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Chapter 17_Fall2011_recovered_student_3slides

Chapter 17_Fall2011_recovered_student_3slides - Chapter 17...

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Chapter 17 Additional Aspects of Aqueous Equilibria 1 1. Common Ion Effect 2. Buffered Solutions 3. Acid-Base Titrations 4. Solubility Equilibria 5. Factors That Affect Solubility 6. Precipitation and Separation of Ions 7. Qualitative Analysis for Metallic Elements (won’t cover this) 2 1. Common Ion Effect ) ( F ) ( O H ) O( H ) F( H - 3 2 aq aq l aq +  → + + Consider an aqueous HF solution, How will the addition of NaF to this solution affect the equilibrium? ) ( F ) ( a N % 100 O H ) ( NaF - 2 aq aq s + + F - concentration increases. HF equilibrium shifts to the left (Le Châtelier). Fluoride is termed a common ion – an ion common to two solutes. Common Ion Effect 3 Common Ion Effect: The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte or A shift in equilibrium caused by the addition of an ion already present in solution
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4 Example: What is the pH of a solution 0.10 M in HCN and 0.20 M in NaCN? Solution : ) ( CN ) ( H ) HCN( - aq aq aq +  → + I 0.10 0 0.20 C -x +x +x E 0.10 - x x 0.20 + x [ ] 10 a 10 6.2 x] [0.10 x] [x][0.20 HCN ] ][CN [H K - - × = - + = = 10 - a 10 6.2 (0.10) x)(0.20) K × ( ] = × [H 10 3.10 x 10 - pH = - log [H + ] = - log (3.10 x 10 -10 ) = 9.51 Compare: The pH of 0.10 M HCN is 5.15. 5 2. Buffered Solutions Buffered solutions (buffers) = solution that resists pH change when small amounts of acid or base is added. Example: 0.10 M HF + 0.15 M Na F 0.05 M NH 3 + 0.10 M NH 4 Cl buffer applications: biological systems, e.g., human blood, saliva soil chemistry A- Buffer composition : a weak acid and its conjugate base , or a weak base and its conjugate acid B. How does a buffer solution work? If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water. Similarly, if acid is added, the F reacts with it to form HF and water. 6
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7 General Case: Assume an HX + NaX buffer. If base (e.g., NaOH) is added, HX (aq) + OH - (aq) → H 2 O( l ) + X - (aq) the base is neutralized. • If acid (e.g., HCl) is added, X - (aq) + H + (aq) → HX (aq) the acid is neutralized. Calculating the pH of a Buffer Consider weak acid HX, [HX] ] ][X O [H K - 3 a = ) ( X ) ( O H ) O( H ) X( H - 3 2 aq aq l aq +  → + + ] [X ] [HX K ] O [H - a 3 = + Transform as follows: (1) Rearrange: (2) Take log 10 of both sides: log 10 [H 3 O + ] = log 10 K a + log 10 [HX] [X - ] 9 (3) Change signs: - - = - + ] [X ] [HX log K log ] O [H log - 10 a 10 3 10 - = ] [X ] [HX log K p H p - 10 a (4) Substitute pH and pK a (5) Invert log argument + = - ] HX [ ] X [ log K p H p 10 a Henderson-Hasselbalch Equation
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10 Example: What is the pH of a buffer prepared from 100 mL each of 0.30 M NH 3 and 0.36 M NH 4 Cl? [ K b (NH 3 ) = 1.8 x 10 -5 ] Solution: [acid] = [NH 4 + ] = 0.18 M (0.36 Mx100mL/200mL) [base] = [NH 3 ] = 0.15 M (0.30 Mx100mL/200mL) 10 - 5 - 14 - b w a 10 5.55 10 1.8 10 1.0 K K K × = × × = = pK a = - log 10 (K a ) = -log(5.55 x 10 -10 ) = 9.25 5 + = [acid] [base] log pK pH 10 a 9.18 M 0.18 M 0.15 log 9.255 pH 10 = + = 11 Adding Acid to (1) Water and (2) Buffer (1) When 5.0 mL of 0.10 M HCl is added to 100.0 mL of water, what is the pH change?
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