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Lect w9 buffers_exercises

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• Lect w9 buffers_exercises

1. 1. Key to Problems Week 9
2. 2. <ul><li>[CH 3 COOH] + H 2 O  [CH 3 COO - ]+ [H 3 O + ] </li></ul><ul><li>initial </li></ul><ul><li>change </li></ul><ul><li>Equilib </li></ul>What is the pH of a buffer that has [CH 3 COOH] = 0.700 M and [CH 3 COO - ] = 0.600 M? CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + K a = 1.8 x 10 -5 Assuming that x << 0.700 and 0.600 , we can use the Henderson-Hasselbalch Equation 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x
3. 3. <ul><li>pH = pK a + log [Base] </li></ul><ul><li> [Acid] </li></ul><ul><li>pK a = -logK a  pK a = -log(1.8 x 10 -5 ) </li></ul><ul><li>pK a = 4.74 </li></ul><ul><li>[Base] = 0.600 M; [Acid] = 0.700 M </li></ul><ul><li>pH = 4.74 + log[0.600] </li></ul><ul><li> [0.700] </li></ul><ul><li>pH = 4.67 </li></ul>
4. 4. <ul><li>You want to buffer a solution at pH = 4.30 and have access to the following acids: </li></ul><ul><li> POSSIBLE ACIDS K a </li></ul><ul><li>HSO 4 - / SO 4 2- 1.2 x 10 -2 </li></ul><ul><li>CH 3 COOH/CH 3 COO - 1.8 x 10 -5 </li></ul><ul><li>HCN / CN - 4.0 x 10 -10 </li></ul>Preparing a Buffer Which one would you choose? If you have a 0.100 M concentration of the base, what concentration of the acid would you need?
5. 5. <ul><li>You want to buffer a solution at pH = 4.30 and have access to the following acids: </li></ul><ul><li> POSSIBLE ACIDS K a pK a </li></ul><ul><li>HSO 4 - / SO 4 2- 1.2 x 10 -2 1.92 </li></ul><ul><li>CH 3 COOH/CH 3 COO - 1.8 x 10 -5 4.74 </li></ul><ul><li>HCN / CN - 4.0 x 10 -10 9.40 </li></ul><ul><li>The acetate buffer would be chosen due to the proximity of the pKa to the desired buffer pH. </li></ul>Preparing a Buffer
6. 6. <ul><li>You want to buffer a solution at pH = 4.30 If you have a 0.100 M concentration of the base, what concentration of the acid would you need? </li></ul><ul><li>pH = pK a + log [Base] </li></ul><ul><li> [Acid] </li></ul><ul><li>pH = 4.30; pK a = 4.74 </li></ul><ul><li>[Base] = 0.100 M; [Acid] = ? </li></ul><ul><li>4.30 = 4.74 + log [0.100] </li></ul><ul><li> [HA] </li></ul><ul><li>-0.44 = log [0.100]  10 -0.44 = [0.100] </li></ul><ul><li>[HA] [HA] </li></ul><ul><li>[HA] = 0.28 M </li></ul>
7. 7. The blood of mammals is an aqueous solution that maintains a constant pH. The normal pH of human blood is 7.40 . Carbon dioxide provides the most important blood buffer. In solution, CO 2 , reacts with water to form H 2 CO 3 , which ionizes to produce H 3 O + and HCO 3 - ions: CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) K a = 4.2 x 10 -7 Use this information to determine the concentration ratio [HCO 3 - ]/[H 2 CO 3 ] in blood.
8. 8. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) K a = 4.2 x 10 -7 Use this information to determine the concentration ratio [HCO 3 - ]/[H 2 CO 3 ] in blood (pH = 7.40). pH = pK a + log [Base] [Acid] pH = 7.40; pK a = 6.38 [Base]/[Acid] = ? 7.40 = 6.38 + log [Base] [Acid] 1.02 = log [ HCO 3 - ]  10 1.02 = [ HCO 3 - ] [ H 2 CO 3 ] [ H 2 CO 3 ] [ HCO 3 - ] = 10.5 [ H 2 CO 3 ]
9. 9. More with Henderson-Hasselbalch pK a (carb. acid) = 2.0 pK a (prot. amine) = 10.5 <ul><li>Draw the structure of the amino acid at </li></ul><ul><li>pH = 1.0, b) pH = 7.0, and c) pH = 12.0 </li></ul><ul><li>Start by using the Henderson-Hasselbalch equation! </li></ul>
10. 10. @pH = 1.0 The acid form (-COOH) is more prevalent. For the carboxylic acid group: The acid form (-NH 3 + ) is more prevalent. For the amine group:
11. 11. C CH 2 O O H H 3 N @pH = 1.0 : : : : + C CH 2 O O H 3 N : : : : + C CH 2 O O H 2 N : : : : @pH = 7.0 @pH = 12.0 pK a (carb. acid) = 2.0 pK a (prot. amine) = 10.5 Key hints: -at low pH’s, the overall molecules will be + or neutral -at high pH’s, the overall charge on the molecules will be neutral or - : - : -
12. 12. C CH 2 O O H H 3 N @pH = 1.0 : : : : + C CH 2 O O H 3 N : : : : + C CH 2 O O H 2 N : : : : @pH = 7.0 @pH = 12.0 : - : - pH = 1.0 pH = 7.0 pH = 12.0 pK a = 2.0 -COOH -COO - pK a = 10.2 -NH 3 + -NH 2 -COO - -NH 3 +
13. 13. Solubility of Salts Consider: AgCl(s)  Ag + (aq) + Cl - (aq) In a saturated solution [Ag + ] =1.34  10 -5 M. What is the concentration of Cl ¯ ions in the system? [Cl - ] = [Ag + ] = 1.34  10 -5 M What is the value of K sp for this salt? K sp =[Ag + ][Cl - ] = (1.34  10 -5 ) 2 = 1.80 x 10 -10
14. 14. Estimate the solubility of the following salt: PbSO 4 (K sp = 1.8  10 -8 ) PbSO 4  Pb 2+ + SO 4 2- I solid 0 0 C -x +x +x E solid x x K sp = [Pb 2+ ][SO 4 2- ] 1.8  10 -8 = [ x ][ x ] solubility = x = 1.3 x 10 -4 M
15. 15. Determine the molar solubility of PbCl 2 ? K sp = 1.7 x 10 -5 PbCl 2  Pb 2+ + 2 Cl - I solid 0 0 C -x +x +2x E solid x 2x K sp = [Pb 2+ ][Cl - ] 2 1.7  10 -5 = [ x ][ 2x ] 2 solubility = x = 0.016 M
16. 16. PbI 2 (s)  Pb 2+ (aq) + 2I - (aq) Calculate K sp for PbI 2 if [Pb 2+ ] = 0.00130 M in a saturated solution. PbI 2  Pb 2+ + 2 I - I solid 0 0 C -x +x +2x E solid x 2x K sp = [Pb 2+ ][I - ] 2 K sp = [ 0.00130 ][ 2(0.00130) ] 2 K sp = 8.79  10 -9
17. 17. <ul><li>Consider the following insoluble salt: CaCO 3 (s) </li></ul><ul><li>Will the solubility of this salt increase, decrease, or remain the same after adding: </li></ul><ul><li>Sulfuric acid (H 2 SO 4 ) </li></ul><ul><ul><li>solubility will  because CO 3 2- is basic: the acid will react with the base, thereby removing a product, shifting the equilibrium to the right </li></ul></ul><ul><li>Hydrochloric acid (HCl) </li></ul><ul><ul><li>solubility will  because CO 3 2- is basic: the acid will react with the base, thereby removing a product, shifting the equilibrium to the right </li></ul></ul><ul><li>Calcium chloride (CaCl 2 ) </li></ul><ul><li>solubility will  because you are adding a common ion (Ca2+) that shifts the equilibrium to the left </li></ul>
18. 18. <ul><li>Consider the following insoluble salt: AgCl(s) </li></ul><ul><li>Will the solubility of these salts increase, decrease, or remain the same after adding: </li></ul><ul><li>Sulfuric acid (H 2 SO 4 ) </li></ul><ul><ul><li>increase; SO 4 2- interacts with Ag + to form an insoluble compound, removing a product and shifting equilibrium to the right </li></ul></ul><ul><li>Hydrochloric acid (HCl) </li></ul><ul><ul><li>decrease; because Cl - is a common ion and shifts the equilibrium to the left </li></ul></ul><ul><li>Calcium Chloride (CaCl 2 ) </li></ul><ul><ul><li>decrease; again, Cl - is a common ion and shifts the equilibrium to the left, thus reducing the solubility </li></ul></ul>
19. 19. Consider the following problem: Imagine you mix 500.0 mL of a solution of AgNO 3 1.5 x 10 -5 M with 500.0 mL of a solution of NaCl 1.5 x 10 -5 M. Will AgCl precipitate? AgCl(s)  Ag + (aq) + Cl - (aq) K sp = 1.8 x 10 -10 Ksp = [Ag + ][Cl - ] First determine actual concentrations. For both: M 1 V 1 =M 2 V 2 M 2 = (1.5  10 -5 mole/L)(0.500 L) (1.000 L) Q = (7.5  10 -6 )(7.5  10 -6 ) = 5.6  10 -11 Q < K, therefore AgCl will not precipitate
20. 20. Consider the case: Hg 2 Cl 2 (s)  Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ]=0.010 M, what [Cl - ] is required to just begin the precipitation of Hg 2 Cl 2 ? At the K sp , there is no precipitate, but is the threshold for when solid will form. K sp = [Hg 2 2+ ][Cl - ] 2 1.1 x 10 -18 = [0.010][Cl - ] 2 [Cl - ] = 1.0 x 10 -8 M
21. 21. Consider a solution containing 0.020 M Cl - and 0.010 M CrO 4 2- ions in which Ag + ions are added slowly. Which precipitates first, AgCl or Ag 2 CrO 4 ? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for AgCl = 1.8 x 10 -10 Solve for [Ag + ]  the one with the lower concentration of Ag + will be the one that precipitates first. K sp = [ Ag + ] 2 [ CrO 4 2- ] 9.0  10 -12 = [ Ag + ] 2 [ 0.010 ] [Ag + ] = 3.0  10 -5 M K sp = [ Ag + ][ Cl - ] 1.8  10 -10 = [ Ag + ][ 0.020 ] [Ag + ] = 9.0  10 -9 M AgCl will precipitate first, because a lower concentration of silver is required to create a precipitate with Cl -
22. 22. 0.020 M Cl - and 0.010 M CrO 4 2- ions in which Ag + ions are added K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for AgCl = 1.8 x 10 -10 Calculate the [Ag + ] when the maximum AgCl is precipitated, but none of the Ag 2 CrO 4 has precipitated. What percent of Cl - remains in solution? [Ag + ] = 3.0  10 -5 M when Ag 2 CrO 4 will be nearly precipitated. At this point, AgCl will have precipitated, but the Ag 2 CrO 4 will not.
23. 23. What percent of Cl - remains in solution? To calculate the [Cl - ] remaining, take the [Ag + ] (at the point where Ag 2 CrO 4 is nearly precipitated) and plug it into the K sp for AgCl and solve for [Cl - ]. [Cl - ] consumed by Ag + : K sp = [Ag + ][Cl - ]  1.8 x 10 -10 = [3.0  10 -5 ][Cl - ] [Cl - ] = 6.0  10 -6 M We know the initial [Cl - ] = 0.020 M [Cl-] remaining 0.020 M Cl - (initially)  6.0  10 -6 M Cl - (consumed by Ag + ) = 0.01999 moles Cl - remaining 0.01999  100% = 99.95% Cl - remaining 0.020
24. 25. <ul><li>What is the pH of a solution of 0.150 M HIO 3 (K a = 0.17) and 0.350 M NaIO 3 ? []/Ka = 0.150/0.17 ≈ 1 </li></ul><ul><li>the approximation doesn’t work!! HIO 3 + H 2 O  IO 3 - + H 3 O + </li></ul><ul><li>I 0.150 0.350 </li></ul><ul><li>C -x +x +x </li></ul><ul><li>E 0.150-x 0.350+x x </li></ul><ul><li>K a = [IO 3 - ][H 3 O + ] 0.17 = [0.350+x][x] </li></ul><ul><li> [HIO 3 ] [0.150-x] </li></ul><ul><li>Quadratic x 2 + 0.52x – 0.255 = 0 </li></ul><ul><li>x = 0.308 = [H 3 O + ] </li></ul><ul><li>pH = 0.511 </li></ul>
25. 26. 2. What is the maximum mass of calcium fluoride that will dissolve in 500. mL of aqueous solution? K sp = 1.46 x 10 -10 MW = 78.08 g/mol CaF 2  Ca 2+ + 2 F - Ksp = [Ca 2+ ][F - ] 2 I 0 0 C +x +2x E x 2x 1.46  10 -10 = (x)(2x) 2 x = 3.32  10 -4 = molar solubility of CaF 2 3.32  10 -4 mol CaF 2  0.500 L  78.08 g CaF 2 = L mol CaF 2 0.0259 g CaF 2 in 500 mL
26. 27. 3. What is the molar solubility of CaF 2 in a solution of 0.10 M NaF? K sp = 1.46 x 10 -10 CaF 2  Ca 2+ + 2 F - Ksp = [Ca 2+ ][F - ] 2 I 0 0.10 C +x +2x E x 0.10+2x 1.46  10 -10 = (x)(0.10+2x) 2 1.46  10 -10 = (x)(0.10) 2 (2x will be small) x = 1.46  10 -8 = molar solubility Even though the soln has F-, some CaF2 is expected to dissolve
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