Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover. -Mark Twain- (just don’t be foolish in the process. )

MORE AND MORE ACIDS AND BASES

MOST OF THE CHEMISTRY IN THE NATURAL WORLD OCCURS IN AQUEOUS SOLUTIONS.

IMPORTANT IN THESE ARE ACID-BASE REACTIONS AND EQUILIBRIA.

BUFFERED SOLUTIONS – SOLUTIONS THAT CAN RESIST CHANGE IN pH.

FOR EXAMPLE, THE pH OF BLOOD IS AROUND 7.4. THE pH OF BLOOD DETERMINES HOW PROTEINS FOLD. THEY NEED TO FOLD IN CERTAIN WAYS TO ACT AS ENZYMES (CATALYSTS) TO CONTROL IMPORTANT REACTIONS.

ALL NATURAL ORGANISMS CAN SURVIVE ONLY WITHIN CERTAIN NARROW pH RANGES.

WHEN WE EXERCISE HARD, OUR MUSCLES PRODUCE LACTIC ACID. THE BODY MUST BE ABLE TO HANDLE THIS AND MAINTAIN pH. THAT’S WHERE BUFFERS COME IN.

LET’S START BY CONSIDERING THE DISSOCIATI0N OF HF, A WEAK ACID.

HF (aq) == H+ (aq) + F- (aq) Ka = 7.2 X 10-4

IF WE ADD NaF, AS A SALT, WHEN IT DISSOLVES, IT BREAKS DOWN INTO Na+ and F-.

THE F- IS COMMON TO THE DISSOCIATION OF BOTH THE NaF and HF.

ACCORDING TO LE CHATILIER’S PRINCIPLE, IT WILL UPSET THE EQUILIBIUM AND FORCE IT TO THE LEFT.

HF (aq) == H+ (aq) + F- (aq)

SO, THERE WOULD BE LESS DISSOCIATION OF THE HF IN A SOLUTION CONTAINING NaF THAN THERE WOULD BE IN A SOLUTION CONTAINING JUST HF.

THIS IS CALLED THE COMMON ION EFFECT.

THE SAME EFFECT WOULD BE SEEN IN A SOLUTION OF AMMONIA IF WE ADDED NH4Cl.

NH3 (aq) + H2O (l) == NH4+ (aq) + OH- (aq)

AND, WE WOULD SEE THE SAME EFFECT WITH POLYPROTIC ACIDS.

1ST IONIZATION: H2SO4 (aq) == H+ (aq) + HSO4- (aq)

2nd IONIZATION: HSO4- (aq) == H+ (aq) + SO4

-2 (aq)

THE HYDROGEN IONS PRODUCED IN THE FIRST DISSOCIATION WOULD INHIBIT THE EXTENT OF HSO4

- DISSOCIATION.

THE CALCULATIONS FOR FINDING THE pH OF A SOLUTION OF A WEAK ACID AND ITS SALT ARE SIMILAR TO THOSE FOR A WEAK ACID.

CALCULATE THE pH OF A 1.0 M HF SOLUTION CONTAINING 1.0 M NaF. Ka = 7.2 X 10-4

MAJOR SPECIES: Na+ HF F- H2O

HF (aq) == H+ (aq) + F-(aq) Ka = [H+][F-]/[HF] = 7.2 x 10-4

[H+] = x [F-] = 1.0 + x (amt from HF dissociation) [HF] = 1.0 – x

Ka = (x)(1.0+x)/(1.0-x) = x(1.0)/(1.0) if x << 1.0

SO, x = 7.2 x 10-4 = [H+] and pH = -log(7.2 x 10-4) = 3.14

HOW DOES THIS COMPARE WITH NO NaF PRESENT?

HF (aq) == H+ (aq) + F- (aq)

[H+] = [F-] = x [HF] = 1.0 – x and, if x << 1.0 [HF] = 1.0

Ka = x2/(1.0) = 7.2 x 10-4 and x = [H+] = 2.7 x 10-2

pH = -log(2.7 x 10-2) = 1.57

SO, THE ADDITION OF NaF PUSHED THE EQUILIBRIUM TO THE LEFT AND CHANGED THE pH FROM 1.57 TO 3.14.

THIS IS WHAT WE WOULD EXPECT FROM LE CHATILIER’S PRINCIPLE.

WE ARE NOW GOING TO APPLY THE COMMON ION EFFECT TO BUFFER SOLUTIONS.

A BUFFER SOLUTION IS A SOLUTION THAT RESISTS CHANGE IN pH.

A BUFFER SOLUTION CAN CONSIST OF A WEAK ACID AND ITS SALT OR A WEAK BASE AND ITS SALT.

HA == H+ + A-

(SIGNIFICANT AMOUNTS)

HERE’S HOW IT WORKS. IF YOU REMOVE H+, SAY BY THE ADDITION OF A BASE, MORE HA CAN DISSOCIATE TO PRODUCE ADDITIONAL H+. IF YOU ADD MORE H+, IT CAN ASSOCIATE WITH A- TO FORM HA AND REDUCE THE ADDITIONAL AMOUNT OF H+.

THIS IS AN APPLICATION OF LE CHATILIER’S PRINCIPLE.

NOW, LET’S CONSIDER A WEAK BASE AND ITS SALT.

B + H2O == HB+ + OH-

(SIGNIFICANT AMOUNTS)

IF YOU ADD H+, IT WILL REACT WITH OH-, RESULTING IN A REDUCTION IN OH-. MORE BASE WILL DISSOCIATE TO PRODUCE ADDITIONAL OH-.

IF YOU REMOVE H+ , YOU PRODUCE MORE OH-. IT WILL REACT WITH HB+ TO REDUCE THE AMOUNT OF OH-

A BUFFERED SOLUTION CONTAINS 0.500 M ACETIC ACID AND 0.500 M SODIUM ACETATE (Ka = 1.8 X 10-

5). CALCULATE THE pH OF THE SOLUTION.

HAc (aq) == H+ (aq) + Ac- (aq) Ka = [H+][Ac-]/[Hac]

MAJOR SPECIES: HAc (HC2H3O2) Ac- Na+ H2O

[H+] = x [Ac-] = 0.500 + x [HAc] = 0.500 – x

Ka = [H+][Ac-]/[HAc] = x(0.500+x)/(0.500-x) = x(0.500)/(0.500)

X = 1.8 x 10-5 (if 0.500>>x)

pH = 4.71

WHAT WOULD BE THE pH IF 0.010 MOLE OF SOLID NaOH IS ADDED TO 1 liter OF THE BUFFER?

AFTER ADDITION OF THE NaOH, AND BEFORE EQUILIBRIUM,

[HAc] = 0.50 – 0.01 = 0.49 M [Ac-] = 0.50 + 0.01 = 0.51 M

HAC (aq) == H+ (aq) + Ac- (aq)

AT EQUILIBRIUM [H+] = x [Ac-] = 0.51 + x [HAc] = 0.49 – x

Ka = x(0.51+x)/(0.49-x) = x(0.51)/0.49 = 1.8 x 10-

5

X = (0.49)(1.8x10-5)/(0.51) = 1.73 x 10-5 = [H+]

pH = - log (1.73 x 10-5) = 4.76

NOTICE THAT THE pH ONLY CHANGED FROM 4.74 TO 4.76

WHAT WOULD THE pH BE IF YOU ADDED 0.01 MOLE NaOH TO 1 liter OF DISTILLED WATER?

NaOH (s) == Na+ (aq) + OH- (aq)

[OH-] = 0.01 M [H+] = Kw/[OH-] = 10-14/10-2 = 10-12

pH = - log (10-12) = 12

BUFFER SOLUTIONS ARE SOLUTIONS OF WEAK ACIDS OR BASES CONTAINING A COMMON ION.

NOTE: WHEN A STRONG ACID OR A STRONG BASE IS ADDED TO A BUFFER SOLUTION, DEAL WITH THE STOICHIOMETRY FIRST AND THEN THE EQUILIBRIUM.

HOW DO BUFFER SOLUTIONS DEAL WITH ADDITIONS OF H+ OR OH- ?

OH- + HA == A- + H2O

H+ + A- == HA

Ka = [H+][A-]/[HA] and [H+] = Ka([HA]/[A-])

IN A BUFFER SOLUTION, THE [H+] IS CONTROLLED BY THE [HA]/[A-] RATIO.

IF THE INITIAL CONCENTRATIONS OF [HA] AND [A-] ARE LARGE COMPARED TO THE AMOUNTS OF OH-OR H+ ADDED, CHANGES TO THE FINAL pH WILL BE SMALL.

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

Taking the log of each side

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

Multiplying by -1

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

And

pH = pKa – log [HA]/[A-]

Or

pH = pKa + log [A-]/[HA]

THIS IS THE HENDERSON-HASSELBACH EQUATION.

OPTIMAL BUFFERING OCCURS WHEN [A-] = [HA].

SO, IN SELECTING A BUFFER FOR USE IN THE LAB, YOU WOULD WANT TO SELECT ONE WHERE pH = pKa.

AGAIN, THE pH OF A BUFFER SOLUTION IS DETERMINED BY THE RATIO [A-]/[HA]. THE BUFFER CAPACITY IS DETERMINED BY THE AMOUNTS OF [A-] AND [HA].

EXAMPLE: A BUFFER SOLUTION CONTAINS 0.25 M NH3 AND 0.40 M NH4Cl. Kb = 1.8 X 10-5 CALCULATE THE pH.

NH3 (aq) + H2O (l) == NH4+ (aq) + OH- (aq)

Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

At equilibrium [OH-] = x [NH4+] = 0.40 + x

[NH3] = 0.25 – x

Kb = (0.4+x)x/(0.25-x) if x<<0.25 Kb = 0.4x/0.25 = 1.8 x 10-5

X = (1.8 x 10-5)(0.25/0.4) = 1.125 x 10-5 = [OH-]

pOH = 4.95 pH = 14-4.95 = 9.05