Buffering Action

1. Buffering Action • If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl. • Calculate the pH of the original buffer solution.

2. Buffering Action • Next, calculate the concentration of all species after the addition of the gaseous HCl. • The HCl will react with some of the ammonia and change the concentrations of the species. • This is another limiting reactant problem.

3. Buffering Action • Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated.

4. Buffering Action • Finally, calculate the change in pH.

5. Buffering Action • If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.

6. Buffering Action • pH of the original buffer solution is 8.95, from above. • First, calculate the concentration of all species after the addition of NaoH. • NaOH will react with some of the NH4Cl. • The limiting reactant is the NaOH. • Calculate the pH using the concentrations of the salt and base and the Henderson-Hasselbach equation. • Calculate the change in pH.

7. Notice that the pH changes only slightly in each case. Buffering Action

8. Preparation of Buffer Solutions • Calculate the concentration of H+ and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. • Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. • NaOH and CH3COOH react in a 1:1 mole ratio. • After the two solutions are mixed, Calculate total volume. • The concentrations of the acid and base are: • Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

9. Preparation of Buffer Solutions • For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. • Calculate the number of moles of solid ammonium chloride, NH4Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15. • Because pH = 9.15, the pOH can be determined. • The appropriate equilibria representations are: • Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia

10. Titration Curves Strong Acid/Strong Base Titration Curves • These graphs are a plot of pH vs. volume of acid or base added in a titration. • As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide. • In this case, we plot pH of the mixture vs. mL of KOH added. • Note that the reaction is a 1:1 mole ratio.

11. Strong Acid/Strong Base Titration Curves • Before any KOH is added the pH of the HClO4 solution is 1.00. • Remember perchloric acid is a strong acid that ionizes essentially 100%. • After a total of 20.0 mL 0.100 M KOH has been added the pH of the • reaction mixture is ___ • After a total of 50.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___ • After a total of 90.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ____ • After a total of 100.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___

12. Strong Acid/Strong Base Titration Curves • We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

13. Weak Acid/Strong Base Titration Curves • As an example, consider the titration of 100.0 mL of 0.100 M acetic acid, CH3 COOH, (a weak acid) with 0.100 M KOH (a strong base). • The acid and base react in a 1:1 mole ratio.

14. Weak Acid/Strong Base Titration Curves • Before the equivalence point is reached, both CH3COOH and KCH3COO are present in solution forming a buffer. • The KOH reacts with CH3COOH to form KCH3COO. • A weak acid plus the salt of a weak acid form a buffer. • Hypothesize how the buffer production will effect the titration curve.

15. Weak Acid/Strong Base Titration Curves • Determine the pH of the acetic acid solution before the titration is begun.

16. Weak Acid/Strong Base Titration Curves • After a total of 20.0 mL of KOH solution has been added, the pH is:

17. Weak Acid/Strong Base Titration Curves • At the equivalence point, the solution is 0.500 M in KCH3COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. • Both processes make the solution basic. • The solution cannot have a pH=7.00 at equivalence point. • Let us calculate the pH at the equivalence point.

18. Weak Acid/Strong Base Titration Curves • After the equivalence point is reached, the pH is determined by the excess KOH just as in the strong acid/strong base example.

19. Weak Acid/Strong Base Titration Curves • We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

20. Strong Acid/Weak BaseTitration Curves • Titration curves for Strong Acid/Weak Base Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted. • Weak Acid/Weak Base Titration curves have very short vertical sections. • The solution is buffered both before and after the equivalence point. • Visual indicators cannot be used.

21. Acid-Base Indicators • The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point. • The point in a titration at which a chemical indicator changes color is called the end point. • A symbolic representation of the indicator’s color change at the end point is: • The equilibrium constant expression for an indicator would be expressed as:

22. Acid-Base Indicators • If the preceding expression is rearranged the range over which the indicator changes color can be discerned.

23. Acid-Base Indicators Color change ranges of some acid-base indicators

24. Solubility Product Constants • Silver chloride, AgCl,is rather insoluble in water. • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. • The equilibrium constant expression for this dissolution is called a solubility product constant. • Ksp = solubility product constant

25. Solubility Product Constants • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

26. Solubility Product Constants • The same rules apply for compounds that have more than two kinds of ions. • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

27. Determination of Solubility Product Constants • One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl. • The molar solubility can be easily calculated from the data:

28. Determination of Solubility Product Constants • The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are: • Substitution of the molar concentrations into the solubility product expression gives:

29. Determination of Solubility Product Constants • One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2. • Calculate the molar solubility of CaF2.

30. Determination of Solubility Product Constants • From the molar solubility, we can find the ion concentrations in saturated CaF2. Then use those values to calculate the Ksp. • Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!

31. Uses of Solubility Product Constants • The solubility product constant can be used to calculate the solubility of a compound at 25oC. • Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.

32. Uses of Solubility Product Constants • The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC. Be careful, do not forget the stoichiometric coefficient of 2! • Substitute the algebraic expressions into the solubility product expression. • Solve for the pOH and pH.

33. The Common Ion Effect in Solubility Calculations • Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (What is the common ion?) Write equations to represent the equilibria. Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x. • The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8M. • The molar solubility of BaSO4 in pure water is 1.0 x 10-5M. • BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate! • Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution! • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

34. The Reaction Quotient in Precipitation Reactions • The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. • We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form? Write out the solubility expressions. • Calculate the Qsp for PbSO4. • Assume that the solution volumes are additive. • Concentrations of the important ions are: • Finally, calculate Qsp for PbSO4 and compare it to the Ksp.

35. The Reaction Quotient in Precipitation Reactions • Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8M? For HgS, Ksp=3.0 x 10-53. • What volume of the solution (1.0 x 10-8M Hg2+ ) contains 1.0 g of mercury? • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

36. Simultaneous Equilibria Involving Slightly Soluble Compounds • If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? • For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5. • Calculate Qsp for Mg(OH)2 and compare it to Ksp. • Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M. • Aqueous ammonia is a weak base that we can calculate [OH-]. • Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp. • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

37. Simultaneous Equilibria Involving Slightly Soluble Compounds • Use the ion product for water to calculate the [H+] and the pH of the solution. • How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+. • Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5M. • Check these values by calculating Qsp for Mg(OH)2.

38. Dissolving Precipitates • Use of an acid in a metathetic reaction. For example, look at the dissolution of Mg(OH)2 in HCl. • A second method is to dissolve insoluble metal carbonates in strong acids. • The carbonates will form soluble salts, carbon dioxide, and water. • A third method is to convert an ion to another species by an oxidation-reduction reaction. • For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur. • A fourth method is complex ion formation. • The cations in many slightly soluble compounds will form complex ions. • This is the method used to dissolve unreacted AgBr and AgCl on photographic film. • Photographic “hypo” is Na2S2O3. • Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+.