1
$\begingroup$

The equivalence point of an acid-base titration can be determined by choosing the appropriate colored indicator. This indicator is a weak acid $\ce{HA}$. The undissociated acid has a different color than it's ion $\ce{A-}$. The $\mathrm{p}K_\mathrm{a}$ value of a colored indicator is given to be $\mathrm{p}K_\mathrm{a}=6$. At what value of the $\mathrm{pH}$ will the color change? Please explain your answer. (Assume that both colors are equally well recognizable).

I've had this question on an exam and I can't seem to figure out what exactly the answer could be.

I would intuitively say that the color change happens at $\mathrm{pH}=6$. If the $\mathrm{pH}$ is lower than 6 then there are less $\ce{A-}$ ions and more $\ce{HA}$ (acid-indicator) in the solution. Hence the solution will take on the color of the undissociated acid-indicator. If the pH is higher than 6 then the $\ce{HA}$ starts to dissociate more to $\ce{H3O+}$ hence there being a higher $\ce{A-}$ concentration and the solution takes the color of the indicator's conjugate base. But I'm not sure if my thought process is right or how to 'prove' it's right.

|improve this question
$\endgroup$
  • $\begingroup$ What ideas do you have? This is a homework (meta.chemistry.stackexchange.com/a/142/6998 ) question, so you should present your current train of thoughts $\endgroup$ – L3ul Mar 28 '16 at 12:03
  • $\begingroup$ Thanks for the answer! Well, I would intuitively say that the color change happens at pH=6. If the pH is lower than 6 then there are less A(-) ions and more HA(acid-indicator) in the solution. Hence the solution will take on the color of the undissociated acid-indicator. If the pH is higher than 6 then the HA starts to dissociate more to H3O+ hence there being a higher A(-) concentration and the solution takes the color of the indicator's conjugate base. But I'm not sure if my thought process is right or how to 'prove' it's right. Thanks again! $\endgroup$ – Michael Mar 28 '16 at 12:44
  • $\begingroup$ @Michael please edit your question and include your thoughts. Voting to leave open $\endgroup$ – M.A.R. ಠ_ಠ Mar 28 '16 at 13:08
  • $\begingroup$ I second @IͶΔ, but I just did that for you. Unfortunately I have no time to answer this, so I give you some hints. You are on the right track. Assume you need about a 1:10 ratio of colour1 to colour2 to actually see a change. Try using the Henderson-Hasselbalch equation to explain it. The answers differ when you increase or decrease the pH. $\endgroup$ – Martin - マーチン♦ Mar 28 '16 at 13:17
  • $\begingroup$ I'll throw this in. Usually the indicator is easier to monitor one way than the other. For example with phenolphthalein it is easier to watch the pink color appear than disappear. It certainly is impossible to judge "maximum pinkness." So you generally titrate with a base when using phenolphthalein as the indicator. $\endgroup$ – MaxW Mar 29 '16 at 5:32
1
$\begingroup$

Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:

The indicator reacts with water according to the following reaction equation: $$\ce{HA + H2O <=> H3O+ + A-}$$ Then using the Henderson-Hasselbach equation, I arrive at the following point: $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$ Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence: $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$ $K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).

Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color.

Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.

|improve this answer
$\endgroup$
  • $\begingroup$ No... you didn't take Martin's hint. Either [HA] = 10*[A-] --or-- 10*[HA] = [A-]. If you've every done a titration with phenolphthalein, then the end point is when the first tinge of pink appears. [HA] = 10*[A-] which is on the acid side of the pKa. $\endgroup$ – MaxW Mar 29 '16 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.