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.