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How do I find the ph without using Hasselbach's equation?

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    hi so in my lab class we made a buffer solution by mixing 20.00 ml of 0.100 M Sodium Hydroxide and 40.00 ml of 0.100 M of acetic acid. We then poured 15ml of this buffer solution into a beaker. So we essentially have 15 ml of this buffer solution in the beaker. We then added 0.15 ml of 1M HCl acid to this beaker. How do i find the ph, he told us we cant use hasselbach's equation


    So what I did was divide :






    2. Relevant equations


    3. The attempt at a solution
    (40ml) (0.1M) = 4mmoles Ch3cooH 4mmoles Ch3cooh
    (20ml) (0.1M) = 2 mmoles NaOH 2mmoles of NaOH
    (.15ml) (1.0M)= 0.15 mmoles HCl .15mmoles HCl

    4/15.15 = 0.246 M Ch3cooH
    2/15.15 = 0.132 M NaOH which is the concentration of the base
    .15/15.15= 0.0099 M
    Now i have to do an ice table and find the PH this problem is driving me crazy pease help i don't even know if my setup is correct. This class is only 1 credit and i have spent 10 hours on this lab report already
     
  2. jcsd
  3. Nov 19, 2015 #2
    So i drew up an ice table im not spending anymore time

    Ch3cooh + h2o <-> ch3coo- + H3o
    I 0.264 0.132 0.0099
    C +0099 -.0099 +X
    E 0.2739 0.1221 X
    ka= 1.75 x 10^-5
    (1.75 x 10^-5 ) = (1221x)/(0.2739)
    X= 3.39 x 10^-5
    log(x)=4.4
    Ph= 4.4

    My original ph of the buffer was 4.76 now it decreased because of the strong acid make it more acidic so my ph is now 4.4 . I don't want to spend anymore time on this so if you can check this and just tell me if it is right i would greatly appreciate it. Thanks to anyone who replies
     
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