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Homework Help: Strong Acid-Strong Base Titration

  1. Mar 10, 2015 #1
    This is an example in my book. I have the answer for it, but I don't know why my answer was wrong. Here's the question:
    Lauren pours 40 mL of .1M NaOH into 100 mL of Megan's .1 M HCl. What is the pH?
    Here's what I did:
    R NaOH+HCl ⇔ NaCl + H2O
    I .004 , .01 ⇔ 0 0
    C -.004 , -.004 ⇔ +.004 , +.004
    E 0 , .006 ⇔ .004 , .004​
    Then, I made another RICE table for the remaining HCl:
    R HCl + H2 O ⇔ H3O++ Cl-
    I .006
    C -.006 -.006 +.006 +.006​
    So then pH=-log(.006/.14)=3.14
    But my teacher's book says pH=1.37.
    UPDATE: I was using WolframAlpha.com to calculate the log and apparently it calls log what I would call ln. When I entered log base 10 (.006/.14) everything worked out fine. If somebody knows how, delete this post please.
    Last edited: Mar 10, 2015
  2. jcsd
  3. Mar 10, 2015 #2


    User Avatar

    Staff: Mentor

    This is just a limiting reagent question.
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