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I have 6 solutions. 3 of them are acids, and 3 of them are bases. The acids and the bases (HCl and NaOH respectively) both have solution with a concentration of $0.1M$, $1M$, and $10M$. I don't know which solution corresponds to which concentration or whether it is an acid or a base. The only tools I have to work with are a titration station and the knowledge that the acids all already contain Phenolphthalein, an indicator that turns purple in a basic solution.

What is the most optimal (least number of times having mixed the solutions with preferably the least number of titrations) way to go about finding which solution is which?

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  • $\begingroup$ are we supposed to assume Phenolphthalein is only like yes/no indicator? because its color also gives other information, such as you can find out the concentration of the mix as well. $\endgroup$
    – Oray
    Feb 24 '18 at 13:24
  • $\begingroup$ Yes, the only indication it gives is it isn't as strong with really weak bases. After a certain point, which it reaches quickly, it stops increasing the shade of purple. $\endgroup$
    – Peregrine Lennert
    Feb 24 '18 at 15:43
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    $\begingroup$ How do you define optimal? Least amount of solution used? Least number of titrations? Least time spent? Something else? $\endgroup$
    – Chris
    Feb 24 '18 at 19:06
  • $\begingroup$ @Chris Least number of times having mixed the solutions with preferably the least number of titrations $\endgroup$
    – Peregrine Lennert
    Feb 24 '18 at 19:47
  • $\begingroup$ So there is one 0.1M HCl, one 1M HCl, one 10M HCl, and one 0.1 NaOH, one 1M NaOH, and one 10M NaOH? $\endgroup$
    – mbjb
    Feb 25 '18 at 5:07
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I think the answer is

7(edition at the end) titrations.

Let's start by choosing any one of the solutions and taking a definite amount of it in 5 different flasks. Now, titrate this with the remaining 5 solutions and record the observations.

Now, if in any one of the titrations, the whole solution in the flask turns purple in the beginning, the chosen solution would be a base, otherwise an acid. In case it is a base, you get to know all acids and their strengths by comparing the amount of their solutions used to obtain a colourless solution and if it's an acid, you get to know the bases and their strengths.

Now, choose any of the known solution, take a definite amount of it in two different flasks and titrate any two of the three unknown ones. You know the concentration of the solution used and you can easily calculate the concentration of the 2 unknown ones. Now the remaining base/acid will have characteristics different than that of all the 5 known ones.

Edit(credit: Chris): After getting to know the characteristics of the first three solutions, you will get to know the concentration of the solution originally used for all 5 titrations. So, you just need one more titration. Final answer:

6 titrations in total

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  • $\begingroup$ The issue with this is that you don't take in to consideration the option of mixing solutions together without using the titration station $\endgroup$
    – Peregrine Lennert
    Feb 25 '18 at 13:04
  • $\begingroup$ After your five titrations you'll also know the concentration of the originally chosen solution since one of your titrations did a colour change after the same volume of the second solution was added. Your original solution is the same concentration as this one. This means you only need one more titration to find one of your two unknowns. $\endgroup$
    – Chris
    Feb 26 '18 at 11:35
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    $\begingroup$ @PeregrineLennert: That isn't an issue unless you can come up with a solution that makes use of that in which case make that an answer and we can vote accordingly. The fact that you are measured on mixings and this is 7 (or 6 after the comment that I've made) I think you would be hard pressed to do better if you mixed solutions together outside the titration station. $\endgroup$
    – Chris
    Feb 26 '18 at 11:42
  • $\begingroup$ I'm a bit confused by your recent edit... You seem to have reintroduced a bunch of spelling errors, taken your answer out of spoiler tags and removed the paragraphs that made the answer much more easy to read... puzzledPig's previous edit seemed much better than this. I'd consider rolling back to his version unless you have pressing reasons not to. I should note that you shouldn't feel bad about other people editing your posts - they are doing it to make them better! $\endgroup$
    – Chris
    Feb 28 '18 at 17:43
  • $\begingroup$ Oh i am so sorry, actually i myself wanted to correct just a single spelling mistake but it seems that i edited previous version of the answer as it was saved for me. Can I undo this? $\endgroup$
    – vaanchit kaul
    Feb 28 '18 at 17:49
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Phenolphthalein is only showing whether if the solution is basic or not and its color becomes darker purple when its ph value is around 14. if pH value is lower than 7, it will not change the color of the solution, so you cannot tell whether if its ph value is close to 0 or 6.

We also know that ph values of 0.1M NaoH and HCI are 13 and 1 respectively. ph value of 1M and 10M NaOH and HCI are 14 and 0 respectively. In other words, you cant tell anything about the mix if the concentration is over 1M.

Lastly, it is given that Phenolphthalein only exists in acidic solutions.

With the information above,

We will take 5 random small samples (let's say 1x amount from each) from 6 solutions and mix them and possible color results will be like the Table 1.

After that,

If the color does not change, it means the last solution $f$ which is not mixed is basic for sure, if its color is purple, it means $f$ is acidic. Then use $f$ with the given amount of sample in the table below on the right hand side, mixing it with each solution ($a,b,c,d,e$) by 1x. For example, if the color of the mix is purple, we will take 0.99x of $f$ and 1x of $a$ to test the color. It will take 6 titrations until now.

As a result, the color table will be like below:

enter image description here

From now on, we need to act for each type of result we get one by one if necessary for some cases. For example (first line and worst case condition):

First one result is no color or yellow in the table. That means $f$ is basic as you see. and there will be 3 purple after mixing it with $a,b,c,d$ and $e$. So we will know that $a,b,c$ are acidic, and $d,e$ is basic. Moreover, by using the table, we can easily conclude that $f$ is 10M NaOH! By using this information, we can determine each acidic solution Molars by adding like very small increment from 0.1x, 0.1000001x of 10M NaOH (which is practically like adding 1x of 1.000001M solution into each acid sample) to each acidic new taken samples and the result becomes like on the table below for acidic solutions and we will find out by color of each acidic solution molars (3 more titrations, actually it is 2 but cant know OP would be okay with that by knowing the solution by just color instead of color difference since one of them is such as 0.000001M and the other one is 0.9M, it will be very light purple for one of them, the other one will be dark purple so I would say 3 for now), lastly for two basics, it is very easy by using the acidic samples, just put 0.5x of 1M of HCI into a random basic solution. If its color changes to purple it is 1M NaOH and its color does not change, it is 0.1M NaoH and the other one will be known automatically. So it is 1 more titration)

Table 2

enter image description here

In total,

it requires $10$ titrations/mixing (9 if OP will allow color difference will be noticable by experimenter) to find each solution if 5 sample solution mixing result is colorless. Actually the second part where the color becomes purple after mixing 5 of them is the same procedure and simpler since we know $f$ is acidic and contains phenolphthalein, will take at most 9 titrations at most so I do not want to exemplify for each condition. I can if you ask.

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  • $\begingroup$ I dont understand the part when you are saying: f is acid and then f is the 10M basic solution. (In the example after table 1) $\endgroup$
    – Untitpoi
    Feb 27 '18 at 7:28
  • $\begingroup$ @Untitpoi there is only one condition where u can get 3 purple (t2,t3,t4) and 2 colorless solution (t5,t6) while the original mix(t1) is colorless as u see in the table. that results f is 10M. $\endgroup$
    – Oray
    Feb 27 '18 at 8:28
  • $\begingroup$ @Untitpoi i dont see where i say f is acid and 10M. $\endgroup$
    – Oray
    Feb 27 '18 at 8:35
  • $\begingroup$ Yeah you are right... must have misread sry! Your answer is valid, even though I think what Asker wants is unclear. $\endgroup$
    – Untitpoi
    Feb 27 '18 at 8:59
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I have a two-steps answer that can be done while mixing:

once the solutions and between 1 to 5 titrations.

In order to be clearer I will call solutions S1 to S6 (no link between concentrations and these names)

STEP1

Identification of the 10M concentration solution.
Start with a volume of 10mL of S1. Add with the titration station 20mL of S2, then 30mL of S3, 40 mL of S4, 50mL of S5 and finally 60mL of S6. Record results. You had at least 1 change of color. Take the last change of color and the volume needed to observe this change.
We know for sure the solution that made that change is a 10M solution. basic if it changes to purple and acidic otherwise. As it was the last change in color, the solution will stay in the state (acid/basic) of this 10M solution. The volume needed to get that last change of color will indicate you which one is the 10M solution of the other state.

Let's make some example:

You get your last change of colors after adding S3 (it went from transparent to purple) after adding 8mL. So, you know S3 is the basic 10M. and you can also understand that S1 was the 10M acid (you may also notice S2 is 1M basic solution but that's not important)

STEP2

Identifying all solutions with one titration
You have now your 210mL solution which is either acid or basic (see the color), you also have identified 10M acid and basic solutions thanks to step1. Titrate the 210mL solution with your 10M solution state contrary to the 210mL (i.e acid if is purple and basic if it is transparent). Get the volume. with that, You must be able to deduce every single solutions concentrations and state.

Example for step2

Let's continue with our example before.
The 210 mL solution was purple, so I titraded that with my S1 solution and got a 18.1mL result. There is 3 unknown solutions so I know there is 6 possibilities: S4,S5 and S6 being 1M acid, 0.1M acid and 0.1M basic. Depending on which is which I could have get 15.9; 16.1; 16.8; 17.1; 17.2; 18.1mL of titration. I am sure the order is then S4 = 1M acid / S5 = 0.1M acid and S6 = 0.1M basic

NOTES

if the last change of colors occured when you put S6, you can know every solutions. But if you have changes of colors every time you put a solution (in case of order 0.1M acid/ 0.1M basic / 1M acid/ 1M basic /10M acid / 10M basic). You will need to record 5 titration.
At the end you have mixed solutions only once (or i didn't get the problem).
Be aware that if the solution turn purple after first drop of the next solution, it means that we only had basic solutions before. We don't really care of that change of color, as there will be another later for sure.

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  • $\begingroup$ How can you be sure that when you add a solution if it is the acid or a base? Both are clear, at least until the other is added, so if you start with a basic solution and add an acid, it will still turn purple. $\endgroup$
    – Peregrine Lennert
    Feb 26 '18 at 11:02
  • $\begingroup$ Let's say the solution turned purple after adding S2. There is a difference between turning purple at the first drop of S2 and turning purple after 5mL of S2. first case we have S2 is acid and S1 basic, second case it means S2 is basic and S1 was acid. I also don't consider the first case (purple after first drop) being a change in color. $\endgroup$
    – Untitpoi
    Feb 26 '18 at 12:32

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