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Acid and Base Equilibrium

Acid and Base Equilibrium. Chapter 18. Strong Electrolytes. Strong electrolytes ionize or dissociate completely Three classes of strong electrolytes 1. Strong Acids 2. Strong Soluble Bases 3. Most Soluble Salts. Strong Electrolytes.

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Acid and Base Equilibrium

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  1. Acid and Base Equilibrium Chapter 18

  2. Strong Electrolytes Strong electrolytes ionize or dissociatecompletely Three classes of strong electrolytes 1. Strong Acids 2. Strong Soluble Bases 3. Most Soluble Salts

  3. Strong Electrolytes Calculation of concentrations of ions in solution of strong electrolytes is fairly easy Ex.1) Calculate the concentrations of ions in 0.050 M nitric acid, HNO3. Ex. 2) Calculate the concentrations of ions in 0.050 M calcium hydroxide, Ca(OH)2, solution.

  4. The water concentration in dilute aqueous solutions is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions: • The water concentration is many orders of magnitude greater than the ion concentrations. • Thus the water concentration is essentially constant.

  5. The Auto-Ionization of Water Pure water ionizes very slightly • less than one-millionth molar • Because the activity of pure water is 1, the equilibrium constant for this reaction is

  6. Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7M at 250C This particular equilibrium constant is called the ion-product for water, Kw so Kw = 1.0 x 10-14

  7. The pH and pOH scales A convenient way to express acidity and basicity is through pH. pH is defined as If we know either [H3O+] or [OH-], then we can calculate pH and pOH and vise versa. [H3O+] = 10^-pH[OH-] = 10^-pOH

  8. Ex. 3) Calculate the concentrations of H3O+ and OH- in 0.050 MHCl and find the pH of the solution. Ex. 4) The pH of a solution is 4.597. What is the concentration of H3O+?

  9. Ex. 5) Calculate [H3O+], pH, [OH-], and pOH for 0.020 MBa(OH)2 solution. Ex. 6) Calculate the number of H3O+ and OH- ions in one liter of pure water at 250C.

  10. Develop familiarity with pH scale by looking at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14M.

  11. Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: The equilibrium constant for this ionization is expressed as:

  12. Ionization Constants for Weak Monoprotic Acids and Bases Values for several ionization constants

  13. Ionization Constants for Weak Monoprotic Acids and Bases From the above table we see that the order of increasing acid strength for these weak acids is: • HF > HNO2 > CH3COOH >HClO >HCN The order of increasing base strength of the anions (conjugate bases) of these acids is: • F- < NO2- < CH3COO- < ClO- < CN-

  14. Ionization Constants for Weak Monoprotic Acids and Bases Ex. 7) Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant.

  15. Calculation of Ionization Constants Ex. 8) In 0.12 M solution, a weak monoprotic acid, HY, is 5.0% ionized. a) Calculate the concentrations of all species in solution. b) Calculate the ionization constant for the weak acid. • Since the weak acid is 5.0% ionized, it is also 95% unionized.

  16. Calculation of Ionization Constants Ex. 9) The pH of a 0.100 M solution of a weak monoprotic acid, HA, is found to be 2.970. What is the value for its ionization constant? pH = 2.970 so [H+]= 10-pH

  17. Calculations Based on Ionization Constants Ex. 10) Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. Ka = 1.8 x 10-5 • Always write down the ionization reaction and the ionization constant expression.

  18. Calculations Based on Ionization Constants Follow these steps • Combine the basic chemical concepts with some algebra to solve the problem • Substitute the algebraic quantities into the ionization expression. • Solve the algebraic equation. If Ka is less than x 10-4 you can cancel out – x or + x , using the simplifying assumption • Complete the algebra and solve for concentrations.

  19. Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

  20. Calculations Based on Ionization Constants Ex. 11) Now calculate the percent ionization for the 0.15 M acetic acid. From Ex. 10, we know the concentration of CH3COOH that ionizes in this solution is 1.6 x 10-3 M. The percent ionization of acetic acid is % ionization = [CH3COOH] ionized x 100% [CH3COOH] original

  21. Calculations Based on Ionization Constants Ex. 12) Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Then find the % ionization. Ka=4.0 x 10-10 for HCN

  22. Let’s look at the percent ionization of two weak acids as a function of their ionization constants for Ex. 11 & 12 The [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

  23. Weak bases work the same way as weak acids Ex. 13) Calculate the concentrations of the various species in 0.25 M aqueous ammonia and the percent ionization. Kb for ammonia = 1.8 x 10-5

  24. Calculations Based on Ionization Constants Ex. 14) The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution. • Use the ionization expression and some algebra to get the equilibrium concentration.

  25. Polyprotic Acids Many weak acids contain two or more acidic hydrogens. • polyprotic acids ionize stepwise • ionization constant for each step • Consider arsenic acid, H3AsO4, which has three ionization constants • K1=2.5 x 10-4 • K2=5.6 x 10-8 • K3=3.0 x 10-13

  26. Polyprotic Acids • The first ionization step is

  27. Polyprotic Acids • The second ionization step is

  28. Polyprotic Acids • The third ionization step is

  29. Polyprotic Acids • Notice that the ionization constants vary in the following fashion: • This is a general relationship.

  30. Polyprotic Acids Ex. 15) Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4, solution. 1. Write the first ionization ionization step and represent the concentrations. 2. Substitute into the expression for K1. 3. Use the quadratic equation to solve for x, and obtain two values. Can’t use assumption. (too close)

  31. 4. ionization and represent the concentrations. 5. Substitute into the second step ionization expression. 6. Now we repeat the procedure for the third ionization step. 7. Substitute into the third ionization expression. 8. Last Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.

  32. Polyprotic Acids A comparison of the various species in 0.100 M H3AsO4 solution follows.

  33. Solvolysis • Solvolysis is the reaction of a substance with the solvent in which it is dissolved. • Hydrolysis refers to the reaction of a substance with water or its ions. • Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

  34. Hydrolysis at 25oC • in neutral solutions: [H3O+] = [OH-] = 1.0 x 10-7M • in basic solutions: [H3O+] < [OH-] > 1.0 x 10-7M • in acidic solutions: [OH-] < [H3O+] > 1.0 x 10-7M for all conjugate acid/base pairs in aqueous solns. Kw = Ka Kb So if we know the value of either Ka or Kb, the other can be calculated.

  35. Salts of Strong Soluble Bases and Weak Acids Note: This same method can be applied to the anion of any weak monoprotic acid.

  36. Salts of Strong Soluble Bases and Weak Acids Ex. 16) Calculate the hydrolysis constants for the following anions of weak acids. a) F-, fluoride ion, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4. b) CN-, cyanide ion, the anion of hydrocyanic acid, HCN. For HCN, Ka= 4.0 x 10-10.

  37. Salts of Strong Soluble Bases and Weak Acids Ex. 17) Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

  38. Salts of Acids and Bases • Aqueous solutions of salts of strong acids and strong soluble bases are neutral. • Aqueous solutions of salts of strong bases and weak acids are basic. • Aqueous solutions of salts of weak bases and strong acids are acidic. • Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic.

  39. Rain water is slightly acidic because it absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth? Ka for H2 CO3 = 4.2 x 10-7

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