Question #765e0

1 Answer
Apr 30, 2017

Answer:

Here's my explanation.

Explanation:

How to choose an indicator

You want to choose an indicator that changes colour at the equivalence point of the titration.

If the equivalence point is at pH 2, you want an indicator that changes colour at pH 2.

An indicator is just a weak acid in which the acid has a different colour from its salt.

For thymol blue, we might write

#underbrace("HTB")_color(red)("red") + "H"_2"O" ⇌ "H"_3"O"^"+" + underbrace("TB"^"-")_color(orange)("yellow") ; K_text(a) = 0.020#

In very acidic solution, the indicator is in the red form, while in a more basic solution the indicator is in the yellow form.

We can write

#K_text(a) = (["H"_3"O"^"+"]["TB"^"-"])/(["HTB"]) = ["H"_3"O"^"+"] × ([color(orange)("yellow")])/([color(red)("red")])#

or

(1) #["H"_3"O"^"+"] = K_text(a) × ([color(red)("red")])/([color(orange)("yellow")])#

Just as #"pH" = "-"log["H"_3"O"^"+"]#, #"p"K_text(a) = "-"logK_text(a)#

We can take the negative logarithm of both sides of equation (1).

#"-log"["H"_3"O"^"+"] = "-log"K_text(a) -log(([color(red)("red")])/([color(orange)("yellow")]))#

#"pH" = "p"K_text(a) + log(([color(orange)("yellow")])/([color(red)("red")]))#

The indicator will change colour when there are equal amounts of the yellow and red forms, i, e, #([color(orange)("yellow")])/([color(red)("red")]) = 1#.

Then,

#"pH" = "p"K_text(a) + log1 = "p"K_text(a) + 0 = "p"K_text(a)#

That is, the indicator changes colour when #"pH" = "p"K_text(a)#.

For thymol blue, #"p"K_text(a) = "-log"0.020 = 1.7#

Thus, thymol blue is a great indicator for a titration with an equivalence point at #"pH ≈ 2"#.