[a_huynh00]

Two drops of indicator HIn (Ka=1.0x10^-9), where HIn is yellow and In- is blue, are placed in 100.0 mL of 0.100M HCl.

A. What is the color of the solution initially? - Answer is Yellow.

B. The solution is titrated with 0.10M NaOH. At what pH will the color change (yellow to greenish yellow) occur? - Answer is pH=8.0.

C. What color will the solution be after 200.9mL of NaOH has been added? - Answer is Blue.


Sooo I'm guessing
A. HIn => H+ + In-
    Ka = [H+][In-] / [HIn] = 1.0x10^-9
   ([H+] = (0.100L)(.100M) = 0.010 moles) <= extra stuff maybe?

    I figured that since the Ka is such a small amount that means the amount of reactants is larger than the amount
    of products so its Yellow.

B. HIn + OH- => H2O + In-
    [H+] = 0.100 M
    [OH-] = 0.100 M
    Ka = [In-] / [HIn][OH-]
    Ka = 1.0x10^-9 / 0.100 = 1.0e-8
    pH = -log(1.0e-8)
    pH = 8
    Did I come to it right?

C. HIn + OH- => H2O + In-
   [OH-] = (.2009L)(0.10M) = 2.009e-2
   Not sure what to do here but with the addition of more OH- according to La Chatelier the reaction goes to the right producing more In- thus blue

Where did I go wrong and how do I do C? Thanks

[Borek]

http://www.titrations.info/acid-base-titration-indicators

&

http://www.titrations.info/acid-base-titration-end-point-detection