chapter 17 - Chapter 17 Additional Aspects of Aqueous...

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Chapter 17 Chapter 17 Additional Aspects of Aqueous Additional Aspects of Aqueous Equilibria Equilibria
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The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. What happens to the pH?
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The 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.”
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The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8 × 10 −4 .
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The Common-Ion Effect Because HCl, a strong acid, is also present, [HF], M [H 3 O + ], M [F ], M Initially 0.20 0.10 0 Change x + x + x At Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x
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The Common-Ion Effect = x 1.4 × 10 −3 = x (0.10) ( x ) (0.20)
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The Common-Ion Effect Therefore, [F ] = x = 1.4 × 10 −3 [H 3 O + ] = 0.10 + x = 1.01 + 1.4 × 10 −3 = 0.10 M So, pH = −log (0.10) pH = 1.00
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Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.
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Which of the following will produce a buffer solution? 1. HCl/NaCl 2. HC 2 H 3 O 2 /NH 3 3. NaH 2 PO 4 /Na 2 HPO 4 4. HNO 3 /Ca(OH) 2 5. KNO 3 / NaOH
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Composition and Action of Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X - ): Buffered Solutions HX( aq ) H + ( aq ) + X - ( aq )
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Buffers 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.
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Buffers If acid is added, the F reacts to form HF and water.
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Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H O H O + A
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Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both sides, we get p K a pH acid base
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Buffer Calculations So p K = pH − log [base] [acid] Rearranging, this becomes pH = p K + log [base] [acid] This is the Henderson–Hasselbalch equation .
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Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3 , and 0.10 M in sodium lactate? K a for lactic acid is 1.4 × 10 −4 .
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Henderson–Hasselbalch Equation pH = p K + log [base] [acid] pH = −log (1.4 × 10 ) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77
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pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a p K a close to the desired pH.
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When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.
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Base to a Buffer 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson–Hasselbalch
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chapter 17 - Chapter 17 Additional Aspects of Aqueous...

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