owlet
  • owlet
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?
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
owlet
  • owlet
I think this one helps:
1 Attachment
owlet
  • owlet
@Rushwr
Rushwr
  • Rushwr
According to the indicators' colour change the pH at perfect orange is around 4 right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mortonsalt
  • mortonsalt
Isn't the "perfect orange" color the moment where Alizarin shifts from its protonated form to its ionized form?
mortonsalt
  • mortonsalt
\[HX \rightarrow H^+ X^-\] \[K_a = \frac{[H^+][X^-]}{[X^-]}\] \[K_a = [H^+]\] Then solve for pH.
mortonsalt
  • mortonsalt
I'm not entirely sure, though, haha! Don't take my word for it.
Rushwr
  • Rushwr
That's not right ! \[K _{a} = \frac{ [H ^{+}] [X ^{-}] }{ [HX] }\]
Rushwr
  • Rushwr
So basically this is a weak acid
Rushwr
  • Rushwr
@owlet Are you given with the concentration ?
mortonsalt
  • mortonsalt
Hang on, \[[HX] = [X^-]\] because it's at equilibrium at the perfect orange point.
Rushwr
  • Rushwr
yeah that's right so Ka = H^+
Rushwr
  • Rushwr
pH = -log {H^+]
mortonsalt
  • mortonsalt
So where did I go wrong? o:
owlet
  • owlet
sorry late response. No, I'm not given any concentrations :
Rushwr
  • Rushwr
@mortonsalt Ur Ka expression is wrong check it again
mortonsalt
  • mortonsalt
You don't need it, really. \[pH = pK_a = -\log(K_a)\]
Rushwr
  • Rushwr
Yeah since u gave it only I said it's not right
mortonsalt
  • mortonsalt
That's the expression for when I've already substituted the value in.
mortonsalt
  • mortonsalt
But yeah, anyway! :) Hope this helps.
owlet
  • owlet
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?
owlet
  • owlet
should I subtract the pH from it from the perfect orange color?
mortonsalt
  • mortonsalt
No, that's not necessary. Look at it this way: Since [H+] = Ka Then pH = pKa
mortonsalt
  • mortonsalt
This is at the equilibrium point, like how we got it previously.
mortonsalt
  • mortonsalt
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?
Rushwr
  • Rushwr
@owlet do u have any way to check the answer?
mortonsalt
  • mortonsalt
STEP 3 This means that \[K_a = [H^+]\] From the equation above, we can say that \[pH = pK_a\]
owlet
  • owlet
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.
mortonsalt
  • mortonsalt
Red + yellow?
mortonsalt
  • mortonsalt
And it says so on the picture too that right at the middle when [HX] is equal with [X^-].
owlet
  • owlet
oh, I missed the red color ! that's why, okay thank you for the both of you :D
Rushwr
  • Rushwr
@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 .
mortonsalt
  • mortonsalt
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.
mortonsalt
  • mortonsalt
This is the equilibrium point.
Rushwr
  • Rushwr
|dw:1446915614764:dw| Therefore we cancel off HX to X^-
Rushwr
  • Rushwr
|dw:1446915733788:dw| Therefore Ka = H^+
Rushwr
  • Rushwr
I hope u got it :)
mortonsalt
  • mortonsalt
Formulas are generally easy to understand. Do you understand it conceptually though?
mortonsalt
  • mortonsalt
The perfect orange is when there is an equal amount of red and an equal amount of yellow. Otherwise, if there's too much or too little of one, it's no longer a perfect orange.
owlet
  • owlet
yes! it makes sense now :D I wish i'm smart like you guys. Thank you so much. From that, the next question is what is the color if pH=13, then from what I understand, it should be red.. which is right. This means I really understand it haha thanks again :P
Rushwr
  • Rushwr
np :) LOL U are smart !

Looking for something else?

Not the answer you are looking for? Search for more explanations.