB = 4.74 so, system point is located at: 14 – 4.74 = 9.26

  32. Logarithmic Concentration Diagram for a Weak Acid and a Weak Base 0.1 M acetic acid and 0.1 M ammonia The logC-pH diagram is made by superimposing the curves for each material on a single diagram. Study Examples 11 to 15

  33. Logarithmic Concentration Diagram for Polyprotic Acids and Bases Take a solution containing 0.01 M carbonic acid (H2CO3) as an example.

  34. Solving the equation for individual carbonic species gives

  35. The diagram was constructed in the same manner as in the case of monoprotic acids and bases except that: • Slope of the line for [CO32-] changes from +1 to +2 when pH drops below pKA1 ([H+] >> KA1) • Slope of the line for [H2CO3*] changes from -1 to -2 when pH becomes greater than pKA2 ([H+] << KA2)

  36. General Procedure for Construction of logC-pH Diagrams for Diprotic Acids and Bases • Draw horizontal line representing log CT • Locate the system point at pH values equal to pKA1 and pKA2 (pKw-pKB1 and pKw-pKB2 for a base) • Draw 45 lines sloping to left and right of each system point to the adjacent system point • The slope of lines changes from -1 to -2 and from +1 to +2 • The procedure for construction of logC-pH diagram for polyprotic acids and bases are similar except that the slope of the diagonal changes from +1 to +2 and then from +2 to +3 as it reaches the adjacent system points.

  37. Problem 4.25 Draw a log C-pH diagram for a 10-2 M solution of hydrogen sulfide. Assume a closed system. From the diagram, determine the pH for solutions that contain the following: • 10-2 M H2S • 10-2 M Na2S • 0.5  10-2 M HS- and 0.5  10-2 M S2- • 0.5  10-2 M H2S and 0.5  10-2 M HS-

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