Question

Alizarin yellow R, \(Ka=7.9×10^{12}\), is yellow in its protonated form (HX) and red in its ionized form \((X^−)\). At what pH will alizarin yellow R be a perfect orange color?

I have no idea how to do this. I tried finding the pH of alizarin yellow, which is 11.1. Then what's next?

Answer 1

I think this one helps:

Answer 2

@Rushwr

Answer 3

According to the indicators' colour change the pH at perfect orange is around 4 right?

Answer 4

Isn't the "perfect orange" color the moment where Alizarin shifts from its protonated form to its ionized form?

Answer 5

\[HX \rightarrow H^+ X^-\]
\[K_a = \frac{[H^+][X^-]}{[X^-]}\]
\[K_a = [H^+]\]

Then solve for pH.

Answer 6

I'm not entirely sure, though, haha! Don't take my word for it.

Answer 7

That's not right !
\[K _{a} = \frac{ [H ^{+}] [X ^{-}] }{ [HX] }\]

Answer 8

So basically this is a weak acid

Answer 9

@owlet Are you given with the concentration ?

Answer 10

Hang on,
\[[HX] = [X^-]\]
because it's at equilibrium at the perfect orange point.

Answer 11

yeah that's right so Ka = H^+

Answer 12

pH = -log {H^+]

Answer 13

So where did I go wrong? o:

Answer 14

sorry late response. No, I'm not given any concentrations :

Answer 15

@mortonsalt Ur Ka expression is wrong check it again

Answer 16

You don't need it, really.
\[pH = pK_a = -\log(K_a)\]

Answer 17

Yeah since u gave it only I said it's not right

Answer 18

That's the expression for when I've already substituted the value in.

Answer 19

But yeah, anyway! :) Hope this helps.

Answer 20

I already did that, pH = -log(7.9x10^-12) = 11.1 , when it is on the yellow interval

but how about when it is already on orange?

Answer 21

should I subtract the pH from it from the perfect orange color?

Answer 22

No, that's not necessary. Look at it this way:
Since [H+] = Ka
Then pH = pKa

Answer 23

This is at the equilibrium point, like how we got it previously.

Answer 24

STEP 1
\[K_a = \frac{[H^+][X^-]}{[HX]}\]

STEP 2
Since we know that we're looking for the "perfect orange", this is the point where the two "colors" are at equilibrium.
Therefore we can say that
\[[HX] = [X^-]\]

We substitute the first equation to the second equation and we get
\[K_a = \frac{[H^+][X^-]}{[X^-]}\]

Are you okay so far?

Answer 25

@owlet do u have any way to check the answer?

Answer 26

STEP 3
This means that
\[K_a = [H^+]\]

From the equation above, we can say that
\[pH = pK_a\]

Answer 27

yeah, i guess. this may sound dumb, but how do you know that the perfect orange is where the colors are at equilibrium?

yeah. it is right 11.1 :)

I just wanted to understand this topic.

Answer 28

Red + yellow?

Answer 29

And it says so on the picture too that right at the middle when [HX] is equal with [X^-].

Answer 30

oh, I missed the red color ! that's why, okay thank you for the both of you :D

Answer 31

@owlet Here there's nothing to do . As we can see in the diagram u attached we can see the perfect orange is obtained when HX is approximately equal to X^- So that's how u get it .

Answer 32

Just think of it this way too:
It's a reversible process. The protonated ones turn into ionized ones and vice versa. At ONE POINT, there is an equal number of protonated ones to ionized ones.

Answer 33

This is the equilibrium point.

Answer 34

Therefore we cancel off HX to X^-