# Buffer Solutions. Buffer Solutions

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1 Chapter 18 Common Ion Effect Buffers and Titration Curves A/B Titrations Salts and Solubility Product The Common Ion Effect and If a solution is made in which the same ion is produced by two different compounds the common ion effect is exhibited. Buffer solutions are solutions that resist changes in ph when acids or bases are added to them. Buffering is due to the common ion effect. 1 A buffer solution is a ecial case of the common ion effect. The function of a buffer is to resist changes in the ph of a solution. Buffer Composition Weak Acid Conj. Base HOAc OAc - H PO - HPO - NH NH The Common Ion Effect and 1. Solutions made of weak acids plus a soluble ionic salt of the weak acid One example of this type of buffer system is: The weak acid - acetic acid CH COOH The soluble ionic salt - sodium acetate NaCH COO The weak acid reacts with bases. CH COOH Na CH COO - CHCOO 0% - - CH COO H O CH COOH OH - Na CH COO H Thesalt anion (a base) reacts with acids. The Common Ion Effect and Example: Calculate the concentration of H and the ph of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate. This is another equilibrium problem with a starting concentration for both the acid and anion. The Common Ion Effect and Compare the acidity of a pure acetic acid solution and the common ion solution. Solution 0.15 M CH COOH [H ] 1.6 x - ph M CH COOH & 0.15 M NaCH COO buffer 1.8 x -5.7 [H ] is 89 times greater in pure acetic acid than in buffer solution

2 The Common Ion Effect and The general expression for the ionization of a weak monoprotic acid is: HA H A The generalized ionization constant expression for a weak acid is: [ H ][ A ] a [ H ] [ HA] [ HA] a [ A ] acid salt 7 The Common Ion Effect and Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation. [ ] [ acid] log H log a log [ salt] multiply by -1 log ph [ ] [ salt] H log a log [ acid] [ salt] pa log [ acid] This version of the Henderson-Hasselbach equation is one method to calculate the ph 8 of a buffer given the concentrations of the salt and acid. Weak Bases plus Salts of Weak Bases. Buffers that contain a weak base plus the salt of a weak base One example of this buffer system is ammonia plus ammonium nitrate. - NH H O NH OH b x b NH NO [ NH ][ OH ] [ NH] [ NH ] [ NH ] 0% [ OH ] NH NO Weak Bases plus Salts of Weak Bases Simple rearrangement of this equation and application of algebra yields the Base form of the Henderson-Hasselbach equation. [ ] [ base] log log log OH multiply by -1 [ salt] [ ] [ salt] log log log OH poh p log b b b [ salt] [ base] [ base] Weak Bases plus Salts of Weak Bases Example: Calculate the concentration of OH- and the ph of the solution that is 0.15 M in aqueous ammonia, NH, and 0.0 M in ammonium nitrate, NH NO. Section 18. HCl is added to pure water. 11 HCl is added to a solution of a weak acid H PO - and its conjugate base HPO -. 1

3 Buffering Action Example: If 0.00 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.0 M in aqueous ammonia and 0.00 M in ammonium chloride, how much does the ph change? Assume no volume change due to addition of the HCl. Buffering Action Example: If 0.00 mole of NaOH is added to 1.00 liter of solution that is 0.0 M in aqueous ammonia and 0.00 M in ammonium chloride, how much does the ph change? Assume no volume change due to addition of the solid NaOH. 1 1 Buffering Action. Original Solution 1.00 L of solution containing 0.0 M NH and 0.00 M NH Cl Original ph 8.95 Acid or base added 0.00 mol NaOH 0.00 mol HCl New ph ph Adding an Acid to a Buffer Problem: What is the ph when 1.00 ml of 1.00 M HCl is added to a)1.00 L of pure water (before HCl,, ph 7.00) b)1.00 L of buffer that has [HOAc[ HOAc] ] M and [OAc - ] M Notice that the ph changes only slightly in each case Preparation of Example: Calculate the concentration of H and the ph of the solution prepared by mixing 00 ml of M acetic acid and 0 ml of 0.0 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. 17 Preparing a Buffer You want to buffer a solution at ph.0. This means [H O ] -ph 5.0 x -5 M It is best to choose an acid such that [H O ] is about equal to a (or ph p a ). then you get the exact [H O ] by adjusting the ratio of acid to conjugate base. [H O [Acid] ] [Conj. base] a 18

4 Preparing a Buffer You want to buffer a solution at ph.0 or [H O ] 5.0 x -5 M POSSIBLE ACIDS a HSO - / SO - 1. x - HOAc / OAc x -5 HCN / CN -.0 x - Best choice is acetic acid / acetate. 19 Preparing a Buffer You want to buffer a solution at ph.0 or [H O ] 5.0 x -5 M [H O ] 5.0 x -5 [HOAc] [OAc - (1.8 x -5 ) ] Solve for [HOAc]/[OAc[ - ] ratio.78 1 Therefore, if you use 0.0 mol of NaOAc and 0.78 mol of HOAc,, you will have ph.0. 0 Preparing a Buffer Preparing a Buffer A final point CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Result: diluting a buffer solution does not change its ph 1 Buffer prepared from 8. g NaHCO weak acid 16.0 g Na CO conjugate base HCO - H O H O CO - What is the ph? Titrations Acid-Base Titrations ph Titrant volume, ml Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the ph increases very slowly. ml

5 Acid-Base Titrations Acid-Base Titrations Additional NaOH is added. ph rises as equivalence point is approached. Additional NaOH is added. ph increases and then levels off as NaOH is added beyond the equivalence point. 5 6 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 0.0 ml of 0.0 M perchloric acid with 0.0 M potassium hydroxide. In this case, we plot ph of the mixture vs. ml of OH added. Note that the reaction is a 1:1 mole ratio. 7 Strong acid titrated with a strong base 8 Acetic acid titrated with NaOH Weak Acid/Strong Base Titration Curves As an example, consider the titration of 0.0 ml of 0.0 M acetic acid, CH COOH, (a weak acid) with 0.0 M OH (a strong base). The acid and base react in a 1:1 mole ratio

6 QUESTION: You titrate 0. ml of of a 0.05 M solution of of benzoic acid with 0.0 M NaOH to to the equivalence point. ph at ph at half-way equivalence point? point? Acid-Base Titration Section 18. QUESTION: You titrate 0. ml of a 0.05 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? What is the ph at the half-way point? HBz NaOH ---> > Na Bz - H O Benzoic acid NaOH ph of solution of benzoic acid, a weak acid 1 C 6 H 5 CO H HBz a 6. x -5 Benzoate ion Bz - b 1.6 x - Acid-Base Reactions QUESTION: You titrate 0. ml of a 0.05 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? What is the ph at the half- way point? Strategy find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate ph. This is a two-step problem 1. stoichiometry of acid-base reaction. equilibrium calculation Figure 18.6 Weak diprotic acid (H C O ) titrated with a strong base (NaOH) Figure 18.7 Weak base (NH ) titrated with a strong acid (HCl( HCl) 5 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: H In H In Color 1 Color [ H ][ In ] a HIn [ ] 6 6

7 Indicators for Acid-Base Titrations Acid-Base Indicators Figure 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. Ag Cl Ag Cl ( aq) ( aq) Solubility Product Constants The equilibrium constant expression for this dissolution is called a solubility product constant. solubility product constant - - [Ag ][Cl ] Solubility Product Constants Solubility Product Constants The solubility product constant,, for a compound is the product of the concentrations of the constituent ions, each raised to the power that correonds to the number of ions in one formula unit of the compound. Consider the dissolution of silver sulfide in water. - Ag S Ag S H O - The solubility product expression for Ag S is: [ ] [ ] Ag S. 9 0% 1 7

8 Solubility Product Constants The dissolution of solid calcium phohate in water is represented as: ( ) ( s ) Ca PO Ca PO H O 0% The solubility product constant expression is: Some Values of Table 18. and Appendix J [ ] [ ] Ca PO. 5 Determination of Solubility Product Constants Example: One liter of saturated silver chloride solution contains g of dissolved AgCl at 5 o C. Calculate the molar solubility of, and for, AgCl. Determination of Solubility Product Constants Example: One liter of saturated calcium fluoride solution contains gram of CaF at 5 o C. Calculate the molar solubility of, and for, CaF. 5 6 Uses of Solubility Product Constants The solubility product constant can be used to calculate the solubility of a compound at 5 o C. Example: Calculate the molar solubility of barium sulfate, BaSO, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 5 o C. For barium sulfate, 1.1 x -. Uses of Solubility Product Constants Example: The solubility product constant for magnesium hydroxide, Mg(OH), is 1.5 x -11. Calculate the molar solubility of magnesium hydroxide and the ph of a saturated magnesium hydroxide solution at 5 o C

9 The Common Ion Effect Adding an ion common to an equilibrium causes the equilibrium to shift back to reactant. The Common Ion Effect in Solubility Calculations Example: Calculate the molar solubility of barium sulfate, BaSO, in 0.0 M sodium sulfate, Na SO, solution at 5 o C. Compare this to the solubility of BaSO in pure water The Reaction Quotient in Precipitation Reactions The reaction quotient, Q, and the of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. Example: We mix 0 ml of 0.0 M potassium sulfate, SO, and 0 ml of 0. M lead (II) nitrate, Pb(NO ) solutions. Will a precipitate form? 51 The Common Ion Effect Calculate the solubility of BaSO in (a) pure water and (b) in 0.0 M Ba(NO ). for BaSO 1.1 x - BaSO (s) Ba (aq) SO - (aq) Solution Solubility in pure water [Ba ] [SO - ] x [Ba ] [SO - ] x x ( ) 1/ 1.1 x -5 M Solubility in pure water 1.1 x -5 mol/l 5 The Common Ion Effect Calculate the solubility of BaSO in (a) pure water and (b) in 0.0 M Ba(NO ). for BaSO 1.1 x - BaSO (s) Ba (aq) SO - (aq) Solution Solubility in pure water 1.1 x -5 mol/l. Now dissolve BaSO in water already containing 0.0 M Ba. Which way will the common ion shift the equilibrium? Will solubility of BaSO be less than or greater than in pure water? 5 The Reaction Quotient in Precipitation Reactions Example: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO ). 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 Na S, is required to reduce the Hg concentration to 1.0 x -8 M? For HgS,.0 x

10 Fractional Precipitation The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation. If a solution contains Cu, Ag, and Au, each ion can be precipitated as chlorides. CuCl Cu Cl AgCl Ag Cl AuCl Au Cl 7 [ Cu ][ Cl ] 1.9 [ Ag ][ Cl ] [ Au ][ Cl ].0 Fractional Precipitation The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation. If a solution contains Cu, Ag, and Au, each ion can be precipitated as chlorides. CuCl Cu Cl AgCl Ag Cl AuCl Au Cl 7 [ Cu ][ Cl ] 1.9 [ Ag ][ Cl ] [ Au ][ Cl ] Fractional Precipitation It is also possible to calculate the amount of Au precipitated before the Ag begins to precipitate, as well as the amounts of Au and Ag precipitated before the Cu begins to precipitate. Fractional Precipitation It is also possible to calculate the amount of Au precipitated before the Ag begins to precipitate, as well as the amounts of Au and Ag precipitated before the Cu begins to precipitate Fractional Precipitation Example: Calculate the percentage of Au ions that precipitate before AgCl begins to precipitate. Use the [Cl - ] from Example 0-9 to determine the [Au ] remaining in solution just before AgCl begins to precipitate. Separating Salts by Differences in A solution contains 0.00 M Ag and Pb. Add CrO - to precipitate red Ag CrO and yellow PbCrO. Which precipitates first? for Ag CrO 9.0 x -1 for PbCrO 1.8 x -1 Solution The substance whose is first exceeded precipitates first. The ion requiring the lesser amount of CrO - ppts. first

11 Separating Salts by Differences in A solution contains 0.00 M Ag and Pb. Add CrO - to precipitate red Ag CrO and yellow PbCrO. PbCrO ppts.. first. (Ag CrO ) 9.0 x -1 (PbCrO ) 1.8 x -1 How much Pb remains in solution when Ag begins to precipitate? 61 11

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