1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

POH and Dissociation Constant

  1. Mar 8, 2006 #1
    1. A bottle of strong monoprotic acid was labelled as having a concentration of 2.040 x 10-1 mol/L. Given that KW = 1.00 x 10-14, determine the p0H of the acid solution.

    [H3O+][OH-]= K_w

    [OH-] = k_W/[H3O+] = (1.00*10^-14)/(2.040*10^-1 M) = 4.90196078E-14

    -log[OH-] = pOH
    -log[4.90196E-14] = pOH = 13.3096 = 13.3

    2. The pH of a 0.2700 molar solution of unknown monoprotic acid was measured and found to be 5.75. Calculate the Ka of this acid.

    pH = -log[H3O+]
    5.75 = -log[H3O+]
    antilog[-5.75] = [H3O+] = 0.000001778 = X

    K_A = [H3O+][A-]/[HA] = X^2/[0.2700 - X] = (0.000001778)/(0.2700 - 0.000001778) = 1.17122166E-11 = 1.17E-11

    Last edited: Mar 8, 2006
  2. jcsd
  3. Mar 10, 2006 #2
    For the first one you could have just found the pH and added that number from 14 to get pOH. But anyway, stick to your way since you probably memorized it already.

    Though your answer is right (you just posted what you did after finding right solution, right?), you forgot to put square 1.778E-6 in the message.
    Last edited: Mar 10, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: POH and Dissociation Constant