Chapter18_students - Chapter 18 Aqueous Ionic Equilibrium The Common-Ion Effect Consider a solution of acetic acid adding common ion CH3COO CH3COOH(aq

Chapter18_students - Chapter 18 Aqueous Ionic Equilibrium...

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Chapter 18 Aqueous Ionic Equilibrium
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The Common-Ion Effect Consider a solution of acetic acid: shifts equilibrium to left, reducing [H 3 O + ] 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.” CH 3 COOH( aq ) + H 2 O( l ) D H 3 O + ( aq ) + CH 3 COO ( aq) adding common ion, CH 3 COO
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The Common-Ion Effect Example Calculate the pH of 0.30 M CH 3 COOH – 0.30 M CH 3 COONa solution. [H 3 O + ] [CH 3 COO ] [CH 3 COOH] K a = = 1.8 10 -5 Solution CH 3 COOH(aq) + H 2 O(aq) D H 3 O + (aq) + CH 3 COO - (aq) K a =1.8 10 −5 (x)(0.30 + x) 0.30 - x = 1.8x10 −5 Assume x << 0.30 M so, 0.30 + x 0.30 0.30 - x 0.30 (x)(0.30) 0.30 = 1.8x10 −5 x = 1.8x10 −5 [H 3 O + ] = 1.8x10 −5 M pH = −log (1.8x10 −5 ) = 4.74 Comparison pH 0.30 M CH 3 COOH 2.64 0.30 M CH 3 COOH – 0.30 M CH 3 COONa 4.74 [CH 3 COOH] M [H 3 O + ] M [CH 3 COO ] M Initially 0.30 0 0.30 Change x + x + x Equilibrium 0.30 − x x 0.30 + x Common-ion effect!
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Buffers A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added.
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What Makes a Buffer? A buffer contains a weak acid and its conjugate base CH 3 COOH/CH 3 COONa a weak acid and its conjugate base CH 3 COOH/CH 3 COONa a weak base and its conjugate acid NH 3 /NH 4 Cl a weak base and its conjugate acid NH 3 /NH 4 Cl or
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What Makes a Buffer? Identify which of the following results in buffer solutions when equal volumes of the two solutions are mixed. (a) 0.1 M KNO 3 and 0.1 M HNO 3 (b) 0.1 M NaNO 2 and 0.1 M HNO 2 (c) 0.1 M HCl and 0.1 M NH 3 (d) 0.2 M HCl and 0.1 M NH 3 (e) 0.1 M HCl and 0.2 M NH 3 The average pH of human blood is 7.4. Blood buffer contains three buffer systems: carbonate buffer: H 2 CO 3 and its conjugate base, HCO 3 - phosphate buffer: H 2 PO 4 - and its conjugate base, HPO 4 2- proteins
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How Does a Buffer Work? Buffer solutions contain significant amounts of the weak acid (HA) and its conjugate base (A - ). HA( aq ) + H 2 O( l ) D A ( aq ) + H 3 O + ( aq ) If a strong base is added to the buffer solution, the added base is neutralized by the weak acid component (HA) in the buffer. HA( aq ) + OH ( aq ) → A ( aq ) + H 2 O( l ) If a strong acid is added, the added acid is neutralized by the conjugate base component (A ) in the buffer. H 3 O + ( aq ) + A ( aq ) → HA( aq ) + H 2 O( l ) The pH change is small.
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Henderson-Hasselbalch Equation HA(aq) + H 2 O(l) D H 3 O + (aq) + A - (aq) [A ] [HA] K a = [H 3 O + ] [A ] [HA] −log K a = −log [H 3 O + ] + −log p K a acid base [H 3 O + ] [A ] [HA] K a = p K a = pH − log [base] [acid] pH = p K a + log [base] [acid] - Henderson-Hasselbalch equation p H The Henderson–Hasselbalch equation is generally good enough when the “ x is small” approximation is applicable. a) The initial concentrations of acid and salt are not very dilute. b) The K a is fairly small.
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