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Chemistry 12 Notes on Unit 4 – Acids, Bases and Salts

Chemistry 12-Unit 4-Notes

Chemistry 12 - Notes on Unit 4 Part 1 – Acid Base

Hydronium Ions and how they are formed First of all, we must apoligize again for something about Chemistry 11. In that

course (which probably seems juvenile now), you learned that acids dissociate in water to form hydrogen ions (H+) and others. For example, when hydrogen chloride gas dissolves in water, you get....

HCl(g) H+(aq) + Cl-(aq)

So you might picture some H+ ions and Cl- ions floating around between water

molecules in the solution. That’s OK in Chemistry 11, but it’s a bit oversimplified! It IS

true that H+ ions are released from the HCl. But they don’t just float around by themselves.

H+ ions are hydrogen ions. Let’s talk about hydrogen atoms. Almost all hydrogen atoms consist of one proton,

no

neutrons and 1 electron. The proton is deep in the center in the nucleus and the electron “buzzes” around the atom in what we call an electron “cloud”.

When a hydrogen atom (H) forms a hydrogen ion (H+), remember, it loses an

electron. When it does this, it also loses it’s electron cloud! So what’s left? Just a very very

tiny nucleus which contains 1 proton!



For this reason, the H+ ion is often called a proton. Because, that’s exactly what it is!

The +1 charge on the H+ ion or proton, is concentrated in a very small volume, much smaller than in any other ion. (All other ions have at least 1 electron, so they have an electron cloud, which makes them thousands of times bigger than H+, which has no electron cloud.)

Reactions in which H+ ions are transferred from one thing to another are called proton transfers.

Because this charge is concentrated in a very small volume, it acts like it is quite

powerful and it is attracted strongly to anything even remotely negative! Remember that in an acid solution, H+ ions (or protons as we also call them) are

surrounded by water

molecules. Let’s take a closer look at a water molecule.

Recall in Chemistry 11, you were introduced to “electron-dot” or “Lewis” diagrams of atoms. (These, again are an oversimplification but that’s another story!)

You might recall that an oxygen atom has 6 valence electrons (6 electrons in the

outer energy level):

O

Hydrogen has one valence electron:

H

When hydrogen and oxygen combine to form water, they share their valence electrons. But, you may also recall that oxygen, being a non-metal has a stronger pull

on electrons (electronegativity) than hydrogen, so the shared electrons are closer to the oxygen atom. This makes water a polar covalent molecule. Since there are more electrons close to the “oxygen end” of the water molecule, that end has a partial negative charge. The “hydrogen end” has less electrons around it, hence has a partial positive charge:

See the diagram on the next page...



O H

H

∂ −∂ +

∂ +

NOTE: The symbol "∂ " means "partial".

Water is a polar covalent molecule with a partial "-" charge near the oxygen end and a partial "+" charge near the hydrogen end.

Now that partial negative charge on the oxygen end looks very attractive to our old friend the H+ ion! And that’s exactly where it goes. It “sits” on one of the “electron pairs” of the oxygen atom. Remember the H+ has no electrons itself, so it doesn’t bring any more electrons into the picture.

It has one proton, though. What that does is bring another + charge into the picture:

O HH

H+

This thing (made up of a proton (H+) added to a water molecule is an ion because it

has a charge.

Its formula is H3O+ and its called the hydronium ion. The hydronium ion always forms when an acid dissolves in water. The H+ from the

acid always goes to the nearest water molecule and forms H3O+. Another way to look at the hydronium ion is to take the point of view of the proton

(H+). Adding water to something is called hydration. (Just like taking water away is called

dehydration.) So if you were a proton, you would have a water molecule “added to you”. For this reason, a hydronium ion could be considered a hydrated proton

.



Whichever way you look at it, just remember that instead of thinking of an acid solution containing H+ ions (as you did in Chem. 11), we now think of acid solutions containing H3O+ (hydronium) ions.

All acid solutions contain hydronium

(H3O+) ions. It is the hydronium ion which gives all acids their properties (like sour taste, indicator colours, reactivity with metals etc. )

Now, recall that in Chemistry 11, when HCl gas dissolves in water, we wrote:

HCl(g) H+(aq) + Cl-(aq)

Now, in Chemistry 12,

we write the following:

HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) The proton (H+) has been transferred

from the HCl molecule to a water molecule, to form a hydronium (H3O+) ion and a Cl- ion.

This type of reaction is called ionization (because ions are being formed) We can look at this using some models:

H

H

Cl ClO

HH

O

HH

+ -

HCl + H O H O + Cl32+ _

+ +

Proton Transfer

Make sure you study the diagram so you can visualize in your mind, what’s going on when you see equations like this.

In this diagram, you must realize that it is NOT an H atom that is moving. The H atom leaves it’s electron behind with the Cl, so it is H+ (a proton) that moves to the water molecule. The Cl-, now having the electron that H left behind, gains a negative charge.



All acids behave similarly in water; they donate (or give) a proton (H+) to the water, forming hydronium ion (H3O+) and the negative ion of the acid.

Another example might be the ionization of nitric acid (HNO3):

HNO3 (l) + H2O (l) H3O+(aq) + NO3-(aq)

Bronsted-Lowry Definition of Acids and Bases You might recall that the definition of an “acid” according to Arrhenius was a

substance that released H+ ions (protons) in water. A couple of fellows called Bronsted and Lowry came up with a theory which is

more useful when dealing with equilibrium and covers a wider range of substances. Our apologies to Mr. Lowry, but from now on we will just refer to “Bronsted”,

when we actually mean both of them. It’s just bad luck that his name came later in the alphabet!

According to Bronsted

(and “What’s his name?”):

An acid is any substance which donates (gives) a proton (H+) to another

substance. A base is any substance which accepts (takes) a proton from another

substance. Or we can also say: A Bronsted Acid is a proton donor A Bronsted Base is a proton acceptor Let’s look at a couple of equations and see how we can identify the acids and the

bases. (We will omit the subscripts (aq) etc. just for simplification here.) eg.) HCl + H2O H3O+ + Cl- Looking at this diagram again:



H

H

Cl ClO

HH

O

HH

+ -

HCl + H O H O + Cl32+ _

+ +

Proton Transfer

We see that the HCl is donating the proton and the water is accepting the proton. Therefore HCl is the Bronsted acid and H2O is the Bronsted base

.

HCl + H2O H3O+ + Cl- acid base Let’s look at another example: NH3 + H2O NH4+ + OH- Now, the NH3 on the left has changed into NH4+ on the right, that means it must

have accepted (taken) a proton. (It has one more H and one more (+) charge.) Since it has accepted a proton it’s called a base

.

The H2O, this time has donated (lost) a proton as it changed into OH-. (It has one less H and one less (+) charge --- one “less (+) charge” than “0” is (-1) or (-).) Since it has donated a proton it’s called an acid

.

So now we can label these: NH3 + H2O NH4+ + OH- base acid Now, you may be a little confused! First we tell you that H2O is a base (see the

reaction near the top of this page), and then we go and tell you H2O is an acid. What’s going on, Bronsted?



Well, both of these statements are correct. Sometimes water acts like a base (takes

a proton) and sometimes it acts like an acid (donates a proton). This is just like you. If you buy something (donate money) you are a buyer. If you

sell something (accept money), you are a seller. I’m sure you have been both at various times.

Animals that can live either in the water or on land are called amphibians. (Yes, this is still Chemistry just in case you’re wondering!)

For things that can be “either / or ”, we can use the prefix “amphi” A substance that can act as either an acid or a base

is called

amphiprotic.

Water (H2O) is an example of an amphiprotic substance. When it was with HCl, it

acted like a base, but when it was with NH3, it acted like an acid. Not only molecules can lose or gain protons. Ions can too. When something loses a proton (acts as an acid), it turns into something with one

less H and one less (+) charge

(which means the same as one more (-) charge.)

When something accepts a proton (acts as a base), it turns into something with one more H and one more (+) charge

(which means the same as one less (-) charge.)

So what you have to do is look at the right side of the equation, as see whether the substance gained or lost a proton.

eg) HCO3- + HSO4- H2CO3 + SO42- HCO3- must have accepted

a proton (1 H and 1 (+) charge) to form H2CO3, so it must be the base.

HSO4- must have donated

a proton (1 H and 1 (+) charge) to form SO42- , so it must be the acid.

so the answer is: HCO3- + HSO4- H2CO3 + SO42-

base acid Read this example again, looking carefully at the charges and # of H atoms, and

how they change from each reactant to it’s product. By the way, there is no rule for



which one comes first in the equation. Basically, each one has a 50/50 chance of coming first. You have to work it out by counting H’s and charges.

Equilibria Involving Acid and Bases So far, we’ve been considering reactions which only go one way. In reality, most

acid-base reactions go forward and in reverse. (They are at equilibrium) If a proton is transferred during the forward reaction, we can also assume there will

be a proton transfer in the reverse reaction. Here’s an example:

HF + SO32- HSO3- + F-

If we consider the reaction going to the right, HF is donating a proton, and is therefore defined as the acid, while SO32- is accepting a proton, and therefore acting as a base.

HF + SO32- HSO3- + F-

acid base Now, when we look at the reverse reaction, in which HSO3- reacts with F- to form

HF and SO32-, we see that HSO32- donates a proton and F- accepts a proton. Thus, HSO3- acts as an acid, while F- acts as a base. So in any acid, base reaction, we start out with an acid and a base on the left and we end up with another acid and base on the right.

HF + SO32- HSO3- + F-

acid base acid base

Looking at this reaction: Conjugate Acid-Base Pairs

HIO3 + NO2- HNO2 + IO3- acid base acid base Notice the HIO3 on the left. We know that it must lose one proton (H+) to become

IO3- on the right. Also notice that HIO3 is acting as an acid while IO3- is acting as a base.

HIO3 and IO3- form what is called a conjugate acid-base pair.



The only difference between these two is the IO3- has one less “H” and one more (-)

charge

than the HIO3. All conjugate acid-base pairs are like this.

The form with one more H

(eg. HIO3) is called the conjugate acid.

The form with one less H

(eg. IO3-) is called the conjugate base.

Out of every acid-base reaction, you always get 2 conjugate pairs For example, in this reaction:

.

HIO3 + NO2- HNO2 + IO3- acid base acid base The two conjugate pairs are: HIO3 & IO3- and NO2- & HNO2 acid base base acid conjugate pair 1 conjugate pair 2 (NOTE

: The “1” and the “2” in “conjugate pair 1” etc. has no special meaning. Pair 1 was just the one we happened to pick first. The NO2

- & HNO2 could just as well have been called “conjugate pair 1”)

************************************************************** One of the things you’ll be required to do is, given an ion or molecule, write the

formula for the conjugate acid of it. Also, given an ion or molecule, write the formula for the conjugate base of it.

If you look carefully at the answers to the preceding question, you can probably

figure out a method on your own. But here is one that works (just in case you can’t) To find the conjugate acid of something:

Something Add one HAdd one + charge

Conjugate Acid



For example, let’s say we want to find the conjugate acid of HSO4-

Add one HAdd one + chargeHSO 4

- H SO2 4

Remember, adding one (+) charge to something that has a (-) charge, brings the

charge to “0”. Now it’s time to find the conjugate bases of things. If you added an “H” and a (+)

charge to get a conjugate acid, I think you can probably guess that to get a conjugate base, you subtract an “H” and add one (-) charge. (or subtract one (+) charge, which means the same thing!)

Something Subtract one H

Add one - charge

Conjugate Base

Let’s say we want to find the conjugate base of the ion H2PO4-. If we use this procedure:

Subtract one H

Add one - charge H PO2 4

- HPO42-

Polyprotic Acids and Amphiprotic Anions After looking at this rather fearsome title, read the following. It’s not that bad! So far, we’ve been looking at acids that only have one proton (H+) to release. These

are acids with one hydrogen in their formulas (eg. HCl, HNO3, HClO etc.)



Acids that release only one proton are called monoprotic acids

.

Believe it or not, acetic acid (CH3COOH) is monoprotic. This is because only the “H” on the end of this acid (The “H” on the “COOH”) comes off in solution. The other three “H”s are bonded directly and strongly to the Carbon atom in the “CH3” and are not

released.

“H”s bonded directly to Carbon atoms (like in “CH3”, “CH3CH2” etc. ) are NOT released in solution and are not considered as “acidic protons”.

In another example, the acid HCOOH and the acid CH3CH2CH2COOH are both

monoprotic

. (Only the “H” on the end of the “COOH” comes off.)

You will also notice that for these organic acids (or more precisely “carboxylic acids”, the “acidic proton” is the “H” on the right end

of the formula.

For inorganic acids (like HCl, HNO3, HClO3 etc)., the acidic proton is always the “H” on the left side

of the formula.

The ionization of a monoprotic

acid is quite simple, as we’ve seen before:

eg.) HNO3(aq) + H2O(l) H3O+(aq) + NO3- nitric acid hydronium nitrate

ion

Notice, that if the name of the acid ends in “ic”, it’s conjugate base is an ion that ends in “ate”.

I would bet that you would be able to correctly guess the name that we give to acids

that release 2 protons (H+’s): “______protic”

Acids that release two protons are called diprotic acids

.

Some examples of diprotic acids are: H2SO4,, H2CO3, H2SO3 etc.



Some acids go as far as releasing 3 (yes, you read it right!) protons:

Acids that release three protons are called triprotic acids

.

Some examples of triprotic acids are: H3PO4,, H3AsO4, H3BO3 etc. As if that’s not enough to remember, chemists like to add one more term: Chemists count in two ways: First, like 1, 2, 3 etc. This is where “mono”, “di” and “tri” are used. They also count like this: 1, many. The prefix, they use for “many” is “poly

”.

So, to a chemists, anything more than one can be called “poly

”

So here’s another term:

Acids that release more than one proton are called polyprotic acids

.

Note here that “polyprotic acids” would include

“diprotic” and “triprotic” acids.

Personally, I don’t know of many acids that give off more than 3 protons. (One is “EDTA” or “ethylenediamine tetraacetic acid”) These are not important in Chemistry 12.

Stepwise ionization of polyprotic acids

It is important to know that polyprotic acids, when mixed with water, do not release all their protons in one step.

They release protons one by one. Each step is a separate equilibrium. Let’s look at an example. Here is the 2 steps in the ionization of sulphuric acid

(H2SO4) Step1. H2SO4 (l) + H2O(l) H3O+(aq) + HSO4-(aq) Step 2. HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq) You’ll probably wonder why there’s a single arrow in step 1 and a double arrow in

step 2. (It’s not a “typo” this time!)



Because H2SO4 is a strong acid, it releases it’s first proton 100%. That is the first step goes to completion. Every

single H2SO4 molecule breaks up into H3O+ and HSO4- ions.

But that doesn’t mean the second proton come off that easily! The HSO4- ion is

NOT a strong acid, so it does not

release all of it’s protons in water. That’s why there is a double arrow in the Step 2 equation.

Here are the equations for the step-wise ionization of a triprotic acid

(phosphoric acid)

Step 1: H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4-(aq) Step 2: H2PO4-(aq) + H2O(l) H3O+(aq) + HPO42-(aq) Step 3: HPO42-(aq) + H2O(l) H3O+(aq) + PO43-(aq) Each one of these is a separate equilibrium. Notice that one “H” and one (+) charge

comes off of the starting acid each time. Now take out your acid chart. (Yes, that means you! If you don’t, you probably

won’t understand the next couple of statements and you will get frustrated etc.) Notice the relative strengths of the acids H3PO4, H2PO4- and HPO42- . You can

find them all on the left side of the table. We can interpret their relative positions by saying that each time a proton is

removed from a polyprotic acid, it gets harder to remove the next one. We can summarize the species formed in the stepwise ionization by leaving out the

H3O+ and the H2O: step 1 step 2 step 3 H3PO4 H2PO4- HPO42- PO43- phosphoric acid dihydrogen phosphate monohydrogen phosphate phosphate Notice naming system for the ions as they form:



Amphiprotic Anions

Let’s go back to our example of phosphoric acid and look at the two ions formed in the middle steps of the process:

step 1 step 2 step 3 H3PO4 H2PO4- HPO42- PO43- phosphoric acid dihydrogen phosphate monohydrogen phosphate phosphate H2PO4- HPO42- Notice that both of these ions have at least one “H” on the left and at least one (-)

charge. Anything with an “H” of the left can act as an acid, and many ions with a (-)

charge can act as bases. Because these ions H2PO4- and HPO42- can act as either acids or bases, they are

called amphiprotic, and since they have at least one (-) charge, they are called anions

.

Putting this all together, we call these amphiprotic anions

.

The dihydrogen phosphate ion (H2PO4-), when mixed with a strong acid (like HCl), will play the role of a base

. eg.)

HCl + H2PO4- H3PO4 + Cl- acid base acid base If the dihydrogen phosphate ion is mixed with a base, or even with water, it will

play the role of an acid

: eg.)

H2PO4- + H2O H3O+ + HPO42- acid base acid base We will be looking more at amphiprotic anions later on in this unit, but for now, it’s

useful to remember that: The ion(s) formed all but the last step of the ionization of a polyprotic

acid are amphiprotic anions.



Strong & Weak Acids & Bases

Strong Acid- An acid which is 100% ionized in a water solution. E.g.) HCl(g) + H2O(l) H3O+

(aq)+ Cl-(aq)

Question: What is the [HCl(g)] in 1 M HCl? Answer: 1M Question: What is [H3O+] in 0.20 M HCl Answer: 0.20M Important:

E.g.) What is [H3O+] in 0.60 M HNO3 Answer: 0.60M Weak Acid: An Acid which is less than 100% ionized in solution. (In Chem 12 WA’s are usually significantly less than 100% ionized.) (Usually < 5% ionized) - In a solution of a weak acid, most of the molecules don’t ionize. E.g.) HF (g) + H2O(l) H3O+

(aq) + F- (aq) ions

(Molecules) (Double arrow)

[H3O+] is only a small fraction of [HF] H2O is omitted in diagram

Single arrow (goes to completion) =Strong acid

In a Strong Acid [H3O+] = [Acid] (to Start with)

HF HF HF HF HF H3O+ HF HF F- HF HF HF HF HF

NOTE: WA’s can be molecules but they might also be + or – ions. E.g.) NH4

+ + H2O H3O+ + NH3 HSO4

- + H2O H3O+ + SO42-

A beaker containing aqueous HF



- Any acid (weak or strong) could have high or low concentration. Weak & Strong refers to % ionization. Concentration the moles of acid dissolved per litre. Eg.) 10.0 M HCl conc. and strong [H3O+] = 10.0 M

0.001 M HCl dilute and strong [H3O+] = 0.001 M 10.0 M HF conc. and weak [H3O+] = low 0.001 M HF dilute and weak [H3O+] = very low

The Acid Table

Strong Acids

HClO4 H+ + ClO4

- HI H+ + I- HBr H+ + Br-

HCl H+ + Cl-

HNO3 H+ + NO3-

H2SO4 H+ + HSO4-

*Note H2SO4 is a SA but diprotic • The first ionization is 100% = H2SO4 + H2O H3O+ + HSO4

- • The second ionization is <100% HSO4

- + H2O H3O+ + SO42

Weak Acids

H3O+ H+ + H2O HIO3 H+ + IO3

-

.

. Most act as weak acids in water

.

H2O H+ + OH-

Concentration of ions would determine conductivity

Notice single arrows for all SA’s

Double arrow



OH- H+ + O2- Bottom 2 on left NEVER act as acids in water NH3 H+ + NH2

- (too weak as acids) Single arrows going backwards ( O2- and H+ can form OH- but OH- cannot form H+ and O2- in water solution.) Strong Base A substance (base) which (ionizes) or dissociates 100% in solution Weak Base A base which is less than 100% ionized in solution. E.g.) NH3(aq) + H2O(l) NH4

+ (aq) + OH-

(aq) - Consists of mostly H2O and NH3 molecules with a few NH4

+ and OH- ions.

Using Acid Table & Periodic Table Bases on Right Side Strong Bases OH- O2- Strong bases (bottom 3 on right side) NH2

- - Any substance which dissociates completely to produce OH-, O2- or NH2

- is a Strong Base

Alkali Metal Hydroxides (Group 1)

LiOH, NaOH, KOH, RbOH, CsOH are all highly (100%) soluble and form OH-, so they are all strong bases.

(Alkaline Earth) Hydroxides (Group 2) Mg(OH)2, Ba(OH)2 , Sr(OH)2 are designated as Strong Bases (even though Sr(OH)2 is the only one called “Soluble” on the Solubility Table. They dissociate to form 2 OH- s each:

Ba(OH)2(s) Ba2+(aq) + 2OH-

(aq)

Forms ions from molecules or atoms Ions in an

ionic solid separate and dissolve in water

A neutral molecule



What is the [OH-] in 0.10 M NaOH?

0.10 M 0.10 M 0.10 M NaOH(s) Na+

(aq) + OH-(aq) [OH-] = 0.10 M

What is the [OH-] in 0.10 M Ba(OH)2 ?

0.10 M _0.10_M _0.20_M Ba(OH)2 Ba2+ + 2OH-

For A Strong Base

Salts which produce O2- and NH2- are definitely strong bases.

E.g.) Quicklime in water: CaO(s) Ca2+(aq) + O2-

(aq) O2- + H2O OH- + OH- (Oxide ion) (100%) Or O2- + H2O 2OH- Find [OH-] in 0.10 M CaO [O2-] = 0.10 M (0.10M) 0.20M

O2- + H2O 2OH-

[OH-] = 0.20M

[OH-] = [Base] x # of OH’s in formula

This is a VERY important equation. Remember it!



Weak Bases Found above OH- on right side of Table. H2O IO3- Most form weak bases in water . Why do I say “most”? . . PO4

3- OH- O2- Strong bases NH2

- Very Weak (non-hydrolyzing Bases) or Spectators These are the top 5 (not 6) “bases” on the right. CIO4

-

They are so weak that they cannot react with H2O to form OH- I- (They do not contribute any OH- to a solution) Br- For this reason, these top 5 on the right are not usually referred to as “bases” in aqueous solution. They are called Spectators! Cl- NO3

-

Conj. Bases of strong acids---- In acid-base reactions they are SPECTATORS In a SA, the bond to H+ is weak H Cl + H2O H3O(aq)

+ +

So weak , it cannot take an H+ from H2O or even H3O+

SA’s have non-hydrolyzing (spectator) ions for conj. Bases.

Cl(aq)-

Easily broken by water

Weak Bond



Amphiprotic Species (ions or molecules) - are found on both sides of the table e.g.) HSO4

- - can act as acids (donate H+’s) or as bases (accept H+’s) - to look at an amphiprotic species as an acid, you must find it on the left side:

e.g.) C6H5OH

HCO3-

H2O2 HCO3

- is a weaker acid than C6H5OH HCO3

- is a stronger acid than H2O2 - to look at an amphiprotic species as a base, you must find it on the right side:

for HCO3- as a base:

e.g.) H+ + Al(H2O)5(OH)2+

H+ + HCO3-

H+ + C6H5O73-

HCO3

- is a weaker base than C6H5O73-

HCO3- is a stronger base than Al(H2O)5(OH)2+

HSO4

- in shaded region on top right will not act as a base in water (Too weak of a base) - However, it is not a spectator! (like NO3

- is) Why not? (HSO4

- is also found on the left side quite a way up, it is a relatively “strong” weak acid.) The Leveling Effect for Acids

What is [H3O+] in 1.0 M H3O+ ? 1.0M

What is [H3O+] in 1.0 M HNO3? 1.0M What is [H3O+] in 1.0 M HCl ? 1.0M

Acids from HClO4 to H2SO4 are 100% ionized in water

only solvent used in Chem 12 (and most Chemistry)

- so even though HClO4 is above HCl on the chart, it is no more acidic in a water

solution. H3O+ is the strongest acid that can exist in an undissociated form in water solution. - all stronger acids ionize to form H3O+

Acid Strength Increases

Base Strength Increases



(NOTE: although H2SO4 is diprotic, the H3O+ produced from the second ionization is very little compared to that from the first) 1st ionization: H2SO4 + H2O H3O+ + HSO4

-

1M(SA) 1M 2nd ionization: HSO4

- + H2O H3O+ + SO42-

~1M (WA) The only way you can tell which strong acid is “stronger” is to react them in a non-aqueous (not H2O) solvent. Eg) HClO4 + H2SO4 H3SO4

+ + ClO4-

(it is found that HClO4 donates a proton to H2SO4, not the other way around, so HClO4 is a stronger acid than H2SO4) This is not important in Chemistry 12. This would not happen in a water solution. (In H2O, they would both form H3O+) Leveling Affects of Bases The strongest base which can exist in high concentrations in water solution is OH- The two stronger bases below it will react with water completely to form OH-. Eg) O2- + H2O OH- + OH- SB

Or O2- + H2O 2OH- What is the final [O2-] in 1.0 M Na2O ? Answer: 0 M - All the O2- will react with water to form OH- 1.0M 2/1 2.0 M

O2- + H2O 2OH- so [OH-] = 2.0 M Acid-Base Equilibria & Relative Strengths of Acids & Bases - Take out your acid table - Mix some H2PO4

- and some CO32-

A very small amount of H3O+

Amphiprotic can donate or accept an H+

Can only act as a base (accept an H+) (doesn’t have an H+ to give

Single Arrow



So, in this case CO3

2- will play the role of base (take H+) and H2PO4- will play the role of

acid (donate an H+). H2PO4

- + CO32- HCO3

- + HPO42-

(A) (B) (A) (B) Consider the 2 acids H2PO4

- and HCO3-

Question: At equilibrium, which will be favoured, reactants or products? They both “want” to donate protons. - look them both up on the left side

H2PO4

- is above HCO3- on LEFT, so H2PO4

- is a stronger acid than HCO3-.

H2PO4- H+ + HPO4

2

H+ + CO32- HCO3

-

So the reaction:

H2PO4- + CO3

2- HCO3- + HPO4

2-

Will have a greater tendency to go right than left and products will be favoured. - so find acid on each side. Equilibrium favors the side with the weaker acid. “Only the weak survive” or “Survival of the weakest”

Strength as acid

Strong “push” to donate a proton

Weaker “push” to donate a proton



“stronger” means a greater tendency to react and change to something else. H2PO4

- + CO32- HCO3

- + HPO42-

SrA WrA Don’t use terms “strong” and “weak”, they have other specific meanings. Question: Will HSO3

- + HCO3- H2CO3 + SO3

2- Favor reactants or products? Mixing 2 amphrotic ions (products not given) -complete rx. and tell which is favoured (r or p) eg.) HSO4

- + H2PO4- ?

Which will play role of acid? (both are capable of being acids or bases) - First, compare these two on LEFT side HSO4

- is higher than H2PO4- on LEFT side so has a greater tendency to act as an acid.

- Complete the equation: (making HSO4

- act as the acid.)

HSO4- + H2PO4

- H3PO4 + SO42-

A B A B

Stronger acid

Weaker acid

HA! HA! I win. I get to give you my proton. You have to act as the base!

Identify A & B on this side

I always knew you were a bigger loser than I am!



Now compare the 2 conjugate acids (Look fo them both on the LEFT side of chart.) HSO4

- is slightly ABOVE H3PO4 on the left side so HSO4- is the SrA and H3PO4 is the WrA.

HSO4

- + H2PO4- H3PO4 + SO4

2- so the products (with WrA,) are favoured! SrA WrA

-Comparing realtive stengths of bases. E.g.) HSO4

- + H2PO4- H3PO4 + SO4

2- Base Base

Compare these on the RIGHT side of table

H2PO4

- is lower on the right side(stronger base) than SO42-

So see: HSO4

- + H2PO4- H3PO4 + SO4

2- SrA SrB WrA WrB -Since this equilm favoured products (H3PO4 is WrA), we can say that equilm favours the side with the weaker conjugate base. NOTICE: The SrA is on the same side as the SrB. [the SrA has the weaker conj. Base] The WrA is on the same side as the WrB (Birds of a feather flock together) or (The weakies hang out together and survive better than the “strongies”.) - So we could compare conj. Acids or conj. Bases. Equilm favors the side with the

weaker conj. Acid and the weaker conj. Base. Starting with “Salts” The amphiprotic ions are often products of the dissociation of salts. - Spectator ions must be discarded. NOTE: All alkali ions Na+, K+, Li+ …etc….. are spectators in Acid-Base reactions. Also top five ions right side of acid chart ( CIO4

-, I-, Br-, Cl-, NO3-) are spectators in Acid-Base

reactions.



E.g.) complete the net ionic reaction between and state whether equilm favors reactants or products

NaHSO3 and K2HPO4

Dissociate (Na+) HSO3

- (K+) HPO42-

HSO3

- + HPO42-

HSO3

- is higher, so it will play the role of the acid. HSO3

- + HPO42- H2PO4

- + SO32-

SrA B WrA B HSO3

- is a stronger acid than H2PO4-, so equilm favors the side with the weaker acid

(H2PO4-) so products are favored!

Relating The Keq to A-B equilibria If products are favored Keq is large (>1) If reactants are favored Keq is small (<1) Eg.) Given: HA + B- HB + A- Keq = 0.003 Which acid is stronger, HA or HB? Keq is small so reactant side is favored. Since equilm favors side with WrA, HA must be the weaker acid, so HB would be the stronger acid. - Which is the stronger base? (the SrB is on the same side as the SrA)

NOTICE THIS Had 2 K’s, so charge is 2- NOT - Spectator ions

DISCARD!

You LOSER!! You get to lose a proton to me!

I WON! I get to be an acid! Look for both

on LEFT side



Ionization constant for water

or ( the weaker acid (HA) has the stronger conj. Base (A-)) Ionization of Water DEMONSTRATION OF CONDUCTIVITY OF TAP WATER AND DISTILLED WATER - Pure distilled water still has a small conductivity. Why? - There are a few ions present. - Almost all the pure water is H2O molecules. - But every once in a while, this happens: H + H -

O + H H O + O

H H O H H A proton is transferred Hydronium Hydroxide From one water molecule to another. Equation: Ionization of Water H2O + H2O H3O+ + OH- Or 2H2O(l) H3O+

(aq) + OH-(aq)

Process is Endothermic 2H2O(l) + 59KJ H3O+

(aq) + OH-(aq)

All water or aqueous solutions

contain these. In neutral water [H3O+] = [OH-] In acidic solutions [H3O+] > [OH-] Know these!! In basic solutions [OH-] > [H3O+] 59KJ + 2H2O(l) H3O+

(aq) + OH-(aq)

Keq = [H3O+] [OH] (liquid water left out) Given a special name for ionization of water—called Kw

So Kw = [H3O+] [OH-]

Always true at any temp!

An equilibrium is established



Since reaction is endothermic: 59KJ + 2H2O(l) H3O+

(aq) + OH-(aq)

At higher temps products are favoured and Kw is higher. At lower temps reactants are favoured and Kw is lower. At 25oC (only) Kw = 1.00 x 10-14 Know this!! At 100C Kw = 0.295 x 10-14 (smaller)

At 600C Kw = 9.55 x 10-14 (larger) So Always: [H3O+] [OH-] = Kw

At 250C only: [H3O+] [OH-] = 1.00 x 10-14 [H3O+] & [OH-] in Neutral Water At 25oC (NOTE: Assume Temp = 25oC unless otherwise noted) [H3O+] [OH-] = 1.00 x 10-14 and [H3O+] = [OH-] if water is neutral. (If “water” is mentioned in a problem, it can be assumed to be NEUTRAL unless otherwise stated!) (substitute. [H3O+] for [OH-]) [H3O+] [H3O+] = 1.00 x 10-14 [H3O+]2 = 1.00 x 10-14 [H3O+] = 1.00 x 10-14 = 1.00 x 10-7 M

Also [OH-] = [H3O+] = 1.00 x 10-7 M

[H3O+] & [OH-] in Acids and Bases 2H2O(l) H3O+

(aq) + OH-(aq)

Add acid, H3O+ increases, so equilibrium shifts LEFT and [OH-] decreases Add base, [OH-] increases, so the equilibrium shifts LEFT and [H3O+] decreases. Finding [H3O+] and [OH-] in Acids and Bases At 250C

For comparison

All acids produce H3O+ in water



[H3O+] is very low in a strong fairly concentrated base

Eg.) Find the [OH-] in 0.0100 M HCl [H3O+] = 0.0100 M [H3O+][OH-] = 1.00 x 10-14 [OH-] = 1.00 x 10-14 = 1.00 x 10-14 = 1.00 x 10-12 M [H3O+] 1.00 x 10-2 Find [H3O+] in 0.300 M NaOH. [H3O+][OH-] = 1.00 x 10-14 [H3O+] = 1.00 x 10-14 = 1.00 x 10-14 = 3.33 x 10-14 [OH-] 0.300 Find [H3O+] in 0.020 M Ba(OH)2 [OH-] = 0.040 M [H3O+] = 1.00 x 10-14 = ___2.5 x 10-12M ( 0.040 ) At Other Temps - you’d be given Kw eg.) Kw at 600C = 9.55 x 10-14 Calculate [OH-] in 0.00600 M HNO3 at 600C. [H3O+][OH-] = Kw SA [H3O+][OH-] = 9.55 x 10-14 [OH-] = 9.55 x 10-14 = 1.59 x 10-11 M 0.00600 pH -Shorthand method of showing acidity (or basicity, alkalinity) If [H3O+] = 0.10 M (1.0 x 10-1 M) pH = 1.00 [H3O+] = 0.00010 M (1.0 x10-4 M) pH = 4.00

[OH-] is less in an acid than in neutral water.

In a Strong Acid ([H3O+] = [acid])

In a STRONG BASE, [OH-] = [base] x # of OH’s



Definition of pH pH= -log10 [H3O+] (assume log = log10) If [H3O+] = 1.0 x 10-7 pH = -log (1.0 x 10-7 ) Regular Scientific Calculator. Enter: 1 EXP 7 +/- LOG +/- and the answer should be 7 For DAL (Sharp) calc. Enter: +/- log 1 Exp +/- 7 = and the answer should be 7 For a TI 83 Enter (-) LOG 1 2nd EE (-) 7 ENTER and the answer should be 7 NOTE: If you are using a DAL or a TI 83 calculator and the number you want to find the pH of is the answer to a an ongoing calculation, leave the answer to your calculation in the calculator and press (-) or +/- log 2nd ANS ENTER ( or =). Practice finding pH’s on your own calculator. You will be doing many of these in the rest of this unit and it’s important that you can do it quickly and easily and accurately! Find the pH of 0.030 M HCl [H3O+] = 0.030 M pH= -log (0.030) = 1.522878745 How to round off?? Sig. Digits in pH start at decimal point!!! 1.52287….. so pH = 1.52 Find the pH of neutral water at 250 C [H3O+] = 1.00 x 10-7

pH = 7.000 Find the pH of 0.00100 M NaOH at 250 C

2 SD’s SA. So [H3O+] = [acid]

No units for pH

Start counting here.

Basic solution

Use the (-) button at the bottom right of the white pad. NOT the “—“ on the far right



[H3O+] = 1.00 x 10-14 = 1.00 x 10-11 M 0.00100 so pH = 11.000 At 25 oC In neutral water pH = 7.0 In acid solution pH < 7.0 In basic solution pH > 7.0

pH Scale (@ 25oC)

More Acidic Neutral More Basic

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Another example: Calculate the pH of 15.0 M NaOH: [H3O+] = 1.00 x 10-14 = 6.67 x 10-16 M 15.0 pH = -log (6.67 x 10-16) pH = 15.176

Converting pH to [H3O+]

pH = - log [H3O+] -pH = log [H3O+] antilog (-pH) = [H3O+]

or [H3O+] = antilog (-pH) eg.) If pH = 11.612 , find [H3O+] [H3O+] = antilog (-11.612) For regular Scientific Calculator: Enter: 11.612 +/- 2nd log The calculator answer should be 2.443430553 x 10-12

3 sig. digits

Put in (-) before you take the antilog

3 SD’s

[OH-]

Very concentrated Base



The original pH had 3 SD’s, so the answer must also have 3 SD’s (Remember the answer is NOT a pH, so digits to the LEFT of the decimal point are also significant!) . Remember that concentration also has a unit!. So the answer would be reported as: [H3O+] = 2.44 x 10-12 M For a DAL or TI 83 enter: 2nd log (-) (or +/-) 11.612 ENTER (or =) Logarithmic Nature of pH

A change of 1 pH unit a factor of 10 in [H3O+] or (acidity) eg.) pH = 3.0 [H3O+] = 1 x 10-3 M pH = 2.0 [H3O+] = 1 x 10-2 M How many times more acidic is pH 3 than pH 7? pH 7 [H3O+] = 1 x 10-7 x 104 = 10,000x pH 3 [H3O+] = 1 x 10-3

or taking antilog of difference in pH 7 – 3 = 4

antilog 4 = 104 = 10,000 times (remember lower pH more acidic)

Natural rainwater pH ~ 6 Extremely acidic acid rain pH ~ 3 diff = 3 & antilog (3) = 103 (1,000) So, the acid rain is 1000 times more acidic than natural rain water!

pOH

pOH = -log [OH-] And [OH-] = antilog (-pOH) Calculate the pOH of 0.0020 M KOH [OH-] = 2.0 x 10-3 M pOH = -log (2.0 x 10-3 ) = 2.70 Find the pH of the same solution: [OH-] = 2.0 x 10-3 M [H3O+] = 1.00 x 10-14 = 5.0 x 10-12 2.0 x 10-3

3 SD’s unit

10 times more acidic pH goes down 1 unit



pH = 11.30 Notice: pH + pOH = 14.00 From Math: If a x b = c Then: loga + logb = logc Eg.) 10 x 100 = 1000 Log(10) + log(100) = log(1000) 1 + 2 = 3 So since [H3O+] [OH-] = Kw log[H3O+] + log[OH-] = log (Kw) or make everything negative -log[H3O+ ] + -log [OH- ] = -log Kw pH + pOH = pKw (relation) where pKw = -log Kw (definition of pKw) Specifically at 250C Kw = 1.00 x 10-14 pKw = -log (1.00 x 10-14) pKw = 14.000 so at 250C

pH + pOH = 14.000 At 250C If pH = 4.00 pOH = 10.00 Or: If pH = 2.963 pOH= 11.037 eg.) Find the pH of 5.00 x 10-4 M LiOH (250C)

plan: [OH-] pOH pH [OH-] = 5.00 x 10-4 pOH = 3.301 pH = 14.000 – 3.301 = 10.699 eg.) Find the pOH of 0.0300 M HBr (250C)

[H3O+] = 0.0300 M (HBr is a strong acid) pH = 1.523

True at all temperatures

Only at 250C



pOH = 14.000 – 1.523 pOH = 12.477

When not at 250C Eg.) At 600C Kw = 9.55 x 10-14 Find the pH of neutral water at 600C. One way: Calculate pKw pKw = -log Kw = -log (9.55 x 10-14) At 600C pKw = 13.020 For neutral water pH = pOH ([H3O+] = [OH-]) pH + pOH = pKw (substitute pH for pOH) pH + pH = 13.020 2pH =13.020 so pH = 13.020 = 6.510 2 Is pH always 7.00 in neutral water? Nope!!!! At higher temp: 2H2O + heat H3O+ + OH- [H3O+] > 1.0 x 10-7 so pH < 7 [OH-] > 1.0 x 10-7 so pOH < 7 Summary: In neutral water pH = pOH at any temp.

pH & pOH = 7.00 at 250C only. At lower temps pH and pOH are > 7

At higher temps pH and pOH are < 7

At any temp: pH + pOH = pKw At 250C: pH + pOH = 14.000

Ka & Kb for Weak acids and Bases

Reminder: pH of SA’s

[H3O+]=[acid] strong means 100% ionized so, to find pH of 0.100 M HCl [H3O+] = 0.100 M pH = 1.000

These are very important? Make sure you study these!



For weak acids [H3O+] << [acid] Eg.) What is pH of 0.10 M HF? Look at equilibrium for Weak Acid HF HF(aq) + H2O(l) H3O+

(aq) + F-

(aq) Keq = [H3O+][F-] for WA’s Keq is called Ka (acid ionization constant)

[ HF] - see acid table for list of Ka’s. higher Ka stronger acid lower Ka weaker acid For SA’s (eg. HCl) Ka = [H3O+] [Cl-] = called “very large!” [HCl] -Discuss Relative Strengths of Oxyacids Calculations Using Ka (Used for Weak Acids )

[H

3O+] from Ka (pH from Ka)

1. [H3O+] from Ka and Original concentration (Co) eg.) Find the [H3O+] in 0.10 M HF 1. Write out equilibrium equation for ionization

HF + H2O H3O+ + F-

Dissolved but not ionized

Pure liquid not included in Keq expression.

Essentially zero molecular HCl

WA



“root” a “coka” (beer) (cola)

2. Ice table HF + H2O H3O+ + F-

[I] 0.10 0 0 [C] -x +x +x [E] 0.10 –x x x

3. Ka expression: Ka = [H3O+][F-] [HF] Ka = (x) (x) 0.10 - x 4. Substitute Ka = x2

0.10 ( Assume 0.10 – x ≅ 0.10 ) 5. Solve for x ([H3O+])

Ka = x2 So x2 = 0.10 Ka 0.100 [H3O+] = x = = [H3O+] = 5.9 x 10-3 M ~ Check assumption (we see that this is quite small compared to 0.10)

Do ex. 74 & 75 Pg. 152

After Questions 74 & 75

Short cut for multiple choice Only!! For WA [H3O+] = Eg.) Find pH of 2.0 M acetic acid (Multiple Choice Question) 1. First [H3O+] =

=

[H3O+] = 6.0 x 10-3 M

2. Find pH = -log (6.0 x 10-3)

pH = 2.22

To avoid quadratic assume x is insignificant compared to 0.10 This can be confirmed later.

Ka from Acid Table

NOTE: In W.R. questions, full solution must be shown including the assumption!

2 SD’s. The “2.0 M” was 2 SD’s and the Ka was 2 SD’s

Try using this for a SA!!

In Chem. 12 with weak acids, always use this assumption - Make sure you state it!

You must state this assumption here!

Co is Original Acid Concentration

Ka from Acid Table

0.10 Ka

0.10 (3.5 x 10-4)

CoKa

CoKa

2.0 (1.8 x 10-5 )



NOTE: Ions which act as acids can come from compounds. Eg.) See table ~ ammonium ion NH4

+ - can be found in NH4NO3, NH4Cl, NH4Br, etc….

ferric ( hexaaquoiron) Fe3+ ( Fe(H2O)63+) could be found in Fe(H2O)6 Br3 (also called

FeBr3) or Fe(H2O)6 (NO3)3 (also called Fe(NO3)3 ) Aluminum (hexaaquoaluminum) Al3+ (Al(H2O)6

3+) could be found in Al(H2O)6Cl3 ( also called AlCl3 )

More Ka Calculations: Ka from pH Eg.) a 0.350 M Solution of the weak acid HA has a pH of 1.620. Find the Ka of HA. 1. First convert pH to [H3O+] [H3O+] = antilog (-pH) = antilog (-1.620) [H3O+] = 2.399 x 10-2 M 2. Write out equilibrium equation for ionization. Make an ICE table:

HA + H2O H3O+ + A- [I] 0.350 0 0 [C] [E] 2.399 x 10-2 Now, you can see that the change in concentration [C] of [H3O+] is + 2.399 x 10-2 M and using the mole ratios (mole bridges) in the balanced equation, you can figure out the [C]’s for the A- and the HA: -2.399 x 10-2M + 2.399 x 10-2M + 2.399 x 10-2M HA + H2O H3O+ + A- [I] 0.350 0 0 [C] - 2.399 x 10-2 + 2.399 x 10-2 + 2.399 x 10-2 [E] 2.399 x 10-2 Now, we can figure out the equilibrium concentrations of HA and A-. There are no “x”s in the table so we don’t need to make any assumptions. It is best to use your calculator to figure out the equilibrium [HA], because the [C] may or MAY NOT be insignificant. Using a calculator 0.350 –2.399 x 10-2 = 0.32601. Don’t round it off too much here. I would keep it in a memory in my calculator. BUT BECAUSE THE “0.350” IS 3 DECIMAL PLACES AND YOU ARE SUBTRACTING, THE [E] OF “HA” CANNOT HAVE MORE THAN 3 DECIMAL PLACES (although you should use 0.32601 in your calculator) JUST REMEMBER THAT IN THE NEXT CALCULATION, THE 3 DECIMAL PLACES IN THE ICE TABLE TRANSLATES TO 3 SD’S, SO YOUR FINAL ANSWER CANNOT HAVE MORE THAN 3 SD’S.

This is the [H3O+] at equilibrium

NOTE: the pH is to 3 SD’s so your final answer cannot have more than 3 SD’s.



-2.399 x 10-2M + 2.399 x 10-2M + 2.399 x 10-2M HA + H2O H3O+ + A- [I] 0.350 0 0 [C] - 2.399 x 10-2 + 2.399 x 10-2 + 2.399 x 10-2 [E] 0.326 2.399 x 10-2 2.399 x 10-2

3. Write Ka expression & substitute values.

Ka = [H3O+][A-] = (2.399 x 10-2) 2 = 1.7653 x 10-3 and expressing in 3SD’s, the answer is: [HA] 0.326

Ka = 1.77 x 10-3

For those that want a short-cut for multiple choice:

Ka from [H3O+] : Ka = [H3O+]2 ( Co –[H3O+] )

Don’t rearrange [H3O+] =

To Calculate Co ( conc. of acid needed) form pH & Ka Eg. Find the concentration of HCOOH needed to form a solution with pH = 2.69 1. First change pH to [H3O+] [H3O+] = antilog (-pH) = antilog (-2.69) [H3O+] = 2.0417 x 10-3 M (notice that the given pH limits us to 2SD’s, but keep more in your calculations.) 2. Write out ionization equilibrium with an ICE TABLE. You can insert 2.0417 x 10-3 for equilibrium

[H3O+]. And since our unknown is the initial [HCOOH], we put in an “Co” for the [I] of HCOOH:

HCOOH + H2O H3O+ + HCOO- [I] Co 0 0 [C] [E] 2.0417 x 10-3

OK for multiple choice ONLY!

Don’t forget this

Was derived using an assumption which may NOT be valid!

CoKa



Now we can see that the change in concentration [C] of [H3O+] is “ + 2.0417 x 10-3 ” and the [C] of HCOO- will be the same. The [C] of HCOOH will be “ - 2.0417 x 10-3 ”

HCOOH + H2O H3O+ + HCOO- [I] Co 0 0 [C] - 2.0417 x 10-3 + 2.0417 x 10-3 + 2.0417 x 10-3 [E] 2.0417 x 10-3

We can now calculate the equilibrium concentrations [E] of HCOOH and HCOO-.

HCOOH + H2O H3O+ + HCOO- [I] Co 0 0 [C] - 2.0417 x 10-3 + 2.0417 x 10-3 + 2.0417 x 10-3 [E] Co - 2.0417 x 10-3 2.0417 x 10-3 2.0417 x 10-3

The next step will be to write the Ka expression and substitute the equilibrium concentrations in: 3. Write Ka expression. Substitute equilibrium concentrations in. Find Ka for HCOOH on the acid table:

Ka = [H3O+] [HCOO-] [ HCOOH]

1.8 x 10-4 = (2.0417 x 10-3)2 ( Co – 2.0417 x 10-3) Now we can solve for Co (the original concentration of the acid): Co- 2.0417 x 10-3 = (2.0417 x 10-3)2 1.8 x 10-4

Co – 2.0417 x 10-3 = 2.3159 x 10-2

Co = 2.3159 x 10-2 + 2.0417 x 10-3 Co = 2.52 x 10-2 M Co = 2.5 x 10-2 M or 0.025 M (remember, we are restricted to 2 SD’s) Now For Bases

Base ionization NH3 very common weak base. It partially ionizes in water to form NH4

+ and OH- :

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

Equilibrium constant – called base ionization constant (Kb)

Find Ka on Acid Table



NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

Kb expression: Kb = [NH4

+] [OH-] [NH3]

NOTE: Ions can also act as a weak bases. The reaction of an ion with water to form OH- is called base hydrolysis. Equilibrium constant is still called Kb.

Eg.) Hydrolysis of CN-

CN- (aq) + H2O(l) HCN(aq) + OH-

(aq)

Kb = [HCN][OH-] [CN-] Ionization of N2H4 (weak base) N2H4 (aq) + H2O (l) N2H5

+ (aq)

+ OH-(aq)

Kb = [N2H5

+] [OH-] [N2H4]

How to Find Kb using Acid Table (not shown directly)

Derivation - Look at hydrolysis of base F-: F- + H2O HF + OH-

Kb (F-) = [HF] [OH-] [F-] - Look at ionization the weak acid HF: HF + H2O H3O+ + F-

Ka (HF) = [H3O+] [F-] [HF] - Multiply Ka[HF] x Kb[F

-]

Ka[HF] x Kb[F

-] = [H3O+] [F-] x [HF] [OH-] = [H3O+] [OH-] (notice that [HF] and [F-] will cancel.)

[HF] [F-]

or Ka(HF) x Kb(F-) = Kw

Already an ion

Kb expression

A neutral molecule

Kb expression

Conj. Acid

Conj. base



Ka(HF) x Kb(F-) = Kw

Or Kb(F

-) = Kw

Ka(HF) In general: Kb(weak base) = Kw Ka(it’s conj. acid) Using Acid Table: 1. Find base on right side ( if amphiprotic -locate base on right side only) 2. It’s conjugate acid will be across from it on the left side. 3. The Ka of it’s conjugate acid is on the far right of the same line. 4. Use equation: Kb(base) = Kw

Ka(conj. acid)

Eg.) Calculate the Kb for HCO3- : ( find HCO3

- on RIGHT SIDE) Line: H2CO3 H+ + HCO3

-………4.3 x 10-7 Base Ka(it’s conj. acid) Kb(HCO3

-) = Kw = 1.0 x 10-14 = 2.3 x 10-8 Ka(H2CO3) 4.3 x 10-7

Find Kb of SO3

2-

Similarly : If Kb (base) given Ka (weak acid) = Kw Kb(it’s conj. Base) Eg.) The Kb for base B- is 2.73 x 10-9 Find the Ka for the acid HB 3 sig. Digs.

Ka(HB) = Kw = 1.00 x 10-14 = 3.66 x 10-6 Kb(B

-) 2.73 x 10-9

It’s conj. acid



NOTES:

Table only states Ka values. For questions like this Kb will have to be calculated if not given. All Ka’s on table are 2 SD’s—limits any calculation using them to 2 SD’s maximum. The larger the Kb, the “stronger ” the weak base - the more OH- produced. The smaller the Ka of an acid, the larger the Kb of its conjugate Base. (Weaker acids have

stronger conjugate bases)

Calculations Involving Kb

Given: Find or find or find [Base] & Kb [OH-] pOH pH eg.) Find [OH-] in a 0.20 M solution of KNO2 (this is a salt, so it must be dissociated into it’s ions first) Dissociation of KNO2 K+ + NO2

- 1. Find Kb of NO2

- Weak base Kb(NO2

-) = Kw Ka(HNO2) = 1.0 x 10-14

4.6 x 10-4 Kb(NO2) = 2.174 x 10-11

2. Hydrolysis ( if ion) or ionization ( if molecule) equation followed by an ICE table:

NO2- + H2O HNO2 + OH-

[I] 0.20 0 0 [C] [E]

We can let “x” equal the moles/L of NO2

- which hydrolyze as the reaction reaches equilibrium. Using the mole ratios in the equation and calculating [E]’s we get:

NO2- + H2O HNO2 + OH-

[I] 0.20 0 0 [C] -x +x +x [E] 0.20 - x x x

3. Kb Expression:

Kb = [HNO2] [OH-] at equilibrium [NO2

-] On the next page, we substitute the [E]’s into the Kb expression (Don’t forget, we are in “baseland” not “acidland” now!)

Neutral spectator (discard)



Substitute: Kb = [HNO2] [OH-] or Kb = x2

[NO2-] (0.20 - x)

Approximation: Kb ≅ x2 0.20 x2 = 0.20 Kb [OH-] = [OH-] =

[OH-] = 2.085 x 10-6 M [OH-] = 2.1 x 10-6 M Both the 0.20 M and the Ka used in calculating Kb limit us to 2 SD’s.

NOTE: In many problems, finding [OH-] is only the first step. Often you have to find the pH. Then you would use the process [OH-] pOH pH

Kb From pH and Concentration

Eg.) At a certain temp, a 0.20 M solution of K2SO3 has a pH of 10.25. Calculate the Kb of SO32- at this

temp. 1. Identify SO3

2- as a weak base (When K2SO3 is dissociated, it yields K+(a spectator) and SO32- )

2. [OH-] can be obtained from pH (pH pOH [OH-] )

pH = 10.25 so pOH = 14.00 – 10.25 = 3.75 [OH-] = antilog (-pOH) = antilog (-3.75) = 1.778 x 10-4 M

3. Write hydrolysis equation and an ICE table. (It is called hydrolysis this time because SO3

2- is an ion.) We know the initial [SO3

2- ] is 0.20M and the equilibrium [OH-] is 1.778 x 10-4 M:

SO32- + H2O HSO3

- + OH- [I] 0.20 0 0 [C] [E] 1.778 x 10-4

So, from this, the change in conc. [C] of OH- is “+ 1.778 x 10-4 ” and using the coefficient ratios we can insert the [C]’s for the other species and calculate the equilibrium concentrations [E]’s:

SO32- + H2O HSO3 + OH-

[I] 0.20 0 0 [C] - 1.778 x 10-4 + 1.778 x 10-4 + 1.778 x 10-4 [E] 0.1998 1.778 x 10-4 1.778 x 10-4

Original conc. of NO2

- (Co)

Assume that 0.20 – x ≅ 0.20

0.20 Kb

0.20 (2.174 x 10-11)



4. Now we write the Kb expression and substitute the values from the [E]’s in our ICE table: Kb = [HSO3

-] [OH-] = ( 1.778 x 10-4)2 = 1.5825 x 10-7 [SO3

2-] (0.199822)

So:

Kb of SO3

2- at the temperature given is = 1.6 x 10-7 Short Cut for Multiple Choice:

Kb = [OH-]2 (CO – [OH-]) Like Weak acids:

Ka = [H3O+]2 (Co – [H3O+])

Hydrolysis: - Reaction between a salt (ion or ions in a salt) and water to produce an acidic or basic solution.

- Net ionic equations for hydrolysis: An ion + water a molecule or ion + H3O+ or OH- SPECTATORS- ions which do NOT hydrolyze (need periodic table and acid table to find these) Spectator Cations (look on per. table) Group 1 (Alkali Metal ions) eg. Li+, Na+, K+, Rb+, Cs+, Fr+ Group 2 (Alkaline Earth ions) eg. Be+, Mg2+, Ca2+, Ba2+, Sr2+, Ra2+ Spectator Anions (look on acid table) - Conjugate bases of strong acids.

- Top 5 ions on the right side of table. - ClO4

- I- Br- Cl- NO3-

(HSO4- is not a spectator – it is amphiprotic – will be dealt with later

- spectators are eliminated in net ionic equations (NIE’s) for hydrolysis! Process – if given salt (dissociate eliminate evaluate) 1. Write dissociation equation 2. Eliminate spectators 3. Remaining ions left side of table – undergo acid hydrolysis is –produce H3O+

right side of table – undergo base hydrolysis – produce OH- amphiprotic – determine Ka and Kb to find dominant hydrolysis.

We round it to 1.6 x 10-7 because the 0.20M and the pH given both had 2 SD’s



Examples: Determining A, B, or N

Is the salt NaF Acidic, basic or neutral in water? Dissociation : NaF Na+ + F- Spectator (alkali cation) Found on right side of acid table-

forms a weak base. - so NaF is basic Is the salt NH4 NO3 acidic, basic or neutral in aqueous solution? Dissociation: NH4NO3 NH4

+ + NO3-

Spectator top 5 on right side of table Found on left side of table – forms a weak acid - so NH4NO3 is acidic.



Is the salt KCl acidic, basic or neutral?

Dissociation: KCl K+ + Cl- - since neither ion undergoes hydrolysis, this salt is NEUTRAL. Cations Which Hydrolyze

- Hydrated cations

- metals from center of the periodic table (transition metals) are smaller ions and have larger charges - this attracts H2O molecules

eg.) Fe3+ (iron (III) or ferric ion)

Hydration: Fe3+ + 6H2O Fe(H2O)63+

This ion acts as a weak acid (see it ~ 13th down on the acid table.) The equation for the hydrolysis of hexaaquoiron or ferric ion is: Fe(H2O)6

3+(aq) + H2O(l) H3O+

(aq) + Fe(H2O)5(OH2+)(aq) 3 Common Hydrated cations (on left of acid chart): iron(III) Fe3+ forms Fe(H2O)6

3+ hexaaquoiron(III) Chromium(III) Cr3+ forms Cr(H2O)6

3+ hexaaquochromium(III) Act as weak acids. Aluminum Al3+ forms Al(H2O)6

3+ hexaaquoaluminum Eg.) AlCl3 is the same as Al(H2O)6Cl3

Another Acidic Cation

NH4+ Hydrolysis equation: NH4

+(aq) + H2O(l) H3O+

(aq) + NH3(aq) Ammonium

Can appear in either form in salts

Spectator (alkali ion)

Spectator (top right of acid table)

Called the hexaaquoiron (III) ion



ANIONS WHICH HYDROLYZE

Looking on the RIGHT side of the ACID TABLE: Strength Base Ka of base

H+ + ClO4

- ................... very large Weak H+ + I- .......................... very large H+ + Br- ....................... very large H+ + Cl- ........................ very large H+ + NO3

- .................... very large H+ + HSO4

- .................. very large H+ + H2O ...................... 1.0 H+ + IO3

- ....................... 1.7 x 10-1 H+ + HOOCCOO- ......... 5.4 x 10-2

H+ + HSO3- ................... 1.7 x 10-2

: : : H+ + PO4

3- .................... 4.4 x 10-13 H+ + OH- ...................... 1.0 x 10-14

H+ + O2- ...................... very small

H+ + NH2- ................... very small Strong

All of the anions in this section from IO3

- down to S2- will undergo base hydrolysis. Anions that are NOT amphiprotic will act as weak bases in water. We will deal with amphiprotic anions (eg. HCOO-) later. Some examples of net-ionic hydrolysis equations for these would be:

IO3-

(aq) + H2O (l) HIO3 (aq) + OH- (aq)

NO2-

(aq) + H2O (l) HNO2 (aq) + OH- (aq)

CH3COO- (aq) + H2O (l) CH3COOH

(aq) + OH- (aq) Salts which contain these anions may also be basic (depending on the cation). When you get a salt, you must dissociate it, eliminate spectators and then look for hydrolysis of any remaining ions. Eg.) Determine whether the salt sodium carbonate (Na2CO3) is acidic, basic or neutral in aqueous solution. First dissociate the salt: Na2CO3 2 Na+

(aq) + CO32-

(aq) The net-ionic equation for the hydrolysis taking place in this salt would be:

CO32-

(aq) + H2O (l) HCO-3

(aq) + OH- (aq) and the salt would act as a weak base in water.

Remember: These five ions DO NOT hydrolyze. They are spectators!

All ions in this section can undergo BASE HYDROLYSIS. Most act as weak bases in water.

These 3 ions act as STRONG BASES. They undergo 100% hydrolysis to form OH- ions

Spectator Cation Eliminate!

Hydrolyzing Anion(weak base)



Remember that “net-ionic” means that any spectator ions have been removed! Hydrolysis When BOTH Cation and Anion hydrolyze Eg. Is the salt ammonium nitrite NH4NO2 acidic, basic or neutral? Of course we start out by dissociating: NH4NO2 NH4

+(aq) + NO2

-(aq)

Remember that NH4

+ produces H3O+ (NH4+ + H2O H3O+ + NH3) (equation 1)

And NO2- produces OH- (NO2

- + H2O HNO2 + OH- ) (equation 2) The Ka for NH4

+ tells how much H3O+ it produces ( The Keq for equation 1 is the Ka of NH4+ )

The Kb for NO2- tells how much OH- it produces ( The Keq for equation 2 is the Kb of NO2

- ) The Ka for NH4

+ is 5.6 x 10-10 (look up NH4+ on the left side of the table and it’s Ka is on the right)

The Kb for NO2

- must be calculated: Kb (NO2) = Kw = 1.0 x 10-14 = 2.2 x 10-11 Ka (HNO2) 4.6 x 10-4 Since the Ka of NH4

+ > Kb of NO2- We can say that this salt is ACIDIC

So, in summary: Hydrolysis of Amphiprotic Anions Amphiprotic Anions Start with “H” and have a “-“ charge. Eg. HSO4

- , HSO3- , H2PO4

- HPO42- HS- etc.

Amphiprotic Anions hydrolyze as acids to produce H3O+ but they also hydrolyze as bases to produce OH- So, how can we tell whether they are acidic or basic or neutral? We need to determine the predominant hydrolysis. See the next page…

If Then the salt is: Ka (cation) > Kb (anion) Acidic Kb (anion) > Ka (cation) Basic Ka (cation) = Kb (anion) Neutral

On left of acid table: Weak acid

On right side of acid table 15th from top: Weak base

5.6 x 10-10 2.2 x 10-11



Find the Ka of the ion. (Look for ion on the LEFT SIDE of the acid table, read Ka on the right.) Find the Kb of the ion. (Look for the ion on the RIGHT SIDE of the table and use: Kb = Kw/ Ka(conj. acid)

Eg. Find the predominant hydrolysis of the hydrogen carbonate ion (HCO3

-) and write the net-ionic equation for it. To find the Ka of HCO3

-, look it up on the left side of table (6th from the bottom) . It’s Ka = 5.6x10-11 To find the Kb of HCO3

-, look it up of the right side of table. (15th from the bottom) ( It’s conjugate acid is H2CO3 and the Ka of H2CO3 = 4.3 x 10-7 )

So we calculate the Kb of HCO3- using : Kb(HCO3

-) = Kw = 1.0 x 10-14 = 2.3 x 10-8

Ka(H2CO3) 4.3 x 10-7

So, since Kb (HCO3-) > Ka (HCO3

-) , the ion HCO3- predominantly undergoes BASE

HYDROLYSIS. ( 2.3 x 10-8 ) (5.6 x 10-11 )

And the net-ionic equation for the predominant hydrolysis is: HCO3

-(aq) + H2O(l) H2CO3 (aq) + OH- (aq)

Read p. 144 – 147 in SW & Do Ex. 69 (a-f) and Ex. 70 (a – j), 71, 72 & 73 on p. 148 SW. Do Worksheet 4-5 (Hydrolysis) & Do Experiment 20-D (Hydrolysis)

Putting it all Together—Finding the pH in a Salt Solution Eg. Calculate the pH of 0.30 M Na2CO3 Step 1: Dissociate and Eliminate any spectators. Identify any ions left as weak acids or weak bases. Na2CO3 2Na+ + CO3

2-

If Then the predominant hydrolysis is:

And, in aqueous solution, the ion:

Ka (the ion) > Kb (the ion) ACID HYDROLYSIS Acts as an Acid Kb (the ion) > Ka (the ion) BASE HYDROLYSIS Acts as a Base

Spectator Found on RIGHT side of acid table 6th from the bottom. Undergoes BASE HYDROLYSIS



Step 2: Write HYDROLYSIS EQUATION ( Don’t forget that CO3

2- undergoes BASE hydrolysis!) And an ICE table underneath it: CO3

2-(aq) + H2O(l) HCO3

-(aq) + OH-

(aq) [I] 0.30 0 0 [C] - x + x + x [E] 0.30 - x x x

Step 3: Since CO3

2- is a WEAK BASE, we need to calculate the value of Kb for CO32-:

Kb (CO3

2-) = Kw = 1.0 x 10-14 = 1.786 x 10-4 (use unrounded value in the next calculation) Ka (HCO3

-) 5.6 x 10-11 Step 4: Write the Kb expression for the hydrolysis of CO3

2-: Kb = [HCO3

-] [OH-] [CO3

2-] Step 5: Insert equilibrium concentration [E] values from the ICE table into the Kb expression. State any valid assumptions: Kb = x2 (0.30 - x) Step 6: Calculate the value of x. Remember in the ICE table, that x = [OH-] Kb ≅ x2 0.30 1.786 x 10-4 = x2 0.30 x2 = 0.30 (1.786 x 10-4 ) [OH-] = x = √ 0.30 (1.786 x 10-4) = 7.319 x 10-3 M Step 7: Calculate pOH (pOH = - log [OH-]) pOH = -log (7.319 x 10-3) = 2.1355 Step 8: Convert to pH ( pH = 14.00 – pOH). Express in the correct # of SD’s as justified by data: pH = 14.00 – 2.1355 = 11.86 Step 9: Make sure your answer makes sense. The salt was a WEAK BASE, so a pH of 11.86 is reasonable!

Assume 0.30-x ≅ 0.30

Both the 0.30 M and the Ka used were 2 SD’s, so our answer cannot have more than 2 SD’s. Remember: In a pH SD’s are AFTER the decimal!



Metal, Non-metal and Metalloid Oxides (also called Anhydrides)

Metal oxides act as bases in aqueous solution.

Non-Metal oxides act as acids in aqueous solution. Explanation: Group 1 and Group 2 Oxides are ionic. They dissociate to form the oxide ion (O2-) Eg. Na2O (s) 2 Na+

(aq) + O2-(aq)

The oxide ( O2- ) ion then undergoes 100% hydrolysis (because it’s a strong base) O2-

(aq) + H2O (l) 2 OH- (aq) Another example: BaO (s) Ba2+

(aq) + O2-(aq)

The oxide ( O2- ) ion then undergoes 100% hydrolysis (because it’s a strong base) O2-

(aq) + H2O (l) 2 OH- (aq) We can also summarize the reactions of group 1 and group 2 metals with water in the form of formula equations: Na2O + H2O 2 NaOH BaO + H2O Ba(OH)2 Non-Metal Oxides act as ACIDS in aqueous solution: Some common examples of non-metal oxides: NO2 , N2O5 , SO2 , SO3 , CO2 , Cl2O These compounds react with water to form ACIDS. The formula equations for some of these are: SO2(g) + H2O(l) H2SO3 (aq) (sulphurous acid) 2NO2(g) + H2O(l) HNO3 (aq) + HNO2 (aq) (nitric and nitrous acids) Once these acids are formed, they can ionize (strong ones 100%, weak ones < 100%) to form H3O+ ions. Eg. H2SO3 (aq) + H2O(l) H3O+

(aq) + HSO3- (aq) ( < 100% ionization since H2SO3 is a weak acid)

Spectator STRONG BASE (2nd from the bottom on the right side of Acid table.)

Spectator STRONG BASE (2nd from the bottom on the right side of Acid table.)

Don’t get these confused with the IONS: NO2

- (nitrite) and SO32-(sulphite)! They

are covalent compounds, not ions!



Eg. HNO3 (aq) + H2O(l) H3O+(aq) + NO3

-(aq) (100% ionization since HNO3 is a strong acid)

Metalloid Oxides (by staircase) Eg. Al2O3 , Ga2O3 , GeO2 These compounds usually have LOW solubility so not many ions are freed to undergo hydrolysis. So very little hydrolysis occurs so they do not act AS acids or bases. These compounds can react WITH acids or bases. Compounds that can do this are called amphoteric. Anhydrides Oxide compounds that react as acids or bases in aqueous solution are also called Anhydrides. (an-hydride translates to “without water”) These are compounds that react WITH water to form acidic or basic solutions. Acidic Anhydride—An oxide (“O” containing) compound which reacts with water to form an ACIDIC SOLUTION. Acidic anhydrides are oxides of elements on the RIGHT side of the periodic table. Some examples of acidic anhydrides are: SO2 , SO3 , Cl2O etc. And some of their reactions with water are: SO3(g) + H2O(l) H2SO4 (aq) (sulphuric acid—a strong acid) 2 NO2(g) + H2O(l) HNO2 (aq) + HNO3 (aq) (nitrous and nitric acids) Cl2O (aq) + H2O(l) 2HClO (hypochlorous acid) (NOTE: You should KNOW these equations!) Basic Anhydride—An oxide (“O” containing) compound which reacts with water to form a BASIC SOLUTION. NOTE: Basic Anhydrides are METAL (LEFT side of Periodic Table) oxides. Some examples are: Na2O , CaO , MgO , CaO ….etc. Formula equations for some Basic Anhydrides reacting with water: Na2O + H2O 2 NaOH (the base is called sodium hydroxide) CaO + H2O Ca(OH)2 (the base is called calcium hydroxide – sometimes called “hydrated lime”)



Oxides of nitrogen are collectively called: NOx

Acid Rain Since our atmosphere naturally contains CO2 (an acidic anhydride), some of this reacts with water (rain) to make the rain slightly acidic: CO2(g) + 2H2O(l) H3O+

(aq) + HCO3-(aq)

So natural rainwater (unaffected by human activities) can have a pH as low as 5.6 (due to the CO2 in air) If rain has a pH < 5.6 it is called ACID RAIN. Acid Rain is caused by Acidic Anhydrides (not counting CO2) in the air. The main human sources of acid rain are: 1. Burning fuels containing sulphur. 2. Car exhaust 1. When burning coal or other fuels containing sulphur, the sulphur burns too forming sulphur dioxide: S(s) + O2(g) SO2(g) (an acidic anhydride) In the atmosphere further oxidation can occur producing sulphur trioxide: 2 SO2(g) + O2(g) 2 SO3(g) (an acidic anhydride) In rainwater or cloud droplets two reactions can then occur: SO2(g) + H2O(l) H2SO3 (aq) (sulphurous acid – a weak acid) SO3(g) + H2O(l) H2SO4 (aq) (sulphuric acid – a strong acid) 2. In the hot cylinders in an internal combustion engine, N2 from the air and O2 from the air react to form nitrogen oxides: N2(g) + O2(g) 2NO (g) (nitrogen monoxide – an acidic anhydride) N2(g) + 2O2(g) 2NO2 (g) (nitrogen dioxide – an acidic anhydride) Nitrogen monoxide can further oxidize in the air to produce nitrogen dioxide: 2NO(g) + O2(g) 2 NO2(g) Nitrogen dioxide reacts with rain water or cloud droplets to produce both nitrous and nitric acids: 2 NO2(g) + H2O(l) HNO2 (aq) + HNO3 (aq) Some natural sources of Acid Rain Volcanoes can produce SO2 into the atmosphere (which produces both H2SO3 and H2SO4 ) Lightning can provide enough energy to cause nitrogen and oxygen in the air to react and form NO2 (which produces HNO2 and HNO3 ).

Nitrous acid (a weak acid)

Nitric acid (a strong acid)



Natural Protection Some areas are not as sensitive to acid rain as others—even when there are major sources of acid rain present! This is because these areas have rocks and soils that are high in “carbonates” or compounds containing carbonate (CO3

2-). The CO32- acts as a weak base and neutralizes the acid rain to a certain extent:

Eg.) H2SO4 (aq) + CaCO3(s) CaSO4(s) + CO2(g) + H2O(l) (This neutralizes the H2SO4 caused by acid rain) In some cases, powdered limestone (CaCO3(s)) is dumped onto lakes that are too acidic and this helps neutralize the acid rain. But the process is expensive. Problems Associated with Acid Rain

• Aquatic life is affected (from bottom of food chain up) • Forests weakened and killed (Quebec, Germany, Scandinavia) • Minerals are leached out of the topsoil to lower levels. Al3+ ions are released which are very toxic

to fish (mucous in gills) and plants (prevents uptake of other important minerals)

• Metal and stone buildings and statues (especially marble (CaCO3)) are damaged. • Acid Rain is carried over large distances (due to high smoke stacks) –can cross international

borders

Possible Solutions to the Problem

• International conferences and agreements to limit sulphur in fuels and NOx in car exhaust. • Alternate, less polluting energy sources being used. (geothermal, solar, wind etc.) • Industrial processes are being modernized (eg. new process for pulp mills involving H2O2 instead

of sulphites.) • Devices to remove gases like SO3 from smoke stacks. (scrubbers). SO3 + CaO CaSO4(s).

The CaSO4 is removed by electrostatic precipitation.

• Recycling

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