1 / 16

chap.15: applications of aqueous equilibria

15-2. . The Common-Ion EffectConsider a solution of HF (Ka = 7.2 x 10

elina
Download Presentation

chap.15: applications of aqueous equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

    1. 15-1 Chap.15: Applications of Aqueous Equilibria Overview Buffer Solutions Acid-Base Reactions, Titrations Solubility Equilibria

    2. 15-2

    3. 15-3 Now consider an aqueous NH3 solution. NH3(aq) + H2O(l) ? NH4+(aq) + OH(aq) Add some solid NH4Cl: NH4Cl(s) ? NH4+(aq) + Cl(aq) The increase in [NH4+] will shift this equilibrium to the left, decreasing [OH]. Conclusion: The addition of a common ion will repress the dissociation of a weak acid or weak base.

    4. 15-4 Buffer Solution: contains a weak acid plus its salt (conjugate base) or a weak base plus its salt (conjugate acid). With both an acidic and a basic species present, buffer solutions can resist changes in pH. Ex 1. A solution contains 0.10 mol HC2H3O2 and 0.20 mol of NaC2H3O2 in 1.0 L total volume. What is the pH of this solution? Always consider complete reactions before equilibria.

    5. 15-5 NaC2H3O2(aq) ? Na+(aq) + C2H3O2(aq) 0 0.20 M 0.20 M neutral HC2H3O2(aq) ? H+(aq) + C2H3O2(aq) I. 0.10 M 0 0.20 M C. -x +x +x E. 0.10x x 0.20+x Now set up the mass-action expression. Note that there are now two contributions to the acetate ion (the salt provides 0.20 M and the acid provides x M).

    6. 15-6 By our usual approximation, (0.10x) 0.10. But then we can also say that (0.20 + x) 0.20. Solving, x = [H+] = 9.0 x 106 M and pH = 5.05

    7. 15-7 In buffer solutions, the initial concentrations of both the weak acid (or base) and the salt are generally >> x. Then, Now, we can derive another useful relationship from this.

    8. 15-8 Taking the logarithm of both sides: And rearranging: This is the Henderson-Hasselbalch equation for buffer solutions.

    9. 15-9 In our example, pKa = log(1.8 x 105) = 4.74 Then, pH = 4.74 + log (0.20 / 0.10) = 4.74 + 0.30 = 5.04 Ex 2: A solution is 0.15 M in NH3 and 0.25 M in NH4Cl. What is the pH of this solution? First, the salt dissociates: NH4Cl ? NH4+ + Cl 0 0.25 M 0.25 M neutral ion

    10. 15-10 Now, the weak base establishes equilibrium: NH3 + H2O ? NH4+ + OH I. 0.15 0.25 0 C. -x +x +x E. 0.15-x 0.25+x x x = [OH] = 1.1 x 105 M pOH = 4.96 pH = 9.04

    11. 15-11 In order to use the Henderson-Hasselbalch equation, we need the Ka for the acidic component of the buffer. In this case, it is NH4+. Since salts do not appear in weak acid tables, we need to calculate Ka. Ka for NH4+ = Kw / Kb for NH3 = (1.00 x 1014) / (1.8 x 105) = 5.6 x 1010 Then, pKa = log (5.6 x 1010) = 9.25

    12. 15-12 Now, pH = pKa + log { [CB]o / [CA]o } = 9.25 + log (0.15 / 0.25) = 9.03 The I.C.E. method and the Henderson-Hasselbalch equation give the same result. Next, we will see how to prepare a buffer solution of a specified pH.

    13. 15-13 Preparation of Buffer Solutions ex. How much solid NaC2H3O2 must be added to 100 mL of 0.10 M HC2H3O2 to make a buffer of pH 3.65? Assume that there is no volume change. We will use the Henderson-Hasselbalch equation. pH = pKa + log { [CB]o / [CA]o }

    14. 15-14 We have seen before that the pKa for acetic acid is 4.74. We know that the initial concentration of the acid is 0.10 M. 3.65 = 4.74 + log { [CB]o / (0.10) } log { [CB]o / (0.10) } = 3.65 4.74 = 1.09 { [CB]o / (0.10) } = 10 1.09 = 8.13 x 102 [CB]o = (0.10) (8.13 x 102) = 8.13 x 103 M (8.13 x 103 mol/L) (0.100 L) (82.0 g/mol) = 0.067 g of NaC2H3O2

    15. 15-14a Summary of the Four Types of pH Problems 1) Strong acid only or strong base only: complete ionization or dissociation. [H+] or [OH] is given by the initial acid or base concentration. Weak acid only or weak base only: the substance is in equilibrium with its ions. The standard I.C.E. setup is used. Salt of weak acid or salt of weak base only. a) Salt dissociates completely. b) One of the ions acts as a weak acid or base. The K value must be calculated.

    16. 15-14b Buffer Solutions: contain weak acid or base and its salt. a) The salt dissociates completely. b) The weak acid or base is in equilibrium. Use the I.C.E. approach, where the initial concentration of the conjugate is due to the salt.

More Related