June 2, 2011

Today we began the day by looking at the previous day's blog. From there, we moved onto pg. 5 where began discussing a half-life and radioactive decay.

A half-life is the amount of time it takes for one-half of a radioactive substance to decay. A half-life can vary in time from a few minutes to millions of years. A majority of radioactive substances decay through radioactive decay which we discussed earlier this weak. Some radioactive decays include: gamma decay, alpha decay, and beta decay. If you wanted to represent this decay in a chemical equation, you do the process that we have covered earlier this weak. For example, in #1, if hydrogen-3 were to decay through beta emission process, it would be represented as:

(3/1)H---->(3/2)He + (0/-1)e

After completing problem 1 and graphing it, we continued on throught the rest of the page. 2A is the same as number one, but 2B is different. Because we know that the half-life is 6, after 6 minutes the percent present is reduced by half, therefore it is 50%. After 12 minutes, the percent present is 25%. This pattern can continue on for ever. Therefore the anser to number 2B is .0976 because after 1 hour of 10, 6 minute half-life's, only .0976% of the substance remains.
For 2C, the equation can be written as N=No*(.5)^n. This can be determined because No represents the mass of the starting substance, while n represents the number of half-life's. By taking .5 to the n'th power, you are accounting for the fact that after each half life you are decreasing the original substance by 50%. Finally, you multiply it by No, in order to find the final mass. Using this equation, we can solve question 3.

3. For the first part, we know that the half-life is 4.5 days long. Since we are given a total of 27 days, you divide 27 by 4.5 to find 6. This is your n value, or the number of half-lifes that took place. The No value is the starting mass which is given to us in the 2nd sentence, 43.8g. From here we can plug in the numbers and solve for N.
-N= 43.8*(.5)^6
-N= .684g
In order to do the second part of question 3, you follow the same steps, except instead o dividing 27/4.5, you do: 33.7/4.5 because you want to find the number of half lifes in 33.7 days not 27 days. The answer for the second part is .243g.

After covering this new material, we proceeded on to a lab, where we were told to find the half-life of a substance by watching it radioactively decay. We performed this by placing cesium-237 into a radioactive detector and recorded its radioactivity every half minute. With this information, we were told to make a scatter plot and determine the substances half-life.

Wednesday, June 1

Today in class we started the topic of Nuclear Bombardment. The main idea that Nuclear transmutations can occur spontaneously or artificially. If it happens artificially there is usually some form of bombardment occurring. There are many examples on page 4. We finished 1-6.

After we finished up with the packet we started on our lab. Our lab was on Radioactivity and Shielding. In it we had see what shields worked best against blocking beta particles and gamma rays. Then once we got that data we had to see which type of radiation did a better job of penetrating through the shields. After we collected all of our data we had an easy reading quiz that consisted of problems one could find on pages 2 and 3 of the packet.

Tuesday May 31, 2011

Today we had a quick reminder that we are in fact still in school and not on vacation yet. We started off class with reading graces blog and going over concepts we learned Friday. Mr. Henderson made some announcements that the webassigns are no longer going to be graded, they are only for learning the material and going over what we learned in class. He also told us we aren't going to be using our lab notebooks anymore. We also got our quizzes back. Then we started on page two and three in our packets, after that a quick lab and the day was over.

On page two we learned a easy new concept, which we used notes on the board to help us with. First of all we learned that when something decays the left side and right side atomic number and mass number have to be equal. So when we do number 7A on page two we use the question to figure out what it decays to. We write out the isotopic symbol for it and obviously the atomic mass and atomic number are not equal. So we determine the difference and start writing a second isotopic symbol. We find out the elemental symbol from the notes on the board. Ex. Uranioum-238 decays to form Thorium 234. We already have uranium there ready for us so we start on Thorium. Thoriums' atomic number is 90 and its atomic mass number is 234. the difference in mass is 4 and the difference for atomic number is 2. Then from using the notes on the board we determine the elemental symbol which is helium and the decay type, which is alpha. we finished the page and moved on to page 3.

For page three Mr. Henderson had us in our lab groups do a certain number then go up to the board and tell the class the answer and how you did it. If you weren't there then sucks for you. But the way you do number one is almost the same way as on page two, except you have the decay type instead of the element it turns into. To figure these our you first write down the elemental symbol for the decay type and set each number (atomic number or mass number) equal and put an X for the missing element we dont know. Then using the PT we figure out the element with the atomic number. For number two its doing everything we just learned but in our heads. All you write down is the isotopic symbol for the element it decays into.

After that we did a quick lab, it was a lab to test if the farther you are away the less radiation you can get. Mr. Henderson had the radiation particle testers out for five minutes before we did our lab to get the background count, for mine it was 123 which we converted to 12 particles every ten seconds or so. We then tested the radation disc from a couple slots under the detector. My data was : 7492, 2385, 1101, 673 and 417. but we then had to subtract the background radiation number which made my data: 7480, 2373, 1089, 661 and 405. We then plotted the data on a little graph and wrote the conclusion.

That was our great day in chemistry. byebye

Friday, May 27

We started off the class with Mr. H telling us a story about how his wife gave him a salad in a bowl giving off radiation. He then proceeded to show us by using this device [I don't know what it's called...] and pointing it at the bowl. Basically, it beeps when it detects it. He pointed it at everyone in the class and the people who were giving off a lot of radiation were Matt, Dmitry, Kevin, and Emma. Mr. H told Emma to make her own Chemistry blog because of the great blog she posted the day before. If you need a good resource, look at Emma's blog.

He then talked about how finals were in 7 days. There is going to be a review day the Tuesday before finals start.

We turned to page 1 to start off the lesson of the day. Mr. H explained to us that radiation is everywhere. Even though when the news talks about radioactive substances, it's not going to be deadly because those amounts wouldn't kill you. He gave us examples like the sun giving off radiation, but the atmosphere absorbs much of them so we aren't harmed by it. He also told us a story about a guy who took a radiation detector on his trip to Hong Kong, Japan, and back to the US and the data showed that the places the radiation detector went nuts on was when he was in the airplane. We can conclude that we are safer on the ground than in the air.

So, page 1, numbers 1, 2 and 3 were all review questions from Unit 2.

Atomic number: number of protons, identifying number

Mass number: number of protons and neutrons

Isotope: form of an atom of an element

Nuclide: nucleus of an atom

2. El is the symbol for the element

A is the mass number

Z is the atomic number or the number of protons.

This was all review so I'm not going to go over it but the answers are as follows for number 3.

Lithium-7: 3, 7, 3, 4

Carbon-14: 6, 14, 6, 8

Lead-210: 82, 210, 82, 138

Uranium-238: 92, 238, 92, 146

#4. a. 1 b. 1.5 c. 83

For number 4, you are supposed to use the chart on the right. The explanation for letter a is that when the proton count is lower(<20), there is a 1:1 ratio between the protons and the neutrons. When the proton count is higher(>20, the neutron count also gets higher. The ratio is about 1:1.5. Then, when the atomic number is at 83 or higher, the element is considered radioactive. In the chart, the elements that are not radioactive are represented by dots. When it is radio active, there are no more dots.

We turned to page 2 and worked on number 6 for the end of the class. Mr. H explained the different types of radioactive decay. Beta decay are high speed electrons and don't get through the skin. Alpha decay also doesn't get through the skin and has a helium nucleus. Positron emission is an anti electron and gamma decay is a high energy light particle. The answers go as follows:

a. beta

b. alpha

c. positron

d. positron

e. gamma

f. alpha

g. gamma

h. beta, alpha

i. alpha

j. electron capture, beta

k. gamma, alpha

l. gamma

We ended the class with that page. Have a great three-day weekend!

Wednesday, May 25

The majority of our day was put into page 12 as well as our only lab for this unit. Of course we asked Mr. H about six flags first and apparently iron wolf is not every mans bestfriend.... hmmm... ANYWAY before flipping to page 12 we spoke about 11 and Mr. H explained that the difference is that there is no picture. So basically for the problems you should first separate the two material into half equations and then using the chart find which one is lower on the chart and is more easily oxidized. That equation will be your cathode. Generally our cathode equations did not have to be flipped. The other half equation is your anode and is usually flipped and the value of the oxidation potential is switched from negative to positive or vice versa. You find the voltage by adding the two potentials together and then you write the standard line notation with the anode farthest to the write and the ions in the middle. The answers are as follows:

a. Cathode half-equation: [Zn]2+ + 2[e]- > Zn

Anode half-equation: Al > [Al]3+ + 3[e]-

Volts: .90

Standard line notation: Al|[Al]3+||[Zn]2+|Zn

b. Cathode half-equation: [Fe]2+ + 2[e]- > Fe

Anode half-equation: Mg > [Mg]2+ + 2[e]-

Volts: 1.93

Standard line notation: Mg|[Mg]2+||[Fe]2+|Fe

c. Cathode half-equation: [Au]3+ + 3[e]- > Au

Anode half-equation: Al > [Ag]+ + [e]-

Volts: .70

Standard line notation: Ag|[Ag]+||[Au]3+|Au

d. Cathode half-equation: [Au]3+ + 3[e]- > Au

Anode half-equation: Mg > [Mg]2+ + 2[e]-

Volts: 3.87

Standard line notation: Mg|[Mg]2+||[Au]3+|Au

e. Cathode half-equation: [Cd]2+ + 2[e]- > Cd

Anode half-equation: Li > [Li]+ + [e]-

Volts: 2.65

Standard line notation: Li|[Li]+||[Cd]2+|Cd

Then we did our lab using juicy fruits to make a battery. You should have ended up seeing Mg most frequently in your negative column and it should have had the highest voltages. The order is Mg, Zn, Al, Pb, Fe, Cu. There is a WebAssign due Friday and a quest tomorrow.

Monday May, 23

Today's class started off with Mr. H giving back our redox quiz we took last Friday. Everyone did really well and it was our best quiz so far. Yay! After this Mr. H talked to us about the lab we are going to do Wednesday which involves using a fruit to make a battery. He explained that someone from your groups of two should bring in a JUICY fruit; oranges, kiwi, lemon and limes are all good choices. Dimitry already brought his fruit for the lab, which he bought from the Soviet market this past weekend. He also bought some borscht from the market too. Once Mr. H finished explaining the lab, he talked about what we are going to do tomorrow. We are going to take a practice Final Exam that consists of 80 questions in order to prepare for the actual final. He said that since there are a lot of problems we are probably not going to finish the whole thing but that's okay. Mr. H did however share some very important information concerning the practice final: DON'T WRITE on the booklet. If one were to write on the booklet they are putting their life in danger because the booklet is pressurized and if you write on it, it will activate an explosive that will blow up the booklet and you. But only you.

After all this blowing up people and borscht, we almost started our lesson for today, until Will found a beetle. Him and Mr. H decided to take the beetle outside to a better place. I think he died (or feigned death five times). Finally we started in our packet on page 9 with a review of the reduction table. The key rules to follow while using a reduction table are:
1. Elements higher on the table are easily reduced, but least likely to oxidize.
2. Elements lower on the table are least likely to reduce, but are easily oxidized.

Then class was interrupted again because Dan lost his packet, he probably sold it. Dan also received a note which means he did something wrong again.

We went on to page 10 where Mr. H taught us how to calculate the volts of a standard reduction potential for any oxidation-reduction reaction. The first problem involved writing two half reactions, listing the standard reduction potentials values, and calculating the overall cell potentials.

Step 1: Write the oxidation and reduction equations:

O: Al(s)----> Al3+(aq) +3e-

R: 2e- +Zn2+(aq)----> Zn(s)

Step 2: Find the standard reduction potentials for the two elements Al and Zn. For the anode, Al, the oxidation potential = +1.66 V. The reason this number is positive is because any oxidation reaction is simply the reverse of the reduction reaction listed in the table; -1.66=+1.66. The reduction potential =-0.76 V for Zn

Step 3: Then you add the two values together to get the cell potential which is +0.90 V.

After this, we went on to page 11 and did problem 8 together. For 8, you had to use standard reduction potentials from the table and the principle that is too long to type. (See page 11 at the top).

Step 1: Find the volt values for Cd and Ag: -0.40 for Cd and +0.80 for Ag.

Step 2: Determine which reaction is found lowest in the table of reduction potentials because this reaction is most likely to occur as an oxidation reaction: Cd.

Step 3: You can now deduce that Cd is oxidized and Ag+ is reduced. Then you can use this information to figure out that Cd is the anode and Ag+ is the cathode.

Step 4: The direction of electron flow is from left to right, and the direction is always from the anode to the cathode.

Step 5: Write the two half equations:

O: Cd(s)----> Cd2+(aq) +2e-

R: 1e- + Ag+(aq)----> Ag(s)

Step 6: Add the two standard reduction potentials together to figure out the cell potential and the answer is +1.20V.

We finished class today by Mr. H giving us a AFKE Quiz (Fake Quiz), numbers 9 and 10 on page 11.

Homework: The redox webassign was moved back to Thursday and their is also a quiz on Thursday

THURSDAY MAY 19, 2011

ELECTROCHEMISTRY

We started the day with a goal. That goal was to be able to improve and perfect our skills in balancing redox equations. We were reminded to always keep the halfsheet in mind when attempting to balance the redox equations. The halfsheet states six steps which are needed to balance those equations. (listed below)

STEP 1: Use oxidation numbers to identify which species is oxidized and which is reduced. STEP 2: Write separate equations for the two half-reactions.
STEP 3: Balance each half-reaction.
a) Balance the elements other than H and O.

b) Balance O by adding H2O.

c) Balance the H by adding H+.

d) Balance charge by adding electrons (e-) to the side that is realtively more positive.

STEP 4: Multiply each half-reaction by an integer so that the electrons (e-) in each equation balance.
STEP 5: Add the two half-reactions, canceling identical specis wherever possible.
STEP 6: CHECK THE EQUATION, making sure both the elements and the charge are balanced.
Using this half sheet, we went over packet page number 4.

Example, letter C:
Zn + [NO3]- --> [NH4]+ + [Zn]2+

First, we have to observe and find out “what” is turning into “what.” For this problem, the Zn would be turning into [Zn]2+ and the [NO3]- would be turning into [NH4]+. After that is decided, you need to separate those elements in to two half-equations. You have now completed steps 1&2 from above. To accomplish the third step above, you need to balance everything including the charges of the two half-equations. For the first half equation (Zn-->[Zn]2+), you need to balance the charge. By doing this, you would add two electrons to the right side which makes both Zinc elements have a charge of 0.

The balancing of the other half-equation ([NO3]- -->[NH4]+) is more challenging. Since there is Oxygen molecules present on the left side, but not the right, you have to add H2O to the right side to balance out of the O molecules. In this case, you would add 3 H2O molecules, because there are 3 O molecules in NO3-. Also, since there are H molecules in NH4+ and H2O, you need to balance by adding H+ molecules to the left side. The 6 H molecules in H2O and the 4 molecules combine to a total of 10 H molecules on the right side, meaning you would balance the left side by adding 10 H+ molecules. The last thing that needs to be done to the second half-equation is balancing out the charge. You would need to add 8 electrons to the left side in order to balance the charge. All of these steps to balance the equation just cover step 3 from the above steps.

Lastly, we need to make sure the top reaction has an equal number of electrons to the bottom reaction. This requires us to multiply the entire top equation by four. After you multiply the top equation by four, you combine both of the equations together. The combined equation should be: 8e- + 10[H+] + 4Zn + [NO3]- --> [NH4]+ + 3H2O + 4 [Zn]2+ + 8e-. If there are “like terms” on opposite sides, you are able to cancel them out. For this instance, you cancel out the electrons, 8e-.

This leaves the equation as: 10[H+] + 4Zn + [NO3]- --> [NH4]+ + 3H2O + 4 [Zn]2+ as your answer for the complete balanced equation.

We finished more problems on this subject, but also moved on to packet page 6 which we were introduced to the topic of salt bridges. Salt bridges are added to provide a permeable membrane which allows a flow of ions between the half cells to prevent a buildup of net charge. Without salt bridges, the Zn in the example would diminish and buildup on the side of the Cu. This would make the battery go dead, becaues there would be no more reactant left (Zn) to keep the ions flowing. A picture of a salt bridge is below:

Tuesday, May 17

Today, we started off class by picking our lab notebooks up from the back table and receiving our grades for the last 3 labs we completed + the new packets. Mr. Henderson announced that the test is on Homelogic as of this morning along with our labs. He told us that if we had any questions about the test to see him in the science office anytime this week in the *MORNING.* Mr. H then began to describe how the rest of the year was going to look like: we're only going to complete about half of the next two packets (1/2 units of ch. 13 & 14), there will be only 4-5 more labs, there will be a quiz for each chapter and quizlets throughout each, and most importantly - IN-SCHOOL FINAL REVIEW SESSIONS. Next Tuesday, Mr. H won't be here so he's going to give us the final exam, which is a standardized test consisting of about 80 multiple choice questions. We will have time to preview it in class and go over it the next day. However, the actual test will be a little different with about 5-6 additional written questions. He reminded us that we will be able to use our note cards once again for the exam. He also said that from this point on reading sheets are optional (but highly recommended). Subsequently, we began the new unit: Electrochemistry.

Essentially, the purpose of the unit is to understand how batteries work. We'll be making a battery out of FRUIT later this week (as a lab). We turned to page 1:
-there are 2 reaction types, acid-base and oxidation-reduction
-oxidation: the imaginary charge which an atom would have if the shared electrons among covalently bonded atoms were considered to be shared equally among bonded atoms.
Mr. H told us the definition (^^) was bad and just remember that it is a number. We used the tablet with "Rules for Assigning Oxidation Numbers" to do number 1A-1H. The answers-

We then turned to page 2:
-oxidation: an increase in oxidation number associated with a loss of electrons
-reduction: a decrease in oxidation number associated with a gain of electrons
-OIL RIG: Oxidation Is [electron] Loss, Reduction Is [electron] Loss
-LEO GIR: Loss Of [electrons] = Oxidation, Gain Of [electrons] = Reduction
-If one substance is losing electrons, another substance must gain the electrons.
-oxidizing agent: substance that is oxidized, gains electrons that are lost
-reducing agent: substance that is reduced, loses electrons
We used the concepts introduced on both pages to completer parts b and c. The answers-

Although there is no homework for points due tomorrow, there is that optional reading sheet.

Thursday, May 12, 2011

Today, class started off with Mr. H giving out a handout that dealt with titration curves. The worksheet was simply a review of what we had learned last Friday in class, but Mr. H felt we needed a review because weren’t quite able to fully understand the concept of titration.

On the first side of the sheet, we learned the concept of titration when a strong acid is mixed with a strong base. Similar to the lab we did the previous day, Mr. H talked to us about the procedure of adding base to an acid with a few drops of phenoplathian. What happens is as base is added to the aqueous solution it slowly reacts away with the H+ inside of the beaker. As soon as there is more OH-, then H+ in the beaker the solution would turn pink. At that point is when titration was reached. This gave a better explanation of what exactly happens when a solution is being titrated.

On the second side of the sheet, we learned about titration when a weak acid is mixed with a strong base. This was also similar to the lab. OH- is constantly added to the beaker as well, but the only difference in this case is that that point of titration is reached earlier. This is because less of the H+ dissolves inside of the phenoplathian. The solution turned to pink faster than when the strong acid and strong base were mixed with each other. Learning these two concepts gave an insight into how to complete the previous labs, and to understand the new lab, Ab7, we were going to do. Lab AB7 required us to find the molarity or concentration of an unknown substance.

During the lab, we had to perform the titration procedures using multiple trials. Almost the whole class finished with time left, so we went to the front of the classroom to discuss the results we got. The point of the lab was to use the basic formula of MaVa=MbVb. The concentration for the base was carried over from the averaged concentrations found in the previous lab. With that, we were able to find the concentration for the unknown substance.

Class ended at that point, and Mr. H reminded us of the 32 point WebAssign we had that was due tomorrow. We only have to get 22 points to get a 100 on the assignment.

Wednesday, May 11

We started class today by going over yesterday's blog. We then spent the first half of class doing page 15 in our packets which was about buffers. A buffer is a solution that resists changes in pH value. That means that when you add an acid or a base to a solution, the pH doesn't change. Then we talked abot what is in a buffer. It is either a weak acid with the conjugate base or a weak base with the conjugate acid. If the buffer was a weak acid with the conjugate base, i would have HA and A-. We then talked about why buffers work. They work becuase the acid won't react with the base. When you add the base, the acid neutralizes so it reacts away what you add. Because of this, the pH doesn't change. The most common buffer is our blood becuase for us to stay alive, the pH has to stay between 7 and 7.8. Most of this is explained in the paragraph at the top of page 15.

We then did problem 28 which was asking if sets of compounds would create a solution that is a buffer. In the first box, the first answer was yes because HNO2 is a weak acid. The other compound listed NaNO2 is a salt containing the conjugate base, NO2-, so it is buffer solution. The next answer is no because HNO3 is a strong acid and a buffer solution has to have a weak acid or weak base, not a strong one. The next solution was a buffer because it has a weak acid with a salt cotaining the conjugate base. The last one in that box is not a buffer because Ca(OH)2 is a strong base. Using those explanations, we did the second box. The answers are yes, yes, no, no.

Next we did problem 29 a. We had to start by wrinting the equation which was H2O+H2CO3<-->HCO3-+H3O+. We were told the initial concentrations of H2CO3 and HCO3- so we put those vlaues in our ICE chart. After adding and subtracting the x's in the middle row of the ICE chart, the bottom row showed the concentrations which were .1-x for H2CO3, .25+x for HCO3- and x for H3O+. We were given the Ka vlaue, 4.4e-7, so we had to set that equal to [HCO3-][H3O+]/[H2CO3]. By plugging in the vlaues we got in the bottom row of the ICE chart, the equation now looks like, 4.4e-7=x(.25+x)/.1-x. By using the 5 percent rule, we can make the problem simpler by eliminating the x's. The equation is now 4.4e-7=.25x/.1. After solving, x=1.76e-7. The problem is asking for pH so we have to recognize that x=[H3O+]. We can then use the equation pH=-log[H3O+] to find that the pH=6.75. After doing this problem we did 29 b on our own. The x value was 2.88e-5. Since this was a base problem, x=[OH-] so we had one more step than we had in the previous problem. Using [OH-] we found the pOH, 4.54, and had to subtract that from 14 to get pH=9.46.

Then we finished doing trials of our lab which was explained in the previous blog. Our homework was to finish the webassign due Friday.

Tuesday, May 10

Today was a very productive class period. In the beginning of the class period, we stayed in the classroom, reviewed Kevin and Kendall's blog, and got our quizzes back.

To do the last problem on the quiz, you had to write the equation consisting of the conjugate base of the weak acid. This would make the equation...

H20 + NO2- ---> HNO2 + OH-

You would then have to set up an ICE table where NO2- would have an initial amount of 2.50 molarity. This is because the problem stated that there were 2.50 M of sodium nitrite (NaNO2). Since NO2- is acting as a base, and we got the Ka value, we have to figure out what the Kb value would be. To do this, one would have to do Kw over Ka which is..

1.00 x 10^-14

7.0 x 10^-4

The Kb value would turn out to be 1.49 x 10^-11. From there it is all algebra (minus the 5% dissociation theory) and the same type of problem we have been doing all unit to find out what the pH value would be.

We spent the remainder of class doing a titration lab. Since it was our first time, our results were not super great as our indicator turned a dark pink instead of an ideal light pink. The procedure of the lab went as the follow. One would have to rinse the buret multiple times to ensure that it was clean. Then, one lab partner would get 30 mL of NaOH solution in a 100 mL beaker and rinse the buret with the NaOH solution. Then, another person would mass out between .30 and 1.00 grams of KHP and would put it in a 250 mL Erlenmeyer flask.

The most important part of the lab comes next...adding three drops of phenolphthalein. If one misses this step, they would have to start all over. After that, one records the initial reading of the buret. Next, the titration of the NaOH solution begins until a light pink color lingers for more than 30 seconds. Now someone records the final reading of the buret.

This is only one titration trial and will have to be repeated at least 2 or 3 more times. It was a very busy day in Mr. Henderson's class. The homework was to finish the labs 1,2, and 3 which we have been working on.

Monday May 9th

Mr. H started off class by announcing that the lab notebooks would be collected on Wednesday, instead of Tuesday. We then turned to page 33 in our unit packets to get the Ch. 14.2-3: Titration and Indicators reading sheet answers. The answers are as follows:
1. C
2. C
3. A
4. D
5. F, A, B
6. 50.0
7. B
8. T, T, F
9. A
10. A, B, C
11. B
12. B
13. A
On the side margin of page 33, we took notes for the class discussion.
1. H+ + OH- ---> H2O
2. HCN + OH- ---> H2O + CN-
3. NH3 + H+ ---> NH4+
We also had a quick discussion on "Mavamibvib". This is the Ma x Va= Mb x Vb (moles of acid= moles of base.) It is important to remember because it came in handy on a "lab" we did later.
We then turned to page 11 of our packets to do problem e. First, we changed the aluminum hyroxide to calcium hydroxide, Ca(OH)2. We then wrote out the major species, which is Ca2+, OH-, H+, and Cl-. We crossed off the Ca2+ and Cl- and wrote out the balanced net ionic equation, which was H+ + OH- ---> H2O. Since we are solving for the mass, this requires some stoichiometry, To find the moles of H+, we converted 20.0 ml to .02 L and multiplied that by .45 M (it is always liters x moles. Then we set up an equation to find the mass of Ca(OH)2. So we step up the equation: .009 mol OH-x 1 mol Ca(OH)2/2 mol OH- x 74 g Ca(OH)2/1 mol Ca(OH)2. Our finals answer was .333 grams of Ca(OH)2.
We then turned to page 13 and did problem 26. This ICE problem is slightly more tricky because it involves a "common anion" and finding the percent dissociation. We wrote out the dissociation equation: HF- <---> H3O+ + F-. After filling out the ICE table (pretty self-explanatory at this point), we set up the Ka equation, which is Ka=7.2x10^-4=(1.5+x)/.5-x. We then simplifying it to 1.5+x/.5=7.2x10^-4. We solved for x and found a value of 2.4x10^-4. To find the pH, we did -log(2.4x10^-4), which equaled a pH value of 3.62.
Next, it was time for our lab. This was unlike most labs because we stayed in our seats and Mr. H gave us the data. We were given two pieces of paper. We filled out the top graph by connecting the points and fmarked the equivalence point (which is when moles of aicd=moles of base). And we found the volume of the base to be 4.1 mL of NaOH. For the bottom graph we also connected the points and found the equivalence point, which was 9.1. The data can slightly vary for that value. We then found the volume of NaOH, which was 25.4 mL.
Next, we did problems 6 and 7 on an analysis sheet. For problem 6 we wrote in the volumes we got for both of the graphs (strong acid with strong base/weak acid with strong base). For problem 7, we used the Mavamibvid formula to find the molarity of the acid for each titration. We solved for the Ma value. The formula looked like this: Ma x 4.1 = .10 x 4.1. I then solved for Ma and got .41. Once again, the values can vary.
We then took ANOTHER very hard quiz to finish our day.

Friday May 6th

Friday started off like every other day in Mr.H's 3rd period class . Began finishing up calculations for LAB AB2. We worked for 5 minutes and then moved on to going over the reading sheet from the past night.
Reading Sheet 13.5-6
1. a. WB
b. N
c.WB
d. SB
e. WB
f. WB
g. N
h. N
2. C
3.D
4. D
5. D
6. A
7. B
8. A
9. C D
10. C
11. FALSE
12 A B C
13. B D
14. B
15. a. Neutral
b. Basic
c. Acidic
16. C

Next we moved on the the main part of the class period where we moved to page 9.

We started with problem 21 where we needed to rank strength of the acids. The key to this was the number closest to 0 was the weakest acid. After this we proceeded to the next problem.

For Problem 22. we needed to examine the aqueous solutions and determine the major species within them. This meant that we first had to determine if something was a strong or weak electrolyte. Then if it was a strong electrolyte then it broke apart into ions and the weak ones did not. So for part a. the solution NaOH is a strong electrolyte meaning that it would break apart into the major species Na+ and OH-. The solution of acetic acid is a weak electrolyte so then the acid does not break apart so the major species would be HC2H3O2.
The next 4 problems followed the similar pattern.
We then moved on the next page.... which is a problem....
To do these problems you need to do some sort of stoichiomestry...ummm....
well...
yah.... um
I think... you start by determining the major species. then for some reason the Na and Cl are insignificant... so you are left with the equation H+ + OH- ===> H2O
Next you convert from .05 L of HCl to the moles of H+ ions. the conversion factor is... (.05 L*.1)/1L= .001 mol H+
This you convert from H+ to OH- which is a 1 mole over 1 mole conversion factor meaning that there are .005 mol OH
then you convert to volume of NaOH. this looked like (.005*1mol*1L)/(1 Mol*.25mol)
this gave us a answer of .02 Liters.
Im sorry for the crummy explaination... my bad...
Next we moved onto the steps if titration for lab AB3
the steps were
1. Clean Buret
2. Fill buret until 0 mark
3 mount buret
4. Clean Flask
5. add acid to flask
6. Add 2 drops of indicator to flask
7. titrate while stirring flask
8. once color changes record buret amount.

After this the period ended.
If your bored, or if your from slovenia you should watch this video.
It has nothing to do with this class but why not. http://www.vimeo.com/23396791

Salts

It's Cinco De Mayo!!!!!!!!!!


We started this class with a look at our packet to look at topics such as conjugate acids and bases to determine whether a solution is
acidic, basic, or neutral.

Let's take this reaction, for instance...
H20(l) + HF(aq) -------- H3O+(aq) + F-(aq)

We know HF is a weak acid and water can be both an acid or a base (though weak). In this case however, we know water would be the base because H3O+ is its conjugate acid - meaning it wants to give up a proton (it has that + charge it really wants to get rid of). We know the F- would be the conjugate base of HF, because F- wants to accept an extra proton (it has that - charge it really wants to get rid of). At this point you might be asking why I'm going on about this basic stuff (ba dum tss) when I could be talking about how salts dissolve. Isn't that what we discussed in class today? Why, yes it is. I am getting there.

I'll give you two examples and go through them step by step:

1. NH4Br
2. KClO4 (both of these are salts)

An aquoeous solution of a salt in water will be either acidic, basic, or neutral due to hydrolysis: either in the cation, anion, or both. Hydrolysis is when an ion reacts with water. These are ions that DO NOT hydrolize:

A. anion of a strong acid (negative charge=anion)
B. cation of a strong base (positive charge=cation)

ions that DO hydrolize:

A. anions and cations of weak acids and bases

Let's look at NH4Br again. It consists of NH4+ and Br-. Br- would not hydrolize because it is the anion of a strong acid (HBr). Therefore, you disregard it. Since NH4+ would react with water, because it is an ion of a weak acid/base, the formula would look like this:

NH4+(aq) + H20(l) -------- NH3(aq) + H3O+(aq)

We still must determine if this salt is acidic, basic, or neither. We now look at H2O. Notice how it is acting as the base in the relationship (because NH4+ is an acid). That means H3O+ would be the conjugate acid of H2O, and the conjugate acid of a weak base (such as water) would mean the salt NH4Br would form an ACIDIC solution in water.

Moving on to the next one, we see
that KClO4 would be neutral, because K+ is the cation of a strong base (KOH) and ClO4- is the anion of a strong acid (HClO4). Therefore, you can think of them as "canceling out".

You may use this logic to solve any other issue of this type you may run into. What I want you to absorb is not only that notecard you've been sleeping on, but also this:

1. Conjugate base of weak acid: basic solution
2. Conjugate acid of weak base: acidic
3. Conjugate base of strong acid or conjugate acid of strong base: neither

This is also helpful (this may be the exact same thing, but I'm not sure):

1. If cation comes from strong bas
e, and anion comes from strong acid, salt is neutral.
2. If cation is from strong base, and anion is from weak acid, salt is basic
3. If cation is from weak base, and anion is from strong acid, salt is acidic
4. If cation and anion are both weak, then you can't tell without knowing relative strengths of acid and base.

It's very much like an arm-wrestling match.


Next, we discussed how to do ICE tables for such salt equations. We were introduced to a new concept called the Kw (which is 1e-14). Basically, to find Ka or Kb of salt, you do Kw/Ka or Kb (given, opposite of what you need). You then complete the ICE table as normal. Remember that notecard with all the important conversions on it! I won't dive to deep into this because this isn't such a new concept (the notecard or the ICE table).

Then, we completed our day with a lab. Basically, this day was all about salts.


Not as unlucky as previously thought.

Wednesday, May 4

Today's class began rather unceremoniously as we were told to open our packets to page 17 and take out a calculator we're comfortable with using. After a quick review of Lindsay's blog post, we dived right into the lesson plan. On the packet page were four important, essential, need-to-know rules on conjugates:

1. The conjugate base of a weak acid forms a basic solution in water

2. The conjugate acid of a weak base forms an acidic solution in water

3. The conjugate base of a strong acid is neither acidic nor basic

4. The conjugate acid of a strong base is neither acidic nor basic

From these rules we can infer that in a reaction, conjugates of weak acids/bases hold significant influence within the reaction, while conjugates of strong acids/bases do not affect the reaction at all. These rules would help with answering the questions and filling out the tables on page 17. However, should a salt be composed of a conjugate of a weak acid and a conjugate of a weak base, the composition of the solution shall be determined by comparing the Ka and Kb values (the "strengths" of each molecule). Page 18 built off of this new understanding by posing a problem involving the familiar ICE table, a novelty by now. Dash through one problem, then off to finish Lab AB2. The remainder of the class period consisted of this lab, so nothing real exciting unless you're really into watching liquids change color from being reacted with testing solutions. And what better way to end such an average day than with a typical homework assignment? The long WebAssign is due Thursday; only 35 points needed to get an A, though it would probably help if you read the heavy textbook if some problems are proving challenging. Until next time, take care and don't mind the sarcasm.

Tuesday May 3, 2011

We started todays class by going over the reading assignment that was due today, which is on pg. 29 in your packet.

1a.WB
b. N
c. WB
d. SB
e. WB
f. WB
g. N
h. N

2. c

3. d

4. d

5. d or b

6. a

7. b

8. a

9. c and d

10. c

11. FALSE—NaF is basic

12. a,b,c

13. d

14. b

15. a. neutral

b. basic

c. acidic

16. c

After going over the reading sheet we turned to pg. 5 in our packet to do #10 a and b. These problems are just like the ones on the previous page except these are for weak bases. As we were doing them we found that it involved the exact same process for both acids and bases. The only difference was that for acids they will ask for the Ka expression and for bases they will ask for the Kb expression.

We briefly went over Will’s blog which talked about what we did on page 4. Afterwards we moved on to page 8 #20. This problem was a little different because this time you are given the X value and told to solve for the Ka expression instead of the other way around.

When we finished with the packet pages for the day, we started Lab AB2 called Indicators. The purpose is to use indicators to determine the approximate pH of serveral acid, base, and salt solutions (5 in total). Here is a picture/example of what we have done so far…

Monday, May 2

We did not do anything today in chemistry that was out of the normal. We learned about chemistry, and then we had a quiz. Because images make any blog post better, I found a picture of an ICE table, since ICE tables are relevant to what we are learning.












Today in chemistry we used our ICE tables to determine the pH of some acids. We started on page 7, and worked through #'s 15 and 16. To determine the pH of the acids in these problems we had to calculate the Ka value, and then use logs to find the pH. Nothing too abstract for our sophomore minds to grasp on to.
But anyway, what you really came here for. Answers to 15 and 16.

15. 5.05
16. 2.036

We flipped back to page 4, but the ICE table problems on this page are review. Nothings difficult. Grace immediately became frustrated with this, and asked if there were going to be any harder problems. This was no problem, and her frustration quickly dissolved when she discovered we were moving on. We flipped to page 8.

Page 8 was nothing special. Plain white sheet of paper, black text, the usual. I'm sure your asking yourself right now "Will, why is page 8 important?" It's problem number 19 on page 8 that's important. It captivated the entire class and left us speechless afterwards. I myself could not help but sit in my seat and gape in awe at the elegant nature of the problem. The problem required us to solve for the pH and the % dissociation of the 0.80 M solution of acetic acid. If you were able to hold off admiring the problem for just long enough to solve it, you would get a pH of 2.420 and a % dissociation of .0047. The % dissociation was much less than %, so we got by using the 5% rule to obtain a fairly accurate answer without the quadratic formula. If it was not less than 5%, than we would most likely be in an AP Chemistry class.

We finished the class today with the recently usual quiz covering current topics in class. It was very enjoyable and I can genuinely say that I enjoyed the quiz's unique option. It was 2 sided, and we were able to choose which side we wanted to do. Mr. Henderson truly is a visionary in the field of quiz making. The quiz definitely sent the class out with a bang.

*Important Disclaimer: Fabrication is evident*

Friday April 29th

Class began with collecting the answers from a reading sheet webassign which had previously been for homework. This can be found on page #25 of your packet.

CHAPTER 13.2-3 pH, pOH and the Ion Product of Water
1. b able to both accpet and donate a proton.

2. a H+ and b OH-

3. d 1.0x10e-14 = [H+] x [OH-]

4. a the dissociation of water into its ions

5. neutral
acidic
basic
basic

6. a increases

7. c 2e-7 M

8. b decreases

9. b basic

10. a acidic

11. b lower, acidic

12. basic
acidic
neutral

13. c 11.0

14. d the conjugate acid of water in an acid is ionization

15.d HNO3(aq) + H2O --> [H3O+](aq) + [NO3-](aq)

16. b pH=1

17. a [Na+]=0.50 M, [OH-]=0.50 M, and [H+]=2.0e-14 M

18. d Use a stoichiometric calculation

*Mr. Henderson pointed out that we should write down the equations/formulas on the right side of the page on a notecard*

equations/formulas:

• [H3O+] x [OH-] = 1.0e-14


• [H+] x [OH-] = 1.0e-14


• pH= -log[H3O+]


• pH= -log[H+]


• [H3O+]=[H+]=10^(-pH)


• pOH= -log[OH-]


• [OH-]= 10^(-pOH)


• pH + pOH = 14 (at 25 degrees C)

We then turned to page #4 where we worked on #9 part a.

a. the first step is to write the equation

H2O + HCN <---> [H3O+] + [CN-]

I 2.0 0 0

C -x +x +x

E 2-x x x

Ka expression= ([CN-] x [H3O+])/[HCN]

Ka=6.2e-10=(x^2)/(2-x)=(x^2)/(2)-->(x^2)=1.24e-9

x=3.5e-5

This is when the 5% approximation comes into play. This says that since the value of x is so small that it can just be ignored in the 2-x part of the equation.

3.5e-5 M=[CN-]=[H3O+]

We then turned to page #6 to finish the table in probblem 12 and work on question #13

12. [H3O+] [OH-] pH pOH acidic or basic?

1.0e-2 1.0e-13 2.00 12.00 acidic

1.0e-11 1.0e-3 11.00 3.00 basic

1.0e-5 1.0e-9 5.00 9.00 acidic

1.0e-3 1.0e-11 3.00 11.00 acidic

1.0e-7 1.0e-7 7.00 7.00 neutral

2.0e-14 5.0e-2 13.70 0.30 basic

6.0 1.67e-15 -0.78 14.78 acidic

2.5e-15 4.0 14.60 -0.60 basic

13. a. [H+]= 2.0 M

pH= -log(2.0)=-0.30

b. [OH-]= 2.0 M

pOH= -log(2.0)= -0.30

pH= 14.30

c. [OH-]= 0.100 M

pOH= -log(0.100)=1.00-->pH=13.00

Review of the six strong basses:Li, Na, K, Ca, Sr, Ba

Thursday April 28th

Todays beautifully short class began with going over the right column of page 5 in the packet. Mr. H. highlighted the water dissociation equation and the fact that the equilibrium of the ions is always 1.0e-14. Also that neutral solutions have equal ion concentrations, acidic solutions have more hydrogen ions, and basic solutions are just written differently and OH dominates. On the back side Mr. H. noted that when discussing powers of ions it really was just referring to the numerical value of the exponent. Below we were introduced to equations we would be using on page 6.

Next we flipped back to page 5. Mr. H. briefly spoke about number 11 which has some useful equations. He also mentioned that on a previous reading webassign there was a study tip regarding those equations:

STUDY TIP:
On a note card, write down the equations/formulas presented in these two sections of reading; these include Equation 13.1, Equation 13.2, Equation 13.3, the equation on the line below Equation 13.3, Equation 13.4, and the first equation on page 356. Keep the note card handy during class, when doing homework and when on FaceBook. Accept it as your friend.

^^^YOU HAVE BEEN WARNED

On page 6 we went over some of the table for number 12. We were able to solve columns 1 and 2 by dividing 1.0e-14 by the given OH or H3O value. The answers were for row1: 1.0e-12 row2: 1.0e-11 row5: 1.0e-7 row6: 2.0e-14 row7: 1.67e-15 row8: 2.5e-15. Next we were easily able to do columns 3 and 4 for 3 and 4. Because the pH and pOH values have to equal 14 row3: 9.00 row4: 3.00. We were also able to do column 3 and 4 for rows 1 and 2. Using our calculator we used the equations from page 26. For row 1 we took the negative log of .o1 with a base of 10 and got 2.00 for pH and 12.00 for pOH. Therefore we knew it was acidic because the pH was less then 7. For row 2 we got 3.00 for pH and 11.00 for pOH and for row 6 we got 13.7 for pH and .3 for pOH. Therefore rows 1, 2, 3, and 4 are acidic and row 6 is basic. That was the end of class.

Unit 12,Day 2

In today’s class, we started out by going over Chapter 4.3 and Chapter 13.1 Reading Sheet, found on page 23 of your unit packet. The answers are as follows:
1. C
2. D
3. True, False, False, True, True
4. D
5. A
6. B
7. A, A, A
8. B
9. A, B
10. False because some substances are amphiprotic
11. Amphiprotic means a proton could either be lost or gained, like with H2O, which can become H3O+ or OH-
12. A, A, B
13. B, C

Then we went over Kon’s blog from Monday, which was about the first day of unit 12.
Next, we began learning about Bronsted and Lowry. They said that you can’t have an acid without having a base. This idea is different than what we learned on Monday, where you would have just split the H from its anion, so keep that in mind.
Then we did problem 6 on page 2. Keep in mind the acid is the proton donator and the base is the proton acceptor.
a. HCN(aq) + H2O(l)  CN- (aq) + H3O+(aq)
b. HF(aq) + H2O(l)  F- (aq) + H3O+(aq)
c. HNO2(aq) + H2O(l)  NO2-(aq) + H3O+(aq)
The H3O+ is called the “Hydronium Ion”
d. *Now we are working with BASES! This means that the base GAINS a proton.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
C5H5N(aq) + H20(l)  C5H6N+(aq) + OH-(aq)

Next we did a few problems on page 3. Keep in mind that Acids donate protons and Bases accepts protons.
Here is the general equation used for an acid-base reaction.
HA + B  A- BH+
(acid) (base) (conjugate base) (conjugate acid)

Note that “the reaction of an acid with a base changes the acid into a conjugate base and changes the base into a conjugate acid.”
a. B, A, CA, CB
b. A, B, CB, CA
f. A, B, CB, CA
g. B, A, CA, CB
F and G are “zingers”. Because HCO3- is and Acid in part F, you would expect it to be an Acid in part G as well. This however is not true, and you have to always look to see what is giving away the proton and what is receiving it.
We ended class by doing Lab AB1.

Monday April 25th

Today, we started class by getting our chemistry lab notebooks from the lab tables in the back of the room. Mr. H then went around the room and gave everyone their rubrics for the labs that we did in our notebook. Mr. H then went on to talk about how if you had a bad test grade, then you should go and see him. He mentioned that he won't be available tomorrow, but that any other day he would be available.

Mr. H then officially began class by introducing to the class another 50 point web assign that will be due in about 3 weeks. If you want to know more about it, here is the link: http://gbschemphys.com/honchem/index.html
Mr. H did tell us though that in order to get 100% on this web assign, you have to get at least 35/50 questions correct. As an addition to this web assign, Mr. H reminded the class that just like the previous unit, there will be frequent quizzes throughout this unit and a test that will be in three weeks.

Mr. H then began his lesson of the day, which was an introduction to Acids & Bases, by letting us know that there will be a lab tomorrow in class, so be sure to bring your lab notebooks. He then proceeded to write important notes on the board which are essential in understanding Acids and Bases. The Notes are as followed:

Acids and Bases


Definitions:

1) observable properties (Lab AB1)

2) Arrhenius (Early 1800s)

Conductor: it needs to have ions in order to be considered a conductor

^^One of the examples that Mr. H gave about conductors was the light bulb with the two rods. He tested to see if multiple liquids were conductors by putting each liquid in a beaker and then placing the beaker in the two rods. If the light bulb lit up, then we knew that that liquid was a conductor. If it didn't, then we knew it wasn't. The liquids that Mr. H experimented with were HNO3, HCl, HC2H3O2. The first two liquids lit the light bulb really bright, so we were able to conclude that those liquids had lots of ions in them. When we experimented with HC2H3O2, (a.k.a. vinegar), we noticed that the light bulb turned on, but the light was very dim. From this, we were able to conclude that vinegar is a weak acid on the pH scale, and therefore is a weak conductor.

Acids: H___ ---> H+  +   ____-

A= anions

Bases= ____OH----> _____ + OH-

^^ Mr. H continued to use the light bulb as a conductor tester, but this time, he used KOH and NaOH, which are solids, and mixed them in a beaker of water. He once again did the same process as before, and after experimenting with both, the class saw how the light bulb became very bright when both of the solutions were tested on the light bulb prongs. This shows that KOH and NaOH are two very strong bases and have plenty of ions in the solid to become a conductor.


Strong Acids: HNO3------------>  H+    + NO3-    (High K) <--------- lot of reactant, little product
                                 <---

Weak Acids: HC2H3O2<------------ H+    + C2H3O2-     (Low K) <------- lot of product, little reactant
                                                --->

Strong Acids to know:                                                                                        

1)HCl

2) HBR

3) HI

4) HNO3

5) H2SO4

6) HClO4



Strong Bases to know:

1) LiOH

2) NaOH

3) KOH

4) Ca(OH)2

5) Sr(OH)2

6) Ba(OH)2




After writing these notes and copying them, Mr. H had the class open up their unit 12 chemistry packets to page 1 and had us work on problem 1. Before doing problem 1, it was essential to know that:

an acid= a substance which produces hydrogen ions when it dissolves in water

a base= a substance which produces hydroxide ions when it dissolves in water

Knowing this, the class was able to answer question 1. The answers are as followed:

A            B           B
N            B           N
B            A           N

The class then went on to do problems 2, 4, and 5. (#3 was a review)

#2

a) H+  + NO3-  (Acid)

b)Na +    + OH-   (base)

c) H+  + CN-          (acid)

d) Ca2+     2OH-       (base)


#4   ( these answers correspond to a table, so the name goes on the left part of the table and the formula goes on the right part)

hydrochloric acid
HBr
hydroiodic acid
nitric acid
H2SO4
ClO4

lithium hydroxide
sodium hydroxide
KOH
calcium hydroxide
Sr(OH)2
BA(OH)2

#5
a) [H+]  = 6.0M

b) [OH-]= 3.0 M

c) [H+]= 1.0 M

d) [OH-]= 2.0M <------ this is because the product is double the reactant

This concluded our lesson for the day. Tonight's homework: 13.2-3 Rdg Sheet.

The Daily Joke: What do you do with dead chemists?
                         Answer: Barium

Tuesday, April 19

Class started off today by Mr. H taking tardies for people who were late (Hannah). He then told us to get out our lab notebooks and write the title and purpose of the lab we will be doing later in the period; Ksp Lab.

After we finished this up, we took a look at Tim's blog from last night which was really classic. "Also, due to the reasons I forgot, the test on Wednesday will be pushed back to Thursday." The wise words of Tim Joo. Anyway, he talked about ICE and some of the Ksp problems which are located on pages 17-19 and 24-25.

Today's lesson focused on continuing with the Ksp problems and learning how to solve for Ksp by using solubility values. The first problem we did was on page 25, problem d. I'm not going to go over it in great detail because we already learned about and I think it is a pretty straightforward topic to understand. But if you want to know the correct answer it is [Cu2+]= 2.29*10^-7 M and [OH-]= 4.58*10^-7 M.

We then moved on to page 26 where Mr. H explained what molar solubility is; a partially souble salt's number of moles which dissolve per liter of aqueous solution. The first problem we solved was 8a and the answer was Ksp= 1.17*10^-10.
Then, we went over to page 27 to solve 8b.
The first step is to write the balanced equation which is Fe(OH)3(s) <---> Fe3+(aq) + 3 OH-(aq). Next, you need to fill out the ICE table. The initial amount for both of these products is 0 because you are given 0 amount of them. The change for Fe3+ is +x because we don't know what the change is and there is no coefficient in front of it. For OH-, the change is +3X because there are 3 moles of OH for every one mole of Fe. The equilibrium amount for Fe is x and 3x for OH. You then have to write the Ksp expression which is Ksp= [Fe3+] [OH-]3, and OH is to the third power because you have to switch the coefficient to a power. but it only matters if the coefficient is greater than 1. After this, you plug in x and 3x for the products values; x*(3x)^3= 27x^4. For x, plug in 1.01*10^-10 and the equation is Ksp= 27*(1.01*10^-10)^4. Finally the last step is to calculate and solve for Ksp= 2.81*10^-39.

After we finished the math portion of the day, Mr. H started off the lab by explaining what we needed to do to complete this lab. The rest of the period everyone worked on the lab.

Homework: Test on Thursday and Lab Notebooks due Thursday.

Monday, April 18

We started our day by getting our quizzes back from Friday. Also, due to reasons I forgot, the test on Wednesday will be pushed back to Thursday.

Shortly, we moved on to page 19 of our packet to work on #7. This was different than the other problems we worked on before because this time, it gave us the equilibrium constant (5.76) and the initial concentrations of H2O (8.2) and CO (8.2). The other concentrations were 0 because it was not given. The mole ratios of the equation were all one, so they all had a change in x. The reaction's equilibrium constant was 8.2-x and the product's equilibrium constant was x. To solve this, all you have to do was set up an equation 5.76= x^2 / (8.2-x)^2. To solve for x, you square both sides, getting 2.4= x/(8.2-x). Multiply both sides by 8.2-x and get 19.68 - 2.4x = x. This equation makes x = 5.78. Remember that the problem is asking for the equilibrium concentration, so the answer is not simply x. The answer would be [H2O]=[CO]=2.42 and [H2]=[CO2]=5.78. Next, we moved on to #8 on the same page. This one was a little bit different because it gave us the concentration on both sides. This is quite similar to #7 except that the direction of reaction must be found out. To find the direction, we have to find Q. If you don't remember what Q is, we did some problems on page 16. The Q value was .5166 and was higher than the equilibrium constant of .175, the equation went to the left. After finding that out, everything was basically the same as #7. The answer to #8 is [HCN]=.85 and [C2N2]=[H2]=.355.

As a reminder, these two type of questions will definitely be on the test. Before we moved on to our next topic, we took a little joke break. I don't really remember any of the jokes so...

We moved on to pages 24-25 of our packet. We worked on a and c. In these types of problems, we were given the Ksp and needed to find the equilibrium ion concentration. In problem a we were given that Ksp=1e-10. Because BaSO4 disassociates into Ba and SO4 their initial is 0. They both have a change in x. The Ksp equation is [Ba]x[SO4], so it is x^2=1e-10. This makes x= 1e-5. The answer to problem a is [Ba 2+]=[SO4 2-]=1e-5. Problem c was basically the same except that one of them had a coefficient of 3. Because of this it becomes +3x and in the equation it becomes 27x^3 because the 3 and the x must be cubed. The rest can be solved with the same method as problem a. The answer to problem c is [OH -]= 3x----> 2.0x10^-9=[Al 3+].

We ended our day with us returning the calculators and Mr.H accusing us of trying to steal them.

Friday, April 15

Today we started off by flipping our packets to page 23. Mr. H said that this would be somewhat difficult, because we would have to use already learned knowledge in order to be able to work out the problems. He said that we would start out easy. The new topic that we would be learning is Ksp. It is the last topic of the unit and we should know it for the major celebration we will be having on Wednesday. Mr. H noted that if we get stuck we should look at the examples listed in our packet, which are in the book. The first thing that we did on page 23 was some review on Ionic compounds. we had to write the formulas and name the compounds. Then we had some practice with solubility and balancing equations. Mr. H also noted that when he said that there are soluble and non-soluble compounds he meant that they are partially soluble. We had some practice with this on the bottom of page 23. We then flipped to page 24, and this is what we would be putting all of those concepts together for. As I said before, Partially soluble salts dissolve only slightly in water. The dissociation of the salts are controlled by a reversible reaction that has a very small K value (much less than one). This is known as a solubility product constant (Ksp). For 4 on page 24 we had to write the dissociation equation, in which we needed to have the solubility rules. If help is needed reference example 16.1 on page 4332 in the book. After that we did some more practice on page 21 with LeChatelier's Principle. Then at the end of class we had another reading quiz which I am sure everyone did well on.

Wednesday, April 13

At the start of class, Mr. H told us to take out our packets and calculators while he was getting Will's blog ready to show to the class. After we went over Will's blog and reviewed the material learned from yesterday's lesson, we turned to page 14.

We began with number 11. It was exactly like yesterday's problems. Step by step: first, you find the ratio of the products to reactants, like always. You plug in the information you have, but we only get the mole values. So, we convert to molarity. That is found by dividing moles by the liters given. Anyways, you plug in the molarity and then solve for x. Simple as that. The answer to number 11 is .00114 for NH3.

Following up with that question, we turned our attention to numbers 13 and 14. The equation is the reciprocal of number 11's equation, so you find the reciprocal of the answer, too. From there, you can figure out the temperature, which is 76.9. Then, Mr. H had us try and figure out number 14. He asked Konstantine and Daniel if the answer to number 14 is the doubled or squ

ared for .0113 or doubled/squared for 76.9. They both said that they think it is not doubled or squared, but kept the same. However, Neil, using his debate skills, countered saying that 76.9 is squared because if you do the ratio for the equation on number 14, it ends up that if you square one side, you have to square the other to keep it equal.

After their debate or whatever, we took the quiz, which was 2 questions. Mr. H said that if we found that easy, we are in good shape.

Finally, we turned to page 15. For number 1, the answer is C. Straight from the packet, for such a situation with a K value greater than 1, the system is described as having an equilibrium which favors the products or lies to the right. So pretty much,

the Kp is a ratio, so the products have to be greater than the reactants if Kp is going to be greater than 1.

When Kp is equal to the ratio, then it is at equilibrium. We learned that Q represents the ratio between products and reactants. When Q is greater than K, there are too much products. If Q is less than K, there is too much reactants. K wants to reach equilibrium, so this is very bad.

Solving is just the same. You find the ratio (products over reactants), raise the power to

the number of the coefficient, plug in the numbers given, and see if it is equal to K (which is given). Like I said before, if Q is greater, then there is too much products (numerator). If Q is less, there is too much reactants (denominator).

We ended with jokes from Mr. H's book. :)

There's a reading assignment due tomorrow so do it!

Tuesday, April 12

Today in class we continued to learn about equilibrium constant expressions. Yesterday we learned how to create Kp expressions, and today we finished learning about Kc expressions. We flipped our packets to page 12 and we worked through the page. To form a Kc expression, I will use letter c on page 12 as an example.

c. HF(aq) + H2O(l) <--> H3O+(aq) + F-(aq)

First we get rid of any substances that aren't gases or aqueous solutions. That would mean getting rid of H20(l). So now we have

HF(aq) <--> H3O+(aq) + F-(aq)

So now taking this, we place it into he equation for the Kc.


[F-] * [H3O+]
Kc = _______________

[HF]

And that is our Kc value.

I found this picture that shows the value of the Kc.

We moved in class, and added math to Kc and Kp. The only thing we changed from before, is that now the values have numbers to them. Using our knowledge of math, we plugged the equations into our calculators and got our answers. The answers o #'s 4-7 on page 13 are as follows.
4. .222
5. .00379
6. 4.315
7. .025

The next thing we did in class, was doing algebra with the constants of equilibrium. We took our equation as shown above, and we plugged in values and solved for the values we didn't have. The answers for page 14 #'s 9 and 10 are as follows.
9. 4.25 * 10^-4
10. 2.6377

Mr. H then told us that we have a quiz tommorow, and that he won't be tellling us about quizzes ahead of time from now on because we stress out less if we don't worry about it. He said we would finish the day with jokes from his jokebook, but we didn't have time. Maybe tomorrow?