Well we had a sub yesterday, but we were given 2 sheets to work on about logs/antilogs related to pH and pOH. There were a couple of formulas in the sheet that we could remember in order to find pH and pOH.
pH = - log[H+] pOH = - log [OH-]
pH + pOH = 14.0 [H+][OH-] = 1.0 x 10-14
[H+] = anitlog (-pH) = 10^-pH [OH-] = antilog (-pOH) =10^ -pOH
Today, we went over the worksheet #2 about acids and bases #’s 1 and 2 where we were to identify which one is an acid, base, conjugate acid and conjugate base. Also, to determine whether they were Arrhenius acids/bases or Bronsted-Lowry acids/bases. We also did the Kw problems #’s 1-10 and the Acid-Base Assignment #6 numbers 1 and 2 on the board today.
Mrs. K assigned numbers 3 and 4 today and this was to be handed at the end of the class.
Dont forget the equilibrium law can be written as:
Kw = [H3O+][OH-] Where the value of Kw is constant at 1.0 x 10-14.
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