If you really pay attention to this fifteen minute brisk, exciting and rigorous overview you may find it will help you realize if you understand the concepts in Chapter 14. Do not take notes and do not look away…you may miss something. Just sit back and appreciate what you see and ask yourself if it makes sense. The presentation will proceed at a fixed pace on its own. Congratulations…for you are traveling down the road of chemical enlightenment!

One of the most important things we did in Chapter 14 was define what an acid and a base is...

The Arrhenius definition of an acid and a base:

Acids - substances that when dissolved in water release H+ ions Bases - substances that when dissolved in water release OH- ions The definition of an Arrhenius acid and base emphasizes the H+ and OH- ions in water

The Brønsted-Lowry definition of an acid and a base:Acid - a substance that can donate a proton to another substance Base - a substance that can accept a proton from another substance These definitions emphasize proton transfer, and can include solvents other than water (aqueous solutions are not part of the definition, proton transfer is the key feature) The definition was developed independently in 1923 by Johannes Brønsted and Thomas Lowry

G.N.Lewis thought about acids and bases in terms of donation and acceptance of unshared pairs of electrons:

A Lewis acid is defined as an electron-pair acceptor A Lewis base is defined as an electron-pair donor

The majority of the work we did in this chapter takes the Bronsted -Lowry perspective:

A substance can only work as a Brønsted-Lowry acid (i.e. donates a proton), if another substance simultaneously acts as a Brønsted-Lowry base (i.e. accepts the proton)

Acid Base

B-L acid-base reactions can be viewed as equilibrium reactions where both the forward and reverse reactions involve proton transfer.

Acid Base AcidBase

In any acid base reaction there will be conjugate acid base pairs

There is an inverse relationship between the strength of an acid and its conjugate base (likewise a base and its conjugate acid)

A strong acid is a molecule that has a strong preference to donate a proton.

Thus, its conjugate base will have an extremely weak tendency to accept a proton The strong acids are:

HCl (hydrochloric acid)

HBr (hydrobromic acid)

HI (hydroiodic acid)

HNO3 (nitric acid)

HClO4 (perchloric acid)

H2SO4 (sulfuric acid)

Weak acids have conjugate bases that have a moderate tendency to be protonated.

Thus, in solution, only a fraction of the molecules of a weak acid will donate a proton. There will be a significant concentration of both the acid and conjugate base forms in solution

H2C6H6O6(aq) + H2O(l) <=> H3O+(aq) + HC6H6O6

-(aq)

ascorbic acid biascorbate the ion

For example:

(lots of this around) (some of this around)

Ascorbic acid is a weak diprotic acid.

Not only does this occur...

H2C6H6O6(aq) + H2O(l) <=> H3O+(aq) + HC6H6O6

-(aq)

ascorbic Acid biascorbate ion

But so does this...

HC6H6O6-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

The conjugate base above can act like an acid below!

biascorbate ion ascorbate ion

Since these are equilibrium reactions, both can be evaluated by K values… and since both involve an acid reacting with water to make hydronium and it’s conjugate base the K value is

called Ka.

H2C6H6O6(aq) + H2O(l) <=> H3O+(aq) + HC6H6O6

-(aq)

Ka1= 6.8 x 10-5

HC6H6O6-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

Ka2= 2.8 x 10-12

A aqueous solution of 0.75 M ascorbic acid then has both reactions (as well as the autoionization of water) taking place inside the water.

H2C6H6O6(aq) + H2O(l) <=> H3O+(aq) + HC6H6O6

-(aq)

Ka1= 6.8 x 10-5

HC6H6O6-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

Ka2= 2.8 x 10-12

...but the size of the Ka value for this reaction indicates that it will dictate the H3O+ concentration at equilibrium.

To determine the pH of a 0.75 M solution of ascorbic acid you would only consider

the reaction associated with Ka1

H2C6H6O6(aq) + H2O(l) <=> H3O+(aq) + HC6H6O6

-(aq)

Ka1= 6.8 x 10-5 = (x)(x) = x2

Init: 0.75 M 0 M 0 M: -x +x +x

Eq: 0.75 M-x x x

0.75-x 0.75-xpH = - (log 0.0071)= 2.15

neglect

x = 0.0071= [H3O+]

So even though both of the reactions occur, the only one that makes a difference is the first one.H2C6H6O6(aq) + H2O(l) <=> H3O+

(aq) + HC6H6O6-(aq)

Ka1= 6.8 x 10-5

HC6H6O6-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

Ka2= 2.8 x 10-12 The ascorbate ion concentration in the second reaction will be dictated by the biascorbate and hydronium produced in the first equation along with the Ka for reaction #2.

To determine the [C6H6O62-] in a 0.75 M

solution of ascorbic acid you would only

consider the reaction associated with Ka2HC6H6O6

-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

Ka1= 2.8 x 10-12 = (0.0071+x)(x) = 0.0071 x

Init: 0.0071 M 0.0071 M 0: -x +x +x

Eq: 0.0071 -x 0.0071+x x

0.0071-x 0.0071neglect

x = 2.8 x 10-12 = [C6H6O62-]

neglect

If the salt Na2C6H6O6 was dissolved in water how do you think it would behave?

HC6H6O6-(aq) + H2O(l) <=> H3O+

(aq) + C6H6O62-

(aq)

Ka= 2.8 x 10-12

Well, based on what you just learned, the solution would be basic because the ascorbate ion, C6H6O6

2- has a parent acid that is weak.

C6H6O62-

(aq) + H2O(l) <=> OH-(aq) + HC6H6O6

-(aq)

Kb= Kw/Ka = 1.0 x 10-14/2.8 x 10-12

The Kb value for the ascorbate ion, C6H6O6

2-, acting as a base can be determined…do you know how?

From the conjugate acid (HC6H6O6-)

Kb= 3.57 x 10-3