Half-Cell aka Ion-Electron Method

Different than the oxidation number method.

Say what?

Different than the oxidation number method.

"But, but, but.... how will I know which method to use?" 


The question will ask you to use one method or the other method.
You will be told!



The method:
-if the reactionis acidic or neutral (neutral = no mention of either acidic or basic conditions)
divide the reaction into two half reactions (an oxidation half and a reduction half) - [at this stage it should be quite obvious about the halves]
Working with one half reaction at a time:
  -balance species except for H and O
  -balance O by adding H2O
  -balance H by adding H+
  -determine charge on reactant/product side of half reaction
  -add electrons (e-) to balance charges
-at this point one reaction should have electrons on the reactant side (reduction half cell - gained electrons) and the other half reaction should have electrons on the product side (oxidation half cell - lost electrons)
   -make electrons equal by multiplying entire half reactions by necessary coefficient
   -add up the two halves, simplify (cancel) as needed
Check the finished reaction by ensuring that the charge is equal on both sides and that the atoms are balanced.

If the reaction is basic, add OH- to both sides equal to the number of H+ in the reaction
Simplify the water: OH- + H+ = H2O (cancel/reduce water that lies on both sides of the reaction)
Check the finished reaction

The above outlined procedure follows along with the Nelson text. There are variations of the procedure out there that also work. Worked out examples can also be found in the Nelson text.

Practice quiz/test type questions?
Try the following Waterloo links. There is a lesson presented and some practice questions that follow - with answers. Some of the questions are are at greater level of difficulty than expected for this course. But everyone likes a challenge, right?

Oxidation States: http://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
(Again some variations  of the oxidation rules have been presented. The oxidation answers do not change.)

Half Reactions: http://www.science.uwaterloo.ca/~cchieh/cact/c123/halfreac.html

Balancing Half Reactions: http://www.science.uwaterloo.ca/~cchieh/cact/c123/balance.html

Not that you are expected to be doing a lot of Chem*Is*Trying over the Winter Break, but....

The final answers to all of the reactions (oxidations, oxidation under A/B conditions and half reaction balancing) have all been moodlized.

The the next lesson (Galvanic Cells) and unit review have also been moodlized. Please note, if you decide to work ahead on the Galvanci cell stuff - the Nelson text does something a wee bit different than the representation style than will be followed in the classroom lessons. The text opted for their own version of "Cell notation", it is not Standard Cell Notation as many others follow. The following link provides examples of the classroom lesson format: pages.towson.edu/ladon/electrochem.doc

EChem in the New Year
-Galvanic Cell
-Galavanic Cell lab/Corrosion Lab
-Corrosion and Corrosion Prevention

All of the organic reaction handouts have also been moodlized.

If someone gets really organized, exam review will also get moodlized.

The moodle may get a bit of revamp over the Winter Break. The ask a question forum may shift to the top of the page so that it becomes an ask a question about any unit. The forum for question question asking is checked on a regular basis. Take advantage of the opportunity. During the break, answers may not appear immediately, but they will still appear. [Unless technical gremlins get in the way.]

May each and everyone of us enjoy our respective Winter Break.

Oxidation Numbering Balancing Day 2

By now
-you know the rules for assigning oxidation numbers
   -the always (elemental state =0;alkali = +1; alkaline earth = +2; fluorine=-1)
   -the usual (O=-2, H=+1)
   -the rest
-you know how to balance a redox reaction using oxidation numbers
    -assign oxidaton numbers to all
    -decide what changed
    -temporary balance of what changed
    -determine electron movement & balance redox portion
    -balance the rest

This lesson will follow all of the above rules but with a few added steps.
Theses redox equations will be under acidic or basic conditions. Whether the equation is under acidic or basic conditions must be stated along with the unbalanced equation.

Acidic Conditions:
-follow all of the steps up to the redox portion being balanced, then
    -determine charge on each side
    -balance charges by adding H+ to one side
    -balance H atoms using H2O
 Basic Conditions:
-follow all of the steps up to the redox portion being balanced, then
     -determine charge on each side
     -balance charges by adding OH- to one side
     -balance H by using H2O

Examples can be found on pages 666-667

The follow-up to this lesson are found on the balancing equations handout

Next up: Half Cell Method (aka ion-electron method)
-balancing equations using this 3rd method are different than using oxidation numbers to balance
       -on a test/quiz the type of balancing to be used will be stated
       -it is expected that equation will be balanced using the prescribed method
       -if the prescribed method is not followed, a mark penalty will result

Is a tree petrified because it is scared? Do librarians use bookworms for bait when they go fishing? When cows play cards, do they pay for big stakes? Is a plumber's favorite vegetable a leek? Is the prettiest vegetable a cute-cumber? When vegetables get married do they wear onion rings?

Balancing Redox Reactions & 3rd Time's The Charm

Follow the rules!
Follow the rules!
Follow the rules!

Rules to Balancing Equations Using Oxidation Numbers:
(1) Assign oxidation numbers to all species
(2) Identify those two reactant species that change
(3) Temporarily balance the species that change [Miss this step and the balancing could be off!]
(4) Determine electron movement. (Oxdn numbers up - lost electrons; oxdn numbers down, gained electrons)
(4) Balance electrons. [make total gained = total lost]
(5) Balance the remainder.

Example in the text - page 664-665

Followup from this lesson:
handout: Balancing Equations Using Oxidation Numbers

Next lesson: Balancing Redox Reactions in Acidic and/or Basic Conditions Using Oxidation Numbers

If you were determining the molar mass of a solid acid.   Basic Instructions have been moodlized.
What concentration of NaOH might you use? ~01.M-0.2M Solid acids can have larger molar masses. 50g/mol to 250g/mol
What volume of that 0.1~0.2M NaOH solution would you prepare....maybe 500mL (or more....)

If you opt to carry out the 3rd Time's The Charm Titration - have all of the calculations completely mapped out. In the lab the only thing that you would then need to do is (1) the titration and (2) finishing the calculations usign the titration values.
Wed Dec 16th = make and standardize base (~30 minutes)
Thurs Dec 17th= determine molar mass of acid day (~30 minutes)

This is not a 'big to do'. There are lots of procedures out there on the www that give instructions for determining the molar mass of a solid acid sample.

A few other tidbits.
(1) You may make and standardize the base with a partner BUT you will each be given your own separate acid for molar mass determination. Partners require more base than individuals.
(2) The complete calculation is submitted before leaving on Thursday. See moodle for expectations of submission.
(3) Not everyone is carrying out this lab, it is not necessary to use your lab book for this 'extra'. You can if you want to, but it is optional. [Adherence to safety rules is not optional :<]

Oxidation Numbers

LEO says GER
Loss of Electons Oxidation Gain of Electrons Reduction

Assigning Oxidation Numbers
Always Oxidation Numbers
-elementary state = 0  [H2(g) H=0)
-alkali metal ion = +1 [Na of NaCl; Na = +1]
-alkaliine earth metal ion = +2 [Ca of CaCl2; Ca = +2]
-flourine ion = -1 [F of NaF; F=-1]

Usually Oxidation Numbers
-hydrogen = +1 [except in hydrides, NaH, where H is -1]
-oxygen = -2 [except in peroxides, H202 where O is -1]

Workout everything else
-the sum of all oxidation numbers must equal the total charrge
(if no charge is indicated then the total charge is zero)

example: Na3PO4   Na = +1 (alkali metal)  O = -2 (oxygen is usually -2)
3(Na) + P + 4(O) = 0
3(+1)  +P + 4(-2) = 0
P= +5

example NO3 1- (nitrate ion)
N + 3(O) = -1
N + 3(-2) = -1  [oxygen usually -2]
N = +5

OIL RIG

Oxidation Is Loss Reduction Is Gain

From the Oxidation Number lesson:
p659 (12-16)
p662 (19)

In honour of LEO (GER!!!!!!!!!), the following:
Why do Lions eat raw meat?  [Because they don't know how to cook.]  Why did the lion loose at the card game?  [Because he was playing with a cheetah.]
A lion spots a monkey walking through the jungle. He grabs him by the neck and roars "Who's the king of the jungle?” The frightened monkey says, "You are, of course, your majesty." The lion does this to several other animals, with the same results. Then the lion goes up to an elephant, grabs him by the trunk and roars, "Who's the king of the jungle?" The elephant picks the lion up with his trunk, bounces him a few times on the ground, grabs his tail, twirls the lion around over his head, and then lets him go flying into a mud puddle. The lion looks up at the elephant and says, "Well, if you don't know the answer, just say so!”

A couple of zoologists decided to give a lion a cell phone in order to keep track of his whereabouts. Unfortunately, whenever they tried to call, the lion was busy!
A hungry lion was roaming through the jungle looking for something to eat. He came across two men. One was sitting under a tree and reading a book the other was typing away on his typewriter. The lion quickly pounced on the man reading the book and devoured him. Even the king of the jungle knows that readers digest and writers cramp.

[Lion jokes from : http://mojolion.com/Lion_Jokes.html]

Electrochemistry

What is it all about?

First up - assigning oxidation numbers & balancing electrochemical equations.

Have you, in the last few weeks encountered a titration lab that left you more mixed up that a well dissolved salt & water solution?
Would you like a chance to redeem yourself?
Block off time after school on Wed Dec 16th and Thurs Dec 17th for an opportunity to standardize a base (Day 1 - Wed Dec 16th) and then determine the molar mass of a weak monoprotic solid acid (Day 2 - Thurs Dec 17th).
Do you require further details?
Additional infomation has been moodlized.
See Acid Base Titrations - 3rd Times the Charm? {or you local Chem*Is*Try teacher}

Test Rvw

Good day Chem*Is*Try-ers

The Acid/Base Equilibrium Questions (The review) have been moodlized. [The solutions were scanned, resized and then loaded up to moodle] There are full solutions to the first 8 questions. The 0.150M questions have numeric answers. The acetic acid/NaOH abcd question have partial solutions.

A reminder:
There are basically 3 question types
(1) those involving pH and strong/weak acid/bases & salts
     -weaks are on the p803 chart
     -%ionization fits in this group as well - go back and look at earlier examples
(2) Buffers - weak and its conjugate ion salt -llbm reactions
    -versions of buffer questions - yet again - go back and look at earlier examples
    -buffer upset with strong acid or base
(3) Neutralization questions - those that combine acid and base  - forward reactions
    -pH at equivalence point (all A and B comsumed, becomes pH of salt)
    -one reactant leftover
            -weak leftover, since conjugate ion also produced, becomes a buffer
            -strong leftover, calc pH of leftover strong

Don't forget about the Buffer/Neutralization/Titration questions - all solutions moodlized. Some images may be PDF - requires Adobe Reader to view
The question forum is available as well.
Mind you, some people get up at 5am, so late at night is earlier for early risers.

Finally, there is always the early morning option for those last minute questions. The early riser arrives just shortly after 7am.

Post 7Up & Test Review

Well that lab write-up is done.

(Well it is complete for most.)

Next up - unit test.
If you cannot wait until next day's class to get cracking on the review, the review sheet has already been moodlized.

Don't forget that the Buffer/Neutralization/Titration questions are all fair game for the test. If you experience technical difficulties opening some of the files, you may need to download Adobe reader. [Adobe Reader is commonly used program to read PDF files.]

 What do you get when you cross a snowman with a vampire?  [Frostbite.]  What do you get from a pampered cow? [Spoiled milk.]  What's brown and sticky?  [A stick.]

7 Up lab

Did anyone google titrating 7Up?

[ Just asking.]

Extra time on your hands? The unit review sheet is on moodle.

Also, the solutions to many of the previously assigned questions have been moodlized.

More A/B stuff

The unit at a glance:

Find the pH of
 - strong acid and/or base
 -weak acid and/or base
 -acidic/basic salt

Find the pH of
 - buffer (no ICE)
 -buffer after being upset with a small amount of strong acid/strong base (ICE needed)

Collection of neutralization type reactions : acid + base --> salt + water
-begin as a neutralization reaction, may evolve into a buffer or pH of a salt type question
for the statements below:
strong = strong acid or strong base
weak = weak acid or weak base

strong + strong --> neutral salt + water
 if only products, pH = 7 (neutral salt and water)
 if excess= strong, determine moles leftover, total voluem, calc conc then pH

strong + weak --> salt and water
-if all reactant consumed (equiv point) only product salt leftover, determine moles of salt, total volume, get conc, then question becomes llbm type for pH of salt (ICE chart required)
-if reactant weak leftover, then weak + salt = buffer question, set llbm for weak A/B buffer, need total volume and resulting con of weak and salt
-if reactant strong leftover, determine moles of strong leftover, total volume, then pH (can ignore weak A/B salt, insignificant compared to strong)

From today - p614 - 6 & 9
Don't forget % ionization and the remaining questions on the buffer/titration/neutalization handout (complete solutions on moodle)

Acid Base Reactions that lead to...

We know that: acid + base --> salt + water
but what if
- the acid and base are completely neutralized, but the salt is not neutral (see lessons on pH of an acidic/basic salt)
- there is leftover strong acid/base reactant?
- there leftover weak acid/base reactant?

Good questions.

Generally the reactions that are under consideration involve 1:1 mole ratios [insert a nice happy face here]

If the leftover reactant is strong, then
-determine the moles of leftover strong
-calc conc using the total volume
-determine pH

If the leftover is weak, then....it becomes a buffer question because there will be both weak acid/base and salt of the weak present. [A solution of weak acid/base and its conjugate salt = BUFFER!]
-determine moles of leftover weak
-determine moles of salt produced (=moles of reactant all used up - that nice 1:1 mole ratio]
-write llbm reaction for acidic/basic buffer
-determine conc of weak and salt using total volume
-use Ka and/or Kb all the way to pH

If there is no leftover reactant, only product salt - and the salt is acidic and/or basic
-determine moles of salt (nice 1:1 mole ratio)
-determine total volume
-write out llbm equation
-set up ICE chart to track changes
-use Ka and/or Kb all the way to pH
***remember that it is sometimes necessary to use the Ka of a weak acid to determine the Kb of its conjugate basic salt KaKb=Kw

From today complete:
p607(4) and p608(6)
A few points about 4 and 6:
4(a) before the titration begins - pH of a weak acid (check your notes)
4(b) the 10.00mL of NaOH is suppose to represent a stage of the titration in progress
4(c) the titration is now over, more than 10.00mLof NaOH was used, but how much? A salt was also produced, a salt with the conjugate ion of the weak acid - a basic salt - need to determine Kb
6(a) before the titration begins - pH of a weak base (check your notes)
6(b) the titration is over, how much acid was required to completely neutralize the weak base? what is the concentration of the salt of the weak base that was formed? {consider the total volume from the weak base and the acid)

Also, all of the remaining questions on the Buffer/Neutralization/Titration handout are now 'fair game'. The remainder of the handout questions are not necessarily like the questions on page 607 & 608. It is not intended that the remaining questions be completed for Monday. There will be more time next week, but not that much!
As I type this post, the game plan is that I will go home and moodlize some of the solutions. If I load them up using the same technology as some of the most recent moodlized solution postings - you may require a current version of Adobe reader 9. It is now several hours later. things have been moodlized -including the aromatic organic outline. If time permits additional organic stuff and buffer/tritration/neutralization solutions will also be moodlized. BufferTitrationNeutralization solutions up to question number 10ab have been moodlized. I am finished moodlizing for today. The solutions are part way down the A/B llbm, just after the BufferTitrationNeutralization Question Sheet. Each of the PDF file names identifies the questions on that particular image.

Next week:
Monday = more calculator work
Tuesday = 7-Up lab
Wednesday = 7-Up follow-up
Thursday = Rvw
Friday = test
Finishing joke: What colour is a burp? [Burple]

AB - another day and another question

Today was a day to re-visit some previously learned concepts.
 In what way did you re-visit earlier concepts?
The following handouts were completed:

(1) Acid-Base Equilibrium Overview (handout on moodle)
-the instructions are on the handout
-you may need to consult your notes
-consider the type of each substance (weak acid, weak base, acidic salt, etc…) before writing the equation and need for an ICE chart
-actually go full out and carry out the calculation to determine pH for the last three on the page
(2) Strong Acid-Strong Base Titration (handout on moodle)
-well laid out example in your text (page 598)
(3) Aromatic Reactions (Organic)

What about a few questions to get you thinking? In each case there will be some acid or base leftover. Notice the type of substances [strong A/B? weak A/B salt?] leftover after the reaction....take into account the total volume as you move beyond the neutralization reaction.
(1) Determine the pH of  the solution that is formed when 15.00mL of 0.100M HCl is combined with 25.00mL of 0.0800M NaOH.
(2) Determine the pH of the solution that is formed when 20.0mL of 0.100M acetic acid is combined with 10.0mL of 0.150M NaOH.
(Will the worked out solutions to these posed questions be moodlized later today? Hmmmm... Further hmmm...why is one question outlined in blue, while the other is outlined in red?)  It is a few hours later as I type these red sentences. The solutions to the above two questions (and a few more things) have been moodlized. Have a question while you moodling? Try the post a question forum.

A few jokes to close out this entry. Why did the child study in the airplane ? [The child wanted a higher education.]  Why was the broom late? [It over swept.] There is room for one more. Whats red and flies and wobbles at the same time? [A jelly copter.]

Acid Base Titrations that lead to another question...

The lesson was about neutralization via an acid/base titration.

In a titration, all of the acid and base are consumed (at the equivalence point) yielding a salt and water. [Thus the reason for writing A/B equations as acid + base --> salt + water]

An interesting aspect fromt these questions is what kind of salt is produced. Is it neutral, acidic or basic?

If the salt is acidic or basic, then what is the pH of the resulting solution.

If the pH of the resulting solution is known, then what is a suitable indicator to use to find the equivalence point of the neutralization (via titration)?

A few questions were left to complete.
For each of the following, determine
(i) the conc of the acid
(ii) the pH at the equivalence point
(iii) a suitable indicator (name, explain why it was selected, expected colour?)
(a) 13.72mL of acetic acid is completely neutralized by 25.00mL of 0.100M NaOH.
(b) 25.00mL of 0.150M NaOH is titrated with 21.44mL of HCl
(c) 8.68mL of HCl completely neutralizes 10.00mL of 0.0750M NH3

The solutions to these questions will posted on moodle within the next 2 hours.
[provided technical issues do not get in the way.]

The solutions to the pH of titration questions (outlined above and in class) are posted on moodle.
{I still need to work on 'sizing' the document...something for another day I suppose.}

I am testing out a new to me moodle feature - post a question forum...currently it is the last item in A/B llbm unit list

A few 'jokes'(?). Ready or not.... Why are gold fish orange? [The water makes them rusty.]  Who held the baby octopus to ransome? [Squidnappers.]  What part of a fish weighs the most?  [It's scales.]

Acid Base Neutralization

Neutralization of acid + base proceeds to the product side, no llbm

Generally:  acid + base --> salt + water

A few examples:
Strong acid/Strong base
   HNO3(aq)  +  NaOH(aq)  -->  NaNO3(aq)  + H2O

Weak acid/Strong base:
   HC2H3O2(aq)  +  NaOH(aq)  -->  NaC2H3O2(aq)  + H2O

Strong acid/Weak base:
-difficult to see "salt + water" products; write strong acid as H3O+
HNO3 + NH3 becomes H3O+  +  NH3  -->  NH4+  +  H2O

Follow-Up
Buffer/Titration/Neutralization handout (#11)
Stuck on the chemical equations? - check moodle.

And now for something a little amusing. On the first day of school, the Kindergarten teacher said, "If anyone has to go to the bathroom, hold up two fingers." A little voice from the back of the room asked, "How will that help

Upsetting Buffers

Dealing with acidic and/or basic buffers.
The buffer is upset by the addition of a small amount of either strong acid or strong base.

Acidic Buffer:
HA  +  H2O  <=>   H3O+   +  A-
  add strong acid (H3O+), llbm shifts to left (reactant) side
    thus [HA] increases, [A-] decreases
  add strong base (OH-), llbm shifts to the right (product) side
     {Why? OH- joins with H3O+ to create H2O}
    thus decreasing [HA], increasing [A-]

Basic Buffer:
Base + H2O  <=>  BaseH+   +   OH-
    add strong acid (H3O+), llbm shifts to right (product) side
           {Why? OH- joins with H3O+ to create H2O}
       thus [Base] decreases,  [BaseH+] increases
    add strong base (OH-), llbm shifts to the left (reactant) side
    thus increasing [Base], decreasing [BaseH+]

Typical Question:
500.0mL of a 0.10M NH3/0.080M NH4Cl solution has 3.00mL of 3.00M NaOH added. Determine the pH of the resulting solution.

-write the llbm reaction for the weak (in this case weak base NH3)
-determine the concentration change of the strong (base in this example) will undergo as it is added to the buffer solution (c1V1=c2V2)**
-organize and ICE chart to track changes to the two components of the buffer (the weak and its salt)
-calculate [OH-] at llbm, (since it is a basic buffer in this example)
-determine pOH
-finally determine pH

**since a small amount of strong base is being added to a much larger volume, the strong base concentration will be greatly impacted.
However the addition of a few mLs to the much larger buffer volume will have negligible impact on the initial buffer concentrations
-buffer concentrations are impacted by the llbm shift that results from the addition of the strong base (or acid)

Homework: From the Buffer/Neutralization/Titration Handout
5,6,7,8
 (do you appreciate?)

Why is it easy to swindle a sheep?  [Because it is so easy to pull the wool over its eyes.] Did you hear about the bungee jumper who shot up and down for 3 hours before they could bring him under control? [He had a yo-yo in his pocket!] Why did the French farmer only keep the one chicken? [Because in France one egg is un oeuf!]

Buffers

Buffers - solution with both a weak (acid and/or base) and its conjugate ion as a salt.

Know the strong acids - if it is not strong it is weak!

Acidic Buffer salt - alkali/alkaline earth metal ion + anion of weak acid
                   (K+ and CH3COO-)
Basic Buffer salt - conjugate molecular base ion + strong acid anion (Cl-, Br-, I- NO3-, ClO4- SO42-)
                   (CH3NH3+ and Br-)

Three types of buffer questions:
(1) Given molar concentrations of both the weak and the salt
(2) Given moles of both the weak and the salt and total volume
          use C=n/V to determine conc of each component
(3) Given conc and volume of weak and conc and volume of salt
          use c1V1=c2V2 to determine conc of each component

Standard question: Determine the pH of....
In the mathematical solution include:
   -llbm equation (use either the acid llbm version or the base llbm version)
   -K expression (read K value from p803 chart)
   -no ICE chart required - assumed that [initial] = [llbm] for buffer questions because of small K value
   -sub in known value, for acids-determine [H3O+], then pH or for bases-determine [OH-], pOH, pH

Homework:
Buffer/Titration/Neutralization handout (available on moodle)
Qs: 1,2,3,4,5a,6a (answers on the last, single sided page)

Next up: Upsetting the buffer - these are definitely cool questions (so cool that ICE charts will reappear)

Anice joke to finish this post: Q: What sits on the bottom of the cold Arctic Ocean and shakes? A: A nervous wreck.
[Question for you, did you read Anice joke as "a nice joke", as "an ice joke" or some other way?]

Salts

What you know:
-the strong acids
-the weak acids
-the weak (molecular) bases - like NH3
KaKb = Kw
cations = +ve ions
anions = -ve ions

Consider the acid HNO3 (aq), the nitrate ion (NO3-) is the anion from a strong acid
Consider the acid HClO(aq), the hypochlorite ion (ClO-) is the anion from a weak acid
Consider the weak molecular base NH3, NH4+ is the conjugate acid of the weak molecular base
the charts on page 803 can help the identification process

General Salt Rules
anions
-anions of strong acids are neutral (if you know the strong acids then you know the six neutral anions)
-anions of weak acids are basic (If it didn't come from a strong acid, then it came from a weak acid....)
cations
-conjugate acids of weak molecular bases are acidic
-alkali/alkaline earth ions are neutral
except Be2+ = acidic and high charge density ions like Al3+; also acidic

Using the above rules, salts can be identified as acidic basic or neutral.
Consider KCl (K+ & Cl-)
    K+ is neutral (alkali metal); Cl- = neutral (anion of strong acid)
    -salt is neutral
Consider NH4NO3
    NH4+ = acidic (conjugate of weak molec base); NO3- = neutral (anion of strong acid)
    -salt is acidic

NaClO = a basic salt (due to the ClO- anion; from a weak acid)

Question: Determine the pH of a 0.100M NaClO solution:
llbm equn: ClO-(aq)  +  H2O  <=>  HClO(aq)  +  OH-(aq)
use ICE chart to organize info
Kb expression** {it is a base afterall....:<}
-solve for x where x = [OH-]
-solve for pOH, then pH
-simple, right.....?
**since the question requires Kb - it must be available somewhere....
back to the top - for a weak and its conjugate, their respective Ka & Kb values are connected: KaKb=Kw
-Kw=1.0x10^-14
-Ka for HClO is on the p803 chart
-via KaKb=Kw, the value for Kb can be determined

Follow-up Questions  P 588(1bc,2,3,4)
-now about the acidity/basicity of a metal oxide...., what about a nonmetal oxide...hmmm - Google anyone?

With all of this examination of K, a knock-knock joke:  Knock, knock     Who's There?     Cow-go    
Cow-go Who?    No, Cow go MOO!!!

Weak, weak day 2

Another day, more weak acid and weak base

The following information is intended to assist you with one
(or two) of the questions assigned from day 2 of A/B llbm

The K value gives an indication about the amount of product.
The larger the K value, the great the amount of product.
In the world of acids and bases, the larger the K,
the greater [H3O+] and/or [OH-]
The greater the [H3O+] the lower the pH.
The greater the [OH-], the lower the pOH, and higher pH.

Ready for more acid base llbm questions?

Assigned Questions:
A few textbook questions to answer:

Page 579:
Weak acids: 4,5,6,7ab(for part 7b – calculate using only HNO3 and HF)
Weak bases: 9ac, 10ab(for part 10b-calculate using only atropine), 16

Relationship between Ka and Kb ? – read pages 559 to 563, then answer:
Page 563 (6)   Page 579 (14)

Finished all of the above? Begin the next organic handout – reactions of alkanes/enes/ynes. Don’t be afraid to look in the textbook.
A few weak jokes. What is the best way to carve wood? [Ans: Whittle by whittle] Don't be afraid of the next couple. If fruit comes from a fruit tree, what kind of tree does a chicken come from? [Ans: A poul-tree] Why did Beethoven get rid of his chickens? [Ans: Because the kept saying "Bach, Bach, Bach."]

Weak acid, weak base

Acids = proton donors.
Weak acids - those that are not strong!
Weak acids do not ionize 100%.
Weak acids set up equilibrium with water to produce the acid's anion and hydronium ion.
If the general representation of a weak acid is HA, then in llbm:
HA(aq) + H2O <=>  H3O+(aq)  +  A-(aq)

Generally two question types are asked
(1) Determine the Ka value for a particular weak acid
   -given llbm information about [H3O+] via pH and/or %ionization information
   -where pH can be used to determine [H3O+] @ llbm
   -and where %ionization = [H3O+]llbm/[HA]  x 100%
(2) Determine pH (or % ionization) given specific acid(useKa chart to determine Ka value)

Complete mathematical solution would include:
-llbm equation
-Ka expression
-ICE chart
-solve for whatever it is that is being asked
     -looking for pH or %ionizn?
     -solve for [H3O+]llbm, then calculate to get requested info
     -looking for Ka, use pH or %ionzn to determine [H3O+]

Weak bases - similar set up as with weak acids
Difference, weak bases involve [OH-], which leads to pOH

If the geneal formula for a weak base is 'Base' and bases are proton acceptors
Generally: Base(aq)  +  H2O  <=>  BaseH+(aq)   +   OH-(aq)
Again given two basic question types
(1) Determine pH of a stated basic solution.
(2) Determine Kb of a stated base.

Complete mathematical solution would include:
    -llbm equation
    -Kb expression
    -ICE chart
    -solve for whatever it is that is being asked
         -looking for pH?
             -solve for [OH-]llbm , then determine pOH, then pH
              @25C pH  + pOH = 14.00
         -looking for Kb use given pH (or pOH) to get to [OH-]llbm

Homework:
p568(8) p570(10)  p574(12,13)  p579(1,3,13)
Next up?
Using the Ka/Kb chart to calculate K values of conjugate ions
KaKb = Kw

Cute joke coming, ready? Why did the chicken cross the playground? [Ans: To get to the other slide.]

Acid- Base: pH/pOH Calculations

Strong Acids
If it is not a stong acid, it is a weak acid.
Know the strong acids (monoprotic and diprotice)
Strong acids ionize virtually 100%

HCl(aq)  + H2O => H3O+(aq)  +  Cl-(aq)

pH = -log[H3O+]; 10^-pH = [H3O+];


Question: Determine the pH of a 0.150M HCl solution
(since strong acid ionizes 100%, [HCl] = [H3O+]=0.150M

pH = -log(0.150M) = 0.824
Notice: 3 sgfigs in concentration 0.150M; 3 decimal places
 in pH value 0.824

Determine the concentration of a monoprotic acid solution if the pH
of the solution is 4.35
10^-4.35 =4.5x10^-5  M
Notice: 2 decimal places n pH value, 2 sigfigs in concentration answer

Only the decimal places of pH values are significant.


In a similar way that pH/[H3O+] connect, so can
pOH and [OH-]

pOH = -log[OH-] and 10^-pOH = [OH-]

Connecting pH and pOH?

@25C:   pH + pOH = 14.00

Question: Determine the pH of a 0.250M NaOH solution
Strong bases dissociate virtually 100%
NaOH is a strong base, dissociates 100%
NaOH(aq) => Na+(aq)  +   OH-(aq)
So, [NaOH] = [OH-] = 0.250M

When working with a base, one cannot determine pH directly from the given base concentration . Determine pOH, then determine pH

pOH = -log(0.250M) = 0.602
pH = 14.00-0.602 = 13.398

there is another way to carry out this calculation
- saving that for another time

A few basic jokes to end this post: What happened when the lion ate the comedian? Ans: It felt funny. Where do sheep go to get a haricut? Ans: To the Baa Baa shop. They are simple, basic jokes. Get it? This post dealt with bases, so it ended with some basic jokes!

llbm Review

-complete the llbm & solubility review handout
-check moodle for introduction to acid/base equilibrum handouts
-check moodle for upcoming quiz/test dates

Will a precipitate occur?

Will 35.0mL of 0.0100M NaCl form a precipitate when combined
with 65.0 mL of 0.0150M AgNO3?

• write the double displacement to identify the slightly soluble salt (the precipitate)
• silver chloride in the example
• determine the concentration of the ions of the slightly soluble salt using c1V1=c2V2
• Cl and Ag in this example
• for Cl: (0.0350L)(0.0100M)=c2(0.100L*);
• * 35mL + 65mL = 100mL = 0.100L
• Write the llbm equation for the slightly soluble salt then the Ksp expression
• Use the calculated concentrations, sub into Ksp to determine a "trial K"
• compare "trial K" to real K (from Ksp chart)
• if trial K less than real K = no precipitate (not enough of the ions present to cause a ppt to form)
• if trial K greater than real K = ppt (too great of an amount of the ions in the solution - excess ppts out)

Remember if you are not part of the solution, you are part of the precipitate.

Lab Follow-Up

The lab involved neutralizing 25.00mL of an unknown concentration of a calcium hydroxide solution with a determined volume (via the buret) of 0.100M hydrochloric acid.

The solution is already at equilibrium when the titrating begins.
With the titration data it is possible to:

• determine molar solubility of the calcium hydroxide in pure water
• determine Ksp value for calcium hydroxide
• determine molar solubility of calcium hydroxide in a calcium chloride solution

There was excess calcium hydroxide in the solution. the mass of calcium hydroxide is not relevant to the lab follow-up calculation. Furthermore, the calcium chloride was added to upset equilibrium. However, the mass of added calcium chloride is not relevant to the lab follow-up calculations.

The titration data is all that is needed to complete the calculations.


Next up - Acid-Base llbm. Two introductory sheets are on moodle. Notice the moodle calendar for upcoming quiz dates.


Talent alone won't make you a success. Neither will being in the right place at the right time, unless you are ready. The most important question is: 'Are your ready?' Johnny Carson Are you ready for acid-base solution equilibrium?

Common Ion Effect

Homework:
p492 (9)
P492 (10,11,12)

Determine the molar solubility of Fe(OH)3 in a 0.300M Fe2(SO4)3 solution.

      Fe(OH)3(s)   <=>     Fe3+ (aq)      +       3 OH-(aq)

I       n/a                           0.600M*                    0
C     n/a                                 +x                       +3x
E     n/a                            0.600 + x                  +3x
                                  ~0.600 (small K)

initial Fe3+ = 0.600M - from the 0.300M Fe2(SO4)3,
where there are twice as many Fe as Fe2(SO4)3 units

Ksp = [Fe3+][OH-]3
2.6x10-39 = (0.600)(3x)3
2.6x10-39 = 0.600(27x3)
5.4x10-14 = x 
molar solubility of Fe(OH)3 in the 0.300M Fe2(SO4)3 solution
 is 5.4x10-14 mol/L

Molar Solubility to Ksp

Molar solubility is measured at llbm.

question: Determine the  Ksp of salt AB2 if the molar solubility is
 0.00015 mol/L

1.Write the llbm dissociation rxn.
2. Write the Ksp expression.
3. Set up the ICE chart.
4. Use molar solubility to determine [ion]llbm
5.Sub into Ksp expression.

    AB2 (s)     <=>    A+(aq)   +   2B-(aq)                        Ksp=[A+][B-]2
I    n/a                      0              0
C  n/a
E   n/a                0.00015*        2(0.00015)
* for every mole of A+ that is present at llbm, one mole of AB2 dissolved

Ksp = (0.00015)(0.00030)2
Ksp = 1.4x10-11 for AB2

Sometimes the molar solubility info can be 'hidden':
for example, a saturated solution contains 0.24g of salt per 500mL.
[Use molar mass, convert the given info to mol/L]

Homework:
p486(4)
p493(7,8,9,12)
 ...not all are molar solblty to Ksp, some are Ksp to molar solblty
What would you get if image 1 and  image 2 were combined.

[Ans: a molar solution.]

Ksp to Molar Solubility

Slightly soluble salts (see data table page 802)

Molar solubility of the slightly soluble salt identifies the degree
 to which the salt actually dissolves in water.
Expressed in mol/L
The concentration of the ions at llbm is related to the molar solubility of
the salt  by the mole relationship in the balanced llbm equation.

example Determine the molar solubility of iron (II) hydroxide.

1.Write the llbm dissociation reaction.
2. Write the Ksp expression.
3.Complete the ICE chart. (not considering solids - llbm reaction, solid concentration will not change)
4.Determine the value of "x"
5.State the molar solubility value.

            Fe(OH)2(s)        <=>         Fe2+ (aq)        +      2OH-(aq)

I            n/a                                    0                           0
C          n/a                                    +x                         +2x
E          n/a                                     x                           2x

     Ksp = [Fe3+][OH-]2
4.9x10-17 = (x)(2x)2
4.9x10-17 = 4x3
2.3x10-6 = x

The molar solubility of Fe(OH)2 is 2.3x10 -6 mol/L.
[According to the balanced llbm equation, for every mole of Fe2+ that shows up at llbm, one mole of Fe(OH)2 had to dissolve.]

Homework:
p486: 1,2,3
p493: 4,5,6

KSCN lab Follow-Up

The nice orange-red colours are now a thing of the past.

A few helpful hints:
Carefully determine the initial concentrations of SCN- and Fe3+

The initial [Fe3+] for test tubes 2 to 5 requires the use of c1V1=c2V2 twice.

When the determination of Kc is done, the results may vary quite a bit.
Generally they are in the 100 to 300 range.

Complete the follow-up to this lab in a timely fashion. Sooner = better.

If you were absent for the lab - see me for some "results" that could be used to work through the various calculations.

Next up:
Solubility (Have you completed the introduction to solubility?)

Equilibrium: The Movie?

If the teeter-totter represents an llbm reaction, what would be its K value?

Small K approximation

The basis of this lesson is the significance of the size of K as it relates to equilibrium position.

K = [product] / [reactant]

A large K value indicates that there is much more product than reactant.

A small K value indicates that there is much less product than reactant.

Since there is very little product actually produced, the change to the amount of reactant is minimal. So minimal that there is approximately no change to the amount of reactant at llbm compared to the amount of reactant initially. This approximation will simplify the calculation.

There is still product made. The amount of product is still represented by some multiple of 'X'.

Initially 0.200mol of A is placed into a 1.00L container, sealed and allowed to reach llbm.
Determine [B]llbm
Consider the generic equation:
    2A      -->        3B      +        2C      K = 3.50x10-5
I  0.200M             0                   0
C -2X                   +3X             +2X
E  0.200-2X*         3X                2X

but since K is small, 'X' is also very small
there 0.200-2X ~ 0.200M

K   =   [B]3[C]2 / [A]2

3.50x10-5 =  [3X]3[2X]2 / [0.200M]2
1.4x10-6 = 108X5
2.65x10-2 =X
0.079M = [B]llbm

Homework: - a few adjustments to the text info
p476 (7) - H2S <=> H2 + S  [ans=9.17X10-4 M]
p476 (8)  2HCl <=> H2 + Cl2 [H2/Cl2 llbm = 3.6x10-17M]
p481 (6)
p640 (15) [ans: O2 llbm = 1.9x10-3M]

llbm Calculations - Day 2

The day 2 lesson of llbm calculations involved the quadratic equation.

The quadratic equation approach was necessary as
the K expression was no longer a perfect square.

A complete solution for these questions will include:
 - completed ICE chart
 - llbm law (the K expression)
 - substitution into the K expression
 - necessary work to get to the quadratic equation
 - the quadratic equation itself
 - both possible values of "x"
[indicate which value is 'invalid' - sub into the llbm expression on the ICE chart]
 - final answer - where various llbm conc have been determined.

 Three tedious questions to complete for homework.
p480(9) and p481(4,5)

The text also works through some examples. Check out the text too.

Have you complete the clock reaction lab yet?

Soon, very soon the lab will be marked and returned.
You may want to be one of those people who gets the lab returned to them.

Soon to be loaded on moodle. The next lab - Quantitative Equilibrium.
There is some pre-lab calculating to do. If you wish to work ahead. The lab
document has three parts to it. Two of the three parts will be loaded up in a
few minutes. The last part rquires scanning - and that will not happen until later
this evening.

Solutions to old homework can also be found on moodle. At this point,
the recent llbm calculations have not been revealed. Something about
having the kiddies work through it first, before seeing the solutions.

Why are gold fish orange? [Ans: The water makes them rusty.] Who held the baby octopus ransome? {Ans: Squidnappers]. Okay, off to upload a few things to moodle.

llbm Calculations Day 1 - a revisit

The lesson was all about calculating llbm concentrations.
(Finding 'x')

The handouts from the day's lesson are up on moodle (llbm calcn's page 1 & page 2)

If you visit moodle, you may also see the homework solutions to the Le Chatelier stuff [pages 457 - 460]

From the day's lesson - homework:
page 466(3,4)  page 472 (5,6)  page480(10) and page 481(3abc)

Most (if not all) of the questions require ICE charts.

Consider the generic reaction A (g)  +  B (g)  <=> 2C (g)  Kc = 37.9
A typical question may be: Determine the llbm concentration of C if 1.80 mol of A and 1.80 mol of B are placed in a 2.00L container and allowed to reach llbm.
 - complete an Initial Change Equilibrium chart and write the llbm law (K= expression)
 - when completing the chart, express values as concentrations
 - since no amount of product is mentioned, it will = 0
 - once you know one change, use the mole ratios of the balanced chemical equation to determine all other changes.
       A      +       B    <=>   2C
I    0.900M    0.900M        0
C    -X              -X           +2X
E  0.900-X   0.900-X        2X

Kc = [C]2 / [A][B] = 37.9 = (2X)2 /(0.900-X)(0.900-X)  
- notice that both numerator and denominator involve 'perfect squares'
- square root both sides

6.156 = 2X / (0.900-X)
5.540 - 6.156X = 2X
5.540 = 8.156X
0.679 = X

Do not leave the answer at 'X',
the question asked for [C]llbm = 2X = 1.36M

Demos were today- :<)

Have you submitted the clock reaction lab yet?

Finally, something (sort of) in honour of the 40th anniversary of Sesame Street. A monster of a cookie joke. Why did the cookie go to the doctor? [Ans: It was feeling crumby.]

Le Chatelier

Le Chatelier Principle:
Ref: pp450-456
When an llbm system is upset, the system responds in such a way as to oppose the change.

Ways to Upset llbm
-change the concentration (add/remove stuff)
   add - llbm shifts to the opposite side to try to 'use up' what was added
   remove - llbm shifts to same side to try to 'replace' what was removed
-change the temperature
  -it sometimes helpful to include the heat term as part of the equation [not deltaH notation], then think of 'heat' like any other substance
   exothermic:
    reactant -->  product + heat
    increase temp (like adding heat), reaction shifts to reactant side(reverse reaction requires heat)
    decreae temp (Like remoivng heat), reaction shifts to product side (forward reaction produces heat)
  endothermic - the opposite of exothermic
-change the pressure/volume
   increase volume (decr pressure) - llbm shifts to side with more moles of gas
        more moles of gas = more pressure, thus trying to increase pressure to be close to previous pressure
   decrease volume (incr pressure) - llbm shifts to side with fewer moles of gas
        fewer moles of gas = less pressure, thus trying to decrease pressure to be close to previous pressure

Variables that DO NOT affect llbm position
-addition of a catalyst - just reach llbm sooner
-addition of an inert gas (gas that does not react with any aspect of the llbm system) - interferes equally affects the probability of collision between both reactant and product particles

Hmwrk: p457(1,4,5,6)    p458(8)   p459(2*,3,5,6)  p460(10,12)  *just predict the shift, no sketch

Why did the chicken cross the road?  According to Le Chatelier: The chicken crossed the road because there were too many moles of chicken on the reactants side of the road equilibrium.