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.