Monday, November 16Acids and Bases

Chem 03 Fall 2009

Brønsted-Lowry Acids • An acid is a proton (H+) donor • Strong acids totally dissociate in water

– hydrochloric acid HCl 12 M – nitric acid HNO3 16 M– sulfuric acid H2SO4 18 M

• Weak acids partially dissociate in water– HA H⇆ + + A–

– pKa measures the strength of the weak acid, pH at which HA = A–, or half of the H+ comes off.

– acetic acid HC2H3O2 pKa = 4.75– carbonic acid H2CO3 pKa= 6.37

strong

weak

(aq)H(aq)HSO(aq)SOH 442

Brønsted-Lowry Acids (proton donors)

Strong Acids: completely dissociateHCl → H+ + Cl-

Weak Acids: only partially dissociate• HA H⇆ + + A- Ka = [products]/[reactants] =

[H+][A-]/[HA]• Identifying (page 654 15-2 and 657 15-3 of Chang)• What do structures of the weak acids above have in common?

• pH scale: -log [H+] elementary texts will say limit is 0-14, but this is not true:

for example: strong acid HCl at 10 M, pH = -1

Brønsted-Lowry Bases • A base is a proton (H+) acceptor • Strong bases totally dissociate in water

– sodium hydroxide NaOH– barium hydroxide Ba(OH)2

– potassium hydroxide KOH

• Weak bases do not totally dissociate– sodium carbonate Na2CO3 – ammonium hydroxide NH4OH– water HOH

weak

strong

Neutralization ReactionsAcid + Base Salt + Water + heat

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) + heat

–Base in burette is added to acid in flask until the moles of base added is equal to the moles of acid, this is called the equivalence point

–moles acid = moles base: MaVa = MbVb

–In practice, an indicator is used to monitor pH, when indicator changes color, the end point of titration is reached, the volume of base added is measured and the assumption is that you are very close to equivalence point

Titration Reaction: often used to calculate M of either acid or base

Figures from: http://www.csudh.edu/oliver/demos/carbnate/carbnate.htm

• Calculate the volume of 0.1500 M NaOH base needed to neutralize 25.00 mL of 0.3000 M HCl

• When the base neutralizes the acid, there is an equivalent number of moles of acid and base– Moles acid = Moles base– MaVa = MbVb – 0.3000 moles/L x 0.3000 L = 0.1500 moles/L x Vb

– Vb =

Example of Titration Reaction I

• Calculate the volume of 0.1500 M NaOH base needed to neutralize 25.00 mL of 0.3000 M HCl

• When the base neutralizes the acid, there is an equivalent number of moles of acid and base– Moles acid = Moles base– MaVa = MbVb – 0.3000 moles/L x 0.3000 L = 0.1500 moles/L x Vb

– Vb =

Example of Titration Reaction I

– 0.05000 L base or 50.00 mL

• Calculate the pH at the start (25.00 mL of 0.3000 M HCl) and the end (after 50.00 mL of base added)

• Calculation of pH– START pH = -log [H+]= -log [0.3000] =– END pH = 7 at equivalence point for a strong acid-strong base titration.

• If you used phenolphthalein as indicator, would you have reached the endpoint? – (endpoint: a color change in indicator)– NO, PPN changes from clear to pink at pH 8.0

Example of Titration Reaction II

• Calculate the pH at the start (25.00 mL of 0.3000 M HCl) and the end (after 50.00 mL of base added)

• Calculation of pH– START pH = -log [H+]= -log [0.3000] =– END pH = 7 at equivalence point for a strong acid-strong base titration.

• If you used phenolphthalein as indicator, would you have reached the endpoint? – (endpoint: a color change in indicator)– NO, PPN changes from clear to pink at pH 8.0

Example of Titration Reaction II

0.5229

Autoionization of Water• H2O H⇆ + + OH-

K = [H+][OH-]/[H2O] but concentration of [H2O] is so HIGH• • Autoionization is just a drop in the bucket, so [H2O] stays ≈ constant at 55.5 M• K x 55.5 = Kw • Kw = [H+][OH-] = 1 x 10-14

• Take [– log] of right and left hand sides of equation:• -log [H+][OH-] = - log [1 x 10-14]

• Right hand side = (– log 1) + (– log 10-14) = 14• Left hand side (– log H+) + (– log OH-) = pH + pOH• pH + pOH = 14 • This is always true, you can always get the concentration of H+ or OH- by knowing just one, and using

the equation above.

• pH = 1 pOH ____ OH-____• pH = 5 pOH ____ OH- ____• pH = 7 pOH ____ OH- ____• • So, at pH = 7, H+ = OH- = 10-7 solution is neutral

Example of Titration Reaction III• In previous example, calculate the pH of the solution if you added

0.50 mL too much base• At equivalence, you have 75.00 mL of salt solution (25.00mL +

50.00 mL) at pH 7.0• The problem now becomes a dilution where you must first find

out the molarity of base in the beaker, and then pOH, and then pH – First find the Molarity of Base in Beaker

• M1V1 = M2V2

• (0.1500 M x 0.50 mL) = (M2 x 75.50 mL) = • 1.000 x 10-3 M NaOH

– Then find the pOH• pOH = -log[OH-] = 3.000

– Finally find the pH• pH = 14 – pOH = • 11

Acid/Conjugate Base Pairs• Table I• Common Name Chemical Name Type Acid/ Conj. Base Ka pKa• Stomach Acid HCl S HCl/Cl- ∞• • Aspirin acetylsalicylic acid W C9H8O4/ C9H7O4

-1 3.0 x 10-4

• • Lemon juice citric acid W C6H8O7/ C6H7O7

-1 7.1 x 10-4

• • Vinegar acetic acid W C2H4O2/ C2H3O2

-1 1.8 x 10-5

• • Vitamin C ascorbic acid W C6H8O6/ C6H7O6

-1 8.0 x 10-5

•

•Baking Soda sodium bicarbonate W HCO3

-/CO3-2 5.0 x 10-7

Buffers resist changes of pH

• pKa = -log Ka• So, if you know the Ka, the acid dissociation constant, you can

calculate a pKa. Go back to Table I and fill in the pKa of the various acids, what do you notice about how the strength varies with the pKa?

Now hold on for something even better: HA H⇆ + + A-

Ka = [H+][A-]/[HA]

-log Ka = -log {[H+][A-]/[HA]} pKa = -log[H+] + {-log[A-]/[HA]}

pKa = pH – log ([A-]/[HA]) 

Buffers

pKa = pH – log ([A-]/[HA]) • What happens when [A-]/[HA] = 1     __________________________________________________________

• If you have a mixture of a weak acid HA and its conjugate base A-, approx equal concentrations, and at a pH close to the pKa, then you have a situation where even if you add a strong acid directly to the solution, the pH won’t change much because the HA/A- ratio will adjust itself and stay close to 1. This is called a BUFFERED solution

• A buffered solution is one which _________________________________________________