Updates Assignment 05 is is due today (in class) Midterm 2 is Thurs., March 15 –Huggins 10, 7-8pm –For conflicts: ELL 221, 6-7pm (must arrange at least one week in advance)

Acids and Bases Chapter 16

SAMPLE EXERCISE Lewis Acids and Bases Describe this reaction according to the Lewis theory of acids and bases: AlCl 3 (s) + Cl - (aq) AlCl 4 - (aq). AlCl 3 would be the Lewis acid because it has a vacant orbital that accepts an electron pair from Cl -. Therefore, Cl - is the Lewis base.

SAMPLE EXERCISE Lewis Acids and Bases Which would be considered a stronger Lewis acid: (a) BF 3 or BCl 3 (b) Fe 2+ or Fe 3+ ? Explain. The more electron deficient the central atom with the vacant orbital, the better the Lewis acid. Three fluorines would pull electron density away from boron more strongly than three chlorines so BF 3 is the better Lewis acid in (a). Fe 3+ is more electron deficient than Fe 2+ so Fe 3+ will be the better Lewis acid.

Acid-Base Equilibria and Solubility Equilibria Chapter 17

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 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 − (aq)

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.”

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

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 Initially Change−x−x+x+x+x+x At Equilibrium 0.20 − x  x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

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)

The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = x =  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

We say that the second solution is able to “buffer” against changes in pH. Why does the solution containing carbononic acid and its conjugate base resist a change in pH? When 0.01 mmol of HCl is added to 1 mL of water at pH 7, the pH drops to When 0.01 mmol of HCl is added to 1 mL of water containing 0.16 mmol of H 2 CO 3 /HCO 3 - at pH 7, the pH drops to 6.92.

HCl H + + Cl - HCl + H 3 CCO 2 - H 3 CCO 2 H + Cl

Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

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.

Buffers If acid is added, the F − reacts to form HF and water.

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 −

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

Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation. When [base] = [acid], pH = pK a

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.

Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = (−0.08) pH = 3.77

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.

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.

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.