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Homework Help: Weak/Strong Acid w/ Strong Base Titrations and pH Indicator Selection Help

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Use the information below to answer the following question(s).

    methyl orange: red at pH < 3.1: orange at pH 3.1-4.4: yellow-orange above pH 4.4
    litmus: red at pH < 4.5: purple at pH 4.5-8.3: blue above pH 8.3
    thymol blue: yellow at pH < 8.0: green at pH 8.0-9.6: blue above pH 9.6
    trinitrobenzene: colorless at pH < 12: yellow at pH 12.0-1: orange above pH 14.0

    Q#1 Weak Acid with Strong Base Titration: Which of the pH indicators from the table would be most appropriate for the titration of 0.30 M acetic acid (Ka = 1.8 × 10-5) with 0.15 M sodium hydroxide?

    a.) Litmus
    b.) Trinitrobenzene
    c.) Methyl Orange
    d.) Thymol Blue
    e.) None of These

    Answer: D Thymol Blue

    Q#2 Strong Acid with Strong Base Titration: Which of the pH indicators from the table would be most appropriate for the titration of 0.30 M hydrochloric acid with 0.15 M sodium hydroxide?

    a.) Thymol Blue
    b.) Methyl Orange
    c.) Litmus
    d.) Trinitrobenzene
    e.) None of These

    Answer: A Thymol Blue

    I already have the answers as per the information above, I just don't know how to get that answer and I don't know where I'm going wrong. Any help is much appreciated.

    2. Relevant equations

    pH = pKa + log [conjugate/weak acid]

    3. The attempt at a solution

    Q#1 Weak Acid and Strong Base Titration:

    HC2H3C2 + OH- ---> C2H3C2- + H2O
    Initial 0.30 M HC2H3C2
    add 0.15 M OH-
    change -015 M HC2H3C2 -0.15 M OH- +0.15 M C2H3C2
    Result (0.30-0.15=0.15 M HC2H3C2) ~0 M OH- 0.15 M C2H3C2

    pH = pKa + log [conjugate/weak acid]
    pH = -log(ka) + log[conjugate/weak acid]
    pH = 4.75 + log [0.15/0.15]
    pH = 4.75 + 0
    pH = 4.75

    If the above calculations are correct, I would have picked C "methyl orange" since it seems to somewhat cover the pH range of what I got (4.75) but obviously something is wrong since the answer is NOT methyl orange but rather thymol blue...?

    Q#2 Strong Acid with Strong Base Titration

    H30+ + OH- -----> 2 H2O
    Initial 0.30 M H30+
    add 0.15 M OH-
    change -0.15 M H30+ -0.15 M OH-
    Result (0.30-0.15=0.15 M H30+) ~0 M OH-

    pH = -log[H30]
    pH = -log[0.15]
    pH = 0.82390874

    Again, I would have picked Methyl Orange....what am I doing wrong? Help please... :(
     
    Last edited: Aug 3, 2010
  2. jcsd
  3. Aug 3, 2010 #2

    Borek

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    Staff: Mentor

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