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Ch. 16: Aqueous Ionic Equilibrium Dr. Namphol Sinkaset Chem 201: General Chemistry II.

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Presentation on theme: "Ch. 16: Aqueous Ionic Equilibrium Dr. Namphol Sinkaset Chem 201: General Chemistry II."— Presentation transcript:

1 Ch. 16: Aqueous Ionic Equilibrium Dr. Namphol Sinkaset Chem 201: General Chemistry II

2 I. Chapter Outline I.Introduction II.Buffers III.Titrations and pH Curves IV.Solubility Equilibria and K sp V.Complex Ion Equilibria

3 I. Last Aspects of Equilibria In this chapter, we cover some final topics concerning equilibria. Buffers are designed to take advantage of Le Châtelier’s Principle. Solubility can be reexamined from an equilibrium point of view. Complex ions are introduced and their formation explained using equilibrium ideas.

4 II. pH Resistive Solutions A solution that resists changes in pH by neutralizing added acid or base is called a buffer. Many biological processes can only occur within a narrow pH range. In humans, the pH of blood is tightly regulated between 7.36 and 7.42.

5 II. Creating a Buffer To resist changes in pH, any added acid or base needs to be neutralized. This can be achieved by using a conjugate acid/base pair.

6 II. How the Buffer Works For a buffer comprised of the conjugate acid/base pair of acetic acid/acetate:  OH - (aq) + CH 3 COOH (aq)  H 2 O (l) + CH 3 COO - (aq)  H + (aq) + CH 3 COO - (aq)  CH 3 COOH (aq) As long as we don’t add too much OH - or H +, the buffer solution will not change pH drastically.

7 II. Calculating the pH of Buffers The pH of a buffer can be calculated by approaching the problem as an equilibrium in which there are two initial concentrations. Since an acid and its conjugate base are both in solution, problem can be solved from a K a or K b point of view.

8 II. Sample Problem A solution was prepared in which [CH 3 COONa] = 0.11 M and [CH 3 COOH] = 0.090 M. What is the pH of the solution? Note that the K a for acetic acid is 1.8 x 10 -5.

9 II. Sample Problem A student dissolves 0.12 mole NH 3 and 0.095 mole NH 4 Cl in 250 mL of water. What’s the pH of this buffer solution? Note that the K b for ammonia is 1.8 x 10 -5.

10 II. A Special Buffer Equation Buffers are used so widely that an equation has been developed for it. This equation can be used to perform pH calculations of buffers. More importantly, it can be used to calculate how to make solutions buffered around a specific pH.

11 II. Henderson-Hasselbalch Eqn.

12 II. The H-H Equation The Henderson-Hasselbalch equation works for a buffer comprised of conjugate acid/base pairs. It works provided the “x is small” approximation is valid.

13 II. Sample Problem A student wants to make a solution buffered at a pH of 3.90 using formic acid and sodium formate. If the K a for formic acid is 1.8 x 10 -4, what ratio of HCOOH to HCOONa is needed for the buffer?

14 II. Sample Problem A researcher is preparing an acetate buffer. She begins by making 100.0 mL of a 0.010 M CH 3 COOH solution. How many grams of CH 3 COONa does she need to add to make the pH of the buffer 5.10 if the K a for acetic acid is 1.8 x 10 -5 ?

15 II. Upsetting the Buffer A buffer resists changes to pH, but it is not immune to change. Adding strong acid or strong base will result in small changes of pH. We calculate changes in pH of a buffer by first finding the stoichiometric change and then performing an equilibrium calculation.

16 II. Illustrative Problem A 1.00 L buffer containing 1.00 M CH 3 COOH and 1.00 M CH 3 COONa has a pH of 4.74. It is known that a reaction carried out in this buffer will generate 0.15 mole H +. If the pH must not change by more than 0.2 pH units, will this buffer be adequate?

17 II. Stoichiometric Calculation When the H + is formed, it will react stoichiometrically with the base. We set up a different type of table to find new equilibrium concentrations. We use ≈0.00 mol because [H + ] is negligible. H + + CH 3 COO -  CH 3 COOH Initial0.15 mol1.00 mol Change-0.15 mol +0.15 mol Final≈0.00 mol0.85 mol1.15 mol

18 II. Equilibrium Calculation We use the new buffer concentrations in an equilibrium calculation to find the new pH. Again, use ≈0.00 M because [H 3 O + ] is negligible. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - Initial1.15 M----≈0.00 M0.85 M Change-x----+x Equil.1.15 – x----x0.85 + x Solve to get [H 3 O + ] = 2.4 4 x 10 -5 M, and pH = 4.61 3.

19 II. How the Buffer Works

20 II. A Simplification for Buffers In buffer problems, # of moles can be used in place of concentration. This can be done because all components are in the same solution, and hence have the same volume.

21 II. Sample Problem Calculate the pH when 10.0 mL of 1.00 M NaOH is added to a 1.0-L buffer containing 0.100 mole CH 3 COOH and 0.100 mole CH 3 COONa. Note that the K a for acetic acid is 1.8 x 10 -5.

22 II. Buffers Using a Weak Base Up until now, we’ve been creating buffers with a weak acid and it’s conjugate base. Can also make a buffer from a weak base and its conjugate acid. Henderson-Hasselbalch still applies, but we need to get K a of the conjugate acid.

23 II. pK a /pK b Relationship

24 II. Sample Problem Calculate the pH of a 1.0-L buffer that is 0.50 M in NH 3 and 0.20 M in NH 4 Cl after 30.0 mL of 1.0 M HCl is added. Note that the K b for NH 3 is 1.8 x 10 -5.

25 II. Making Effective Buffers There are a few parameters to keep in mind when making a buffer:  The relative [ ]’s of acid and conjugate base should not differ by more than factor of 10.  The higher the actual [ ]’s of acid and conjugate base, the more effective the buffer.  The effective range for a buffer system it +/- 1 pH unit on either side of pK a.

26 II. Buffer Capacity Buffer capacity is defined as the amount of acid or base that can be added to a buffer without destroying its effectiveness. A buffer is destroyed when either the acid or conjugate base is used up. Buffer capacity increases w/ higher concentrations of buffer components.

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