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.