The following acid-base indicators are available to follow the titration. Which of them would be most appropriate for signaling the endpoint of the titration? EXPLAIN
------- COLOR CHANGE ------
INDICATOR <-> Acid form <-> Base Form <-> pH Transition Interval
Bromphenol blue <-> yellow <-> blue <-> 3.0-5.0
Bromthymol blue <->yellow <-> blue <-> 6.0-7.6
Thymol blue <-> yellow <-> blue <-> 8.0-9.6
PLEASE ANSWER IT!!!!!!!!!!!!!!!
6 answers
What are you titrating? It makes a difference. Look up or calculate the approximate pH for the equivalence point; pick the indicator from that.
it says the titration of 0.145g of a weak acid with 0.100M NaOH as the titrant
A weak acid titrated with a strong base has an equivalence point of about 8.7 or so; therefore, you want to use an indicator that changes color as close to 8.7 as possible. (The average of 8.0 and 9.6 is 8.8) There is only one indicator in your list that comes close. Given a choice, I would choose phenolphthalein which has a range of about 8 to 10 for a mid point of ab out 9.
THANX. but did u guess 8.7 or did u calculate it? if so could u show howw???
I did not guess. You said a weak acid and NaOH so I picked acetic acid (which I will call HAc) with NaOH. I used 0.1M HAc and titrated with 0.1M NaOH. At the equivalence point we will have 0.05M NaAc which is hydrolyzed in the water solution.
...........Ac^= + HOH ==> HAc + OH^-
initial....0.05M...........x.....x
change.....-x..............x......x
equil....0.05-x..........x.......x
Kb for the Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05)
Ka is 1.8E-5, Kw is 1E-14 and I obtained 5.27E-6 = x = (OH^-). I converted that to pOH and stuck it into pH + pOH = pKw = 14 and pH = 8.72.
By the way, we're glad to help but I've answered quite a few questions for you gratis; it's time you start showing some of your work. After all, this is a home work HELP site and not some place for you to dump all of your home work. The idea here is for you to learn how to do these yourself.
...........Ac^= + HOH ==> HAc + OH^-
initial....0.05M...........x.....x
change.....-x..............x......x
equil....0.05-x..........x.......x
Kb for the Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05)
Ka is 1.8E-5, Kw is 1E-14 and I obtained 5.27E-6 = x = (OH^-). I converted that to pOH and stuck it into pH + pOH = pKw = 14 and pH = 8.72.
By the way, we're glad to help but I've answered quite a few questions for you gratis; it's time you start showing some of your work. After all, this is a home work HELP site and not some place for you to dump all of your home work. The idea here is for you to learn how to do these yourself.
thank u and next time i will show some work
1 answer