Lecture 6 - Acids and Bases (Part 3)

Lecture Recording:

Lecture Summary/Notes:

pH of Salts!

Perhaps I didn't provide the best explanation - you can find the information relevant to this section on on page 725 of the textbook.

**See p.672 for a basic chart of strong acids and bases**

Textbook Example - P.E.A. and P.E.B. (page 725)

P.E.B. - Na+ is a spectator ion since it is the conjugate acid of the strong base, NaOH. Thus it is very weak and has no effect on the pH of the solution.

P.E.B. - H2PO4 is a polyprotic acid, and is has already been deprotonated once. At this point, it can gain one proton and become H3PO4, or deprotonate again and form HPO42-. In the question we are given that the pH of the solution is 4.7, and thus is acidic. Therefore the first reaction will occur to the greater extent!

Textbook Example - P.E.B. (page 727)

P.E.B. - NaCN + H2O <=> NaOH + HCN. Thanks to the common ion effect, we can ignore the Na, thus leaving us with the reaction of CN- + H2O <=> HCN + OH-. Initially, we have a y amount of CN- (from NaCN), and final conditions we have y-x and an x gain of HCN and OH-. Since x = OH-, we need the pOH. We can get that from subtracting the given pH from 14. We get the OH- concentration from the pOH, and plug it into equilibrium equation. We found the Kb from the example problem on p.727. We solved for y to find the concentration of CN ion.

Acidity and Molecular Structure

Acidity has to do with the dissociation of a molecule to release a proton in a solution! For example, let's say we have HX, where X is an element. The acidity has to do with the H-X bond! The WEAKER the bond, the stronger the acid!

There are two factors to do with molecular structure of the element X that affect acidity:

1. Electronegativity - the higher the EN of X, the stronger the acid (weaker bond). 
2. Size - the larger the radius of X, the stronger the acid (weaker bond).

Example)
HF < HCl < HBr < HI in terms of the strength of the acid, since the sizes of the atoms are increasing.


Electronegativity and Divorce, a short story.
My highschool teacher used to teach us mechanics based on stories. When she talked about electronegativity, she would say that a bond between elements was like a marriage. If one element was more electronegative than the one it's bonded to, then that element is like the greedy one between the partners. So in the case of X-H, where X was highly electronegative, X would be very greedy, and H would be unhappy. The greedier X is, the more unhappy H is, and it is more likely that they will get a divorce (weak bond). When they do get divorced, since X is greedy, it will take everything (the house, the dog, the kids, etc.. i.e. the electrons). So think of EN as the greed level of X. The higher the greed, the more likely the divorce :). And for size - out of sight, out of mind :C.

Oxoacids

Oxoacids are acids that are in the general form of : HOXOn, where X is an element, and n is the number of terminal oxygens. H2SO4 is an oxoacid.

Factors affecting acidity in oxoacids:

• The number of Oxygens - A higher n is always more acidic.
• i.e. H2SO4 - Ka > 10^3
• i.e. H2SO3 - Ka = 1.3 x 10^-2
• If the number of Oxygens is the same, then EN is used to compare. A higher EN makes the compound more acidic. In the following special cases, there are no terminal Oxygens!
• i.e. HOCl - Ka = 2.9 x 10^-8
• i.e. HOBr - Ka = 2.1 x 10^-9

Organic Acids

99.9% of organic acids are CARBOXILIC acids. They have the general form:

This notes the difference between a carboxylic acid and an alocohol. Acetic acid and ethanol (though they have the same number of carbons) have a difference of 10^11! Why???? See below.

What makes acetic acid and ethanol different by 11 orders of magnitude???
It has to do with the stabilizing effect of electron delocalization (RESONANCE).

Factors Affecting Acidity (Ka)

We covered three possible factors, one of which we've already discussed (resonance).

What is the effect of chain length on acidity??
It is negligent. This is displayed by the following example:

• i.e. CH3(CH2)6COOH - Ka = 1.3 x 10^-5
• i.e. CH3COOH - Ka = 1.8 x 10^-5

What is the effect of the inductive effect on acidity??

As you substitute the alpha carbon group (i.e. the carbon next to the branch that makes the molecule acidic) with higher electronegative elements, the inductive effect makes the acidity go up!

• i.e. ClCH2COOH - Ka = 1.4 x 10^-3
• i.e. Cl3CCOOH - Ka = 0.16 (illustrated below - recall that high EN elements are greedy and they divorce easily - makes a weaker bond, thus a stronger acid. The C circled in pink is the alpha carbon group).

The Chlorines are pulling on the electrons of the carbon.

Textbook Example - P.E.A. and P.E.B. (page 731)

cith the first comparison, we are comparing two oxoacids. HClO4 as one more terminal Oxygen, so it wins. In the second example, F is more electronegative than Br, so that molecule is more acidic. (They have the same number of Oxygens, so we use EN to differentiate).

P.E.B. - H3PO4 wins since it has one more Oxygen than H2SO3 (they're both oxoacids), and The second molecule has an F replacing a Cl atom, and thus is more acidic (more electronegative).

Factors Affecting Basicity (Kb)

Basically, the main question to ask when determining the relative strength of a base, you have to ask - how available is the lone pair???. See the following:

We will use NH3 for comparing. In the yellow cloud is NF3. That would be a weaker base than NH3 since F is highly EN (remember that high EN makes a strong acid, and thus, high EN makes a weak base). Basically what's happening there is that the F's are sucking the lone pair into themselves (super greedy element), and thus makes the lone pair less available for bonding. In the pink cloud is N(CH3)3. This is a stronger base than NH3 since CH3 is more polarizable than H (it's basically sending the inductive effect in the opposite direction, making the lone pair more available for bonding).

The second factor affecting basicity is resonance. Resonance - since resonance makes a strong acid, it would conversely make a weak base! Easy! A good example of this is Aniline. It has 3 different resonance structures, which delocalizes the lone pair of electrons from the Nitrogen (see below) to the ring. This makes the pair much less available for bonding.

The pair of electrons is delocalized from the N in the first molecule.

Lewis Acids and Bases

Recall that a Bronsted-Lawry base is any substance that accepts a proton, and an acid is one that donates a proton. Basically, the Lewis acid and base go beyond that, generalizing beyond protons. To not get confused, remember that the B-L theory, the proton acceptors (bases) have an extra pair of electrons because they are picking up the electron-free Hydrogen (H+). When you donate a Hydrogen (base), you're kicking the H out without any electrons (i.e. when you dissociate HCl, you get H+ and Cl-. The Cl is the conjugate base because now it wants a proton again!

A Lewis base is any substance that is nucleophilic 

(an electron pair donator).
<NH3 is the Lewis base

A Lewis acid is any substance that is electrophilic 

(an electron pair acceptor)

<BH3 is the Lewis acid

Let's look at the Valence Bond Theory's perspective on things! Notice that after bonding, B is left with a pure 2p orbital! That orbital is HUNGRY for electrons! This makes it electrophilic, thus making it a Lewis BASE!

Common Ion Effect

Let's look at the general acid dissociation reaction:
HAc <=> H+ + Ac-
Where HAc is acetic acid!

If we add HCl to this solution, or any other acid containing Hydrogen ion, we are increasing the [H+] in this reaction. Therefore, Le Chatalier's Principle tells us that if we add HCl, and increase the [H+], the equilibrium will shift to the left.

Textbook Example - P.E.A. and P.E.B. (page 747)

P.E.A. - In the first part, we are finding just the [H+] when we have 0.500M of HF. In the second part, we apply the common ion effect, and start with an initial [H+] of 0.100M (from the HCl). We lose an x amount of HF, and gain an x amount of H+. We apply the simplifying assumption to both the numerator and the denominator and solve for x to find [H+] and [HF].

P.E.B. - We first found out the inital concentration of HOAc (CH3COOH) when we add HCl by finding its concentration at equilibrium before the perturbation. We found that it doesn't change by very much at all, and thus can keep the concentration at 0.01M. We made the final conditions of the equilibrium the initial conditions of before adding the HCl. So since x = 1.34 x 10^-3, it = [H+] and [OAc-]. After we add some amount of HCl, we are given that the [OAc-] is 1.0 x 10^-4, so we put that as the final condition for [OAc-]. We then write out the equilibrium reaction for those conditions, where the Ka is 1.8 x 10^-5 (found in the example on p.747) and solve for x. x is = [H+], so we find the change in H+ by subtracting the final condition from the initial condition. Then we found the moles of H+ by dividing it by the volume of the solution (1.00L). Then we found the volume of the H+ by dividing it by the molarity of HCl (where we got the H+ ions from). We then divided that value by 0.050mL to get the number of drops of HCl we need. PHEW!