# Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier.

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Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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. CH 3 COOH (aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO − (aq)

Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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.”

Aqueous Equilibria What is the PH of a solution made by adding 0.3 mol of acetic acid and 0.3 mol of soduim acetate to enough water to make 1L of solution.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4

Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect = x 1.4  10 −3 = x (0.10) (x) (0.20) 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10)

Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers Similarly, if acid is added, the F − reacts with it to form HF and water.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH pH = 3.77 pH = 3.85 + (−0.08) pH

Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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 pK a close to the desired pH.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. 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.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Addition of Strong Acid or 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 equation to determine the new pH of the solution.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol

Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0.200) pH = 4.74 + 0.06pH pH = 4.80

Aqueous Equilibria © 2009, Prentice-Hall, Inc. *What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOK aq (0.52 M ) → K + aq + HCOO - aq ( 0.52 M ) HCOOH ↔ H + aq + HCOO - aq initial : 0.30M 0.00 0.52M Change : -x +x +x Equilibrium: 0.30-x x 0.52+x, 0.30-x ≈ x, 052+x ≈ 0.52 pH = pK a + log [HCOO - ] / [HCOOH] = 3.77 + log [0.52] / [0.30] = 4.01 ----------------------------------------------------------------------------------------------------- * Which of the following are buffer system ? a) KF /HF (buff.) b) KBr/HBr ( not buff.) c) Na 2 CO 3 /NaHCO 3 (buff.) d) KH 2 PO 4 / H 3 PO 4 ( buff.) e) NaClO 4 / HClO 4 (not buff.) f) C 5 H 5 N/C 5 H 5 NH + (buff. ) g) K 2 HPO 4 / H 3 PO 4 ( not buff.). ---------------------------------------------------------- Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH After the addition of 0.01 mol of NaOH to 1.0 L of buffer solution ?. Assume that the volume of the solution does not change when the NaOH is added. NH 4 + (aq) ↔ H + (aq) + NH 3 (aq), pH = pK a + log [NH 3 ] /[NH 4 + ] pH = 9.25 + log [0.30] / [0.36] = 9.17 : NH 4 + (aq) + OH aq → H 2 O (l) + NH 3(aq) start : 0.36 M 0.01 M 0.30 M end : 0.35 M 0.00 0.31 M pH = 9.25 + log [0.31] / [0.35] = 9.20 --------------------------------------------------------------------------------------------------------------

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