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Acid-Base Equilibrium Part II: Ionization Constants K a and K b

Acid-Base Equilibrium Part II: Ionization Constants K a and K b. Jespersen Chap. 17 Sec 3. Dr. C. Yau Fall 2014. 1. Dissociation Constants of Acids. Strong acids: HCl H + + Cl - HCl + H 2 O H 3 O + + Cl - Weak acids:

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Acid-Base Equilibrium Part II: Ionization Constants K a and K b

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  1. Acid-Base EquilibriumPart II: Ionization Constants Ka and Kb Jespersen Chap. 17 Sec 3 Dr. C. Yau Fall 2014 1

  2. Dissociation Constants of Acids Strong acids: HCl H+ + Cl- HCl + H2O H3O+ + Cl- Weak acids: CH3CO2H CH3CO2- + H+ CH3CO2H+ H2O CH3CO2- + H3O+ Keq = ? Ka = ?

  3. Dissociation of Weak Acids Using HA as general formula of an acid, the chemical equation for the dissociation if HA is a weak acid would be.... Its corresponding equilibrium law would be...

  4. Nitrous acid, HNO2, is a weak acid that's formed in the stomach when nitrite food preservatives encounter stomach acid. There has been some concern that this acid may form carcinogenic products by reacting with proteins. Write the chemical equation for the equilibrium ionization of HNO2 in water and the appropriate Ka expression. 4

  5. Acids that are not molecular Unlike in Gen Chem I, you will be seeing some acids that are charged: CH3CH2NH3+ + H2O ? The general equation would be written as: BH+ + H2O Why do you suppose we use the letter B? What is the equilibrium law?

  6. Write the equation and the equilibrium law for the ionization in water of N2H5+. Do Practice Exer 17.9 & 17.10 p. 781

  7. pH of Weak Acids In the previous lecture we see that... If [HCl] = 10-3 M [H+] = 10-3 M pH = 3 If [CH3CO2H] = 10-3 M pH is not 3. Why? Would you expect it to be larger or smaller than 3? To determine the pH, we must know its Ka

  8. Ka and pKa Remember pH = - log [H+] Similarly, pKa = - log Ka e.g. What is the pKa of acetic acid which has a Ka of 1.8x10-5?

  9. Note the inverse relationship of Ka and pKa. The stronger acid has a larger Ka but smaller pKa

  10. Acid Strength vs. Ka and pKa Acid strength Weaker acid Stronger acids Ka smaller larger pKa larger smaller Remember: Larger [H3O+] means more acidic. Smaller pH means more acidic. Larger Ka means stronger acid. Smaller pKa means stronger acid. STRONG ACID has LARGE Ka and SMALL pKa Ka = 10-15 Ka = 10-3 (pH = 3 is more acidic than pH =4.)

  11. Using the Ka and pKa Table Which is the stronger acid? • formic acid (in ant stings) • barbituric acid (used in production of barbiturate drugs) Phenol has a pKa of 9.89 And butyric acid has a pKa of 4.82. Which is the stronger acid? • phenol B. butyric acid Do Pract. Exer17.11 & 17.12 p. 781

  12. Dissociation Constants of Weak Bases You are already familiar with NH3 being a weak base: NH3 + H2O NH4+ + OH- Kb = Hydrazine, N2H4 is a weak base. Write the equation for the reaction of hydrazine with water and write its expression for its Kb.

  13. Bases that are not molecular Remember that weak acids make strong conjugate bases? Acetic acid is a weak acid. Give the formula of its conjugate base. Now write the equation for its reaction with water:

  14. Solutions of chlorine bleach such as Clorox contain the hypochlorite ion, OCl¯ which is a weak base. Write the chemical equation for the reaction of OCl¯ with water and the appropriate expression for Kb for this anion. Do Pract Exer 17.13 & 17.14 p. 782 14

  15. General Equations ACIDS HA + H2O H3O+ + A− BH+ + H2O H3O+ + B BASES B + H2O BH+ + OH− B− + H2O BH + OH− Note the conservation of charges in these equations.

  16. pKb = - log Kb Again, we see the inverse relationship of Kb and pKb. The stronger base has the larger Kb and smaller pKb. B + HOH BH+ + OH- (Strong bases grab protons more greedily.)

  17. Relationship of Ka and Kb Write the general chemical equation and Ka expression for HA. Write the general chemical equation and Kb expression for the conjugate base of HA. Using the Ka and Kb expressions you just wrote, write the expression for Ka x Kb.

  18. Summary HA + H2O H3O+ + A- pKa = - log Ka B + H2O BH+ + OH- pKb = - log Kb If B = conjugate base of A =A- , A-+H2O HA + OH- Ka x Kb = 1.0x10-14 pKa + pKb = 14.00

  19. What is the Kb value for CN¯? Examine Table 17.2 for Kb values. Note that Table 17.1 and 17.2 are for neutral molecules. You will not find Ka or Kb values for species with a charge. What are you going to do in order to determine Kb for CN¯? Look up Ka of HCN. Calculate Kb from Ka.

  20. With the use of the tables for Ka and Kb, calculate the Kb of OCN¯. Calculate the Ka of the ammonium ion. Ans. Kb = 5x10-11 Ka = 5.6x10-10 Do Pract Exer 17.15 & 7.16 p. 783

  21. Stronger acids give weaker conjugate bases. CONJUGATE BASES ACIDS Fig. 17.3 p. 784

  22. Ka x Kb = 1.0x10-14 If Ka is large, Kb must be small. If Ka is small, Kb must be large. Ka and Kb are INVERSELY PROPORTIONAL. When comparing 2 acids, the weaker acid produces the stronger conjugate base.

  23. Henderson-Hasselbalch Eqn If we can get [A-] = [HA], how would it simplify this equation? This allows us to experimentally determine pKa easily.

  24. How Ka can be determined experimentally: • Prepare a solution of the weak acid and divide it into two halves by volume. • Neutralize one half with a strong base (such as NaOH) and HA become A¯. • Add the un-neutralized half to the neutralized solution and we have "half-neutralized" solution. [HA] = [A-] • Measure the pH. (pH = pKa, calculate Ka from pKa.)

  25. For example, to determine the ionization constant of barbituric acid: 1) Prepare 1000 mL of 0.10 M barbituric acid. 2) Pour 500 mL of this solution into a flask and titrate it to the end point with an aqueous solution of NaOH. 3) Add the other 500 mL (un-neutralized) barbituric acid solution and measure the pH. If the pH is 4.01, what is the Ka of barbituric acid? HA + NaOH  H2O + Na+A– 500 mL Combine and get 0.050 mol A–/1000mL = 0.050M A–0.050 mol HA/1000mL= 0.050M HA [A–] = [HA] 0.050 mol HA • 0.050 • molA– 1000mL 500 mL 0.10 M HA 1.00Lx0.10M M=0.10mol HA 0.050 mol HA

  26. Determination of Ka Another method of getting pKa = pH (and then calculating Ka from pKa) is from a titration curve: • Record the pH while adding a base to the acid and plot pH vs. Volume of Base Added. • Determine on the graph what the volume of base is at the equivalence point (the exact point when all acid has reacted). • Divide this volume by two (the point when exactly half of acid has reacted). • Locate the pH on the graph at this volume of base. This is the point when [HA] = [A-]! Think about this! This is, therefore, the point when pH = pKa.

  27. HA + NaOH Na+A- + HOH Half of 25.00 mL (12.50 mL): Exactly half of acid has reacted Fig. 17.8 p. 815 Titration of Weak Acid with Strong Base This is when phenolphthalein would turn pink. pH= 4.8 Ka = ?

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