Acids and Bases Lecture 2

1. Acids and Bases Lecture 2

2. Homework Ch 7 due Wednesday Oct 23 1,4,6,9,18,19,22,29,34,35

3. What is the pKa of a weak acid if a 0.100 M solution has a pH of 4.68?

4. What is the pKa of a weak acid if a 0.100 M solution has a pH of 4.68? • Let HX represent the weak acid. The chemical equation describing its ionization is HX < == > H + + X – and Ka = [H +] [ X –] / [HX]

5. What is the pKa of a weak acid if a 0.100 M solution has a pH of 4.68? • Let HX represent the weak acid. The chemical equation describing its ionization is HX < == > H + + X – and Ka = [H +] [ X –] / [HX] • Note that [H +] = [ X –] since both come from the ionization reaction; let x = [H +] = [ X –] = 10-pH = 10-4.68 = 2.09 x 10 -5.

6. What is the pKa of a weak acid if a 0.100 M solution has a pH of 4.68? • Let HX represent the weak acid. The chemical equation describing its ionization is HX < == > H + + X – and Ka = [H +] [ X –] / [HX] • Note that [H +] = [ X –] since both come from the ionization reaction; let x = [H +] = [ X –] = 10-pH = 10-4.68 = 2.09 x 10 -5. • Substitution into the Ka expression, Ka = (2.09 x 10 -5)2 / 0.100 = 4.4 x 10 -9

7. Weak Bases Weak bases are treated much like weak acids. The chemical equation for their ionization is B + H2O< == > BH + + OH – and the equilibrium constant Kb is defined as Kb = [BH +] [OH –] / [B]

8. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5?

9. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5? • Let B represent the weak base. The chemical reaction for its ionization is B + H2O< == > BH + + OH –

10. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5? • Let B represent the weak base. The chemical reaction for its ionization is B + H2O< == > BH + + OH – • Let x = [OH –]; x also = [BH +] since both come from the ionization of B. Substitution into the Kb expression, Kb = 7.2 x 10 -5 = X2 / (0.500 – X)

11. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5? • Let B represent the weak base. The chemical reaction for its ionization is B + H2O< == > BH + + OH – • Let x = [OH –]; x also = [BH +] since both come from the ionization of B. Substitution into the Kb expression, Kb = 7.2 x 10 -5 = X2 / (0.500 – X) • Assume that X << 0.500; (this allows us to obtain a solution without solving the quadratic equation that would result.) It is an assumption that we will need to confirm after we have a solution. X2 = (7.2 x 10 -5)(0.500) = 3.6 x 10 -5; X = 6.0 x 10 -3

12. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5? • Let B represent the weak base. The chemical reaction for its ionization is B + H2O< == > BH + + OH – • Let x = [OH –]; x also = [BH +] since both come from the ionization of B. Substitution into the Kb expression, Kb = 7.2 x 10 -5 = X2 / (0.500 – X) • Assume that X << 0.500; (this allows us to obtain a solution without solving the quadratic equation that would result.) It is an assumption that we will need to confirm after we have a solution. X2 = (7.2 x 10 -5)(0.500) = 3.6 x 10 -5; X = 6.0 x 10 -3 • Before we continue, we need to check the above assumpution. Is 0.006 << 0.500? Yes, less than 2% ionization, so the assumption was okay.

13. What is the pH of a 0.500 M solution of a weak base whose Kb = 7.2 x 10 -5? • Let B represent the weak base. The chemical reaction for its ionization is B + H2O< == > BH + + OH – • Let x = [OH –]; x also = [BH +] since both come from the ionization of B. Substitution into the Kb expression, Kb = 7.2 x 10 -5 = X2 / (0.500 – X) • Assume that X << 0.500; (this allows us to obtain a solution without solving the quadratic equation that would result.) It is an assumption that we will need to confirm after we have a solution. X2 = (7.2 x 10 -5)(0.500) = 3.6 x 10 -5; X = 6.0 x 10 -3 • 694 we continue, we need to check the above assumpution. Is 0.006 << 0.500? Yes, less than 2% ionization, so the assumption was okay. • Since X = [OH –], pOH = 2.22 and pH = 11.78

14. Very dilute solutions of weak acids (bases) or weak acids (bases) with relatively large Ka values If the ratio of [HA] : Ka > ~ 1000, {for bases, [B] : Kb > 1000}, then we can obtain reasonable values without solving the second-order equation (quadratic equation).

15. Very dilute solutions of weak acids (bases) or weak acids (bases) with relatively large Ka values In the previous problem the ratio of [B] : Kb = 0.500 / 7.2 x 10 -5 = 6940. Another way to determine is (X << [ Y ]initial ) is to look at the fraction (percent) dissociation. If the % dissociation < ~ 10%, the contribution of the X term in ([ Y ]initial - X) does not affect the answer.

16. Very dilute solutions of weak acids (bases) or weak acids (bases) with relatively large Ka values When it is determined by either of the two methods significant ionization has occurred, you may use either the quadratic formula or successive approximations to solve for the value of X. Both methods are described in detail in Box 7-1, pages 152-3 of your textbook.

17. Review of Conjugate Acids and Bases The conjugate base of a weak acid is a weak base, and the conjugate acid of a weak base is a weak acid. The relationship between their acidity/basicity is given by the express Ka X Kb = Kw, or in p functions. pKa + pKb = pKw = 14.0

18. Chapter 8 – pH Buffers

19. Chapter 8 – pH Buffers pH buffer solutions resist changes in pH whenever small amounts of acids or bases are added, or when the buffer solution is diluted.

20. Chapter 8 – pH Buffers Buffer solutions consist of a mixture of a weak acid (or base) and its conjugate member.

21. Chapter 8 – pH Buffers The pH region that a specific buffer is most effective is in the region of the pKa of the acid form.

22. Chapter 8 – pH Buffers The concentrations of the weak acid and conjugate base forms (or weak base and conjugate acid) are approximately equal in this ‘best’ pH region.

23. Chapter 8 – pH Buffers The chemical reaction that describes the action of the buffer is the same as that for the weak acid, that is HX < === > H+ + X – The Ka is also as met before, namely, Ka = [H+ ] [X –] / [HX]