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

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Homework Statement



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.

Homework Equations



pH = pKa + log [conjugate/weak acid]

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:

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