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.