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Topic: Final volume of the titrated solution  (Read 1717 times)

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Offline cezarrrr

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Final volume of the titrated solution
« on: January 16, 2013, 03:13:28 PM »
Hello.
Last month, I was taking a selective test in which there was a question involving titration. The question itself is very common. It said that, in order to discover the concetration of a 10 mL solution of Mg(OH)2, the mixture was titrated with 12.5 mL of a HCl solution 0,5 mol/L. Then, it required the concentration of the solution of Mg(OH)2. However, I became very confused due to this specific excerpt of the problem: "The concentration of a sample of 10 mL of Mg(OH)2 that was titrated with 12,5 mL..." Well, maybe I'm too caught up in details, but I noticed that they'd used a restrictive clause in there, as if they wanted to know the concentration of Mg(OH)2 AFTER the titration. Well, the difference is that, in my accounts, instead of using 10 mL as the volume, I used 22.5 mL (10 mL + 12.5 mL of the HCl solution). That led me to the wrong answer, as I should've used 10 mL, apparently, as the volume.
Well, now I don't really know if it was their error or carelessness or if, because the mixtures react with each other, the volume really doesn't change at all. I mean, both of the solutions have water in them, they're not only HCl and Mg(OH)2, so I don't see how the volume won't change, especially in those low concentrations of solutes. I'd like to know, then, if it was my mistake and if there is any way to determine the final volume of the mixture of titrant + titrate in order to determine the final concentration in the mixed samples of one of the components.
Thanks in advance  :)

Offline Borek

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Re: Final volume of the titrated solution
« Reply #1 on: January 16, 2013, 03:55:19 PM »
After the titration there is no Mg(OH)2 in the solution, don't you think?

Final volume is around 22.5 mL, not exactly that amount, as volumes are not additive.
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Offline cezarrrr

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Re: Final volume of the titrated solution
« Reply #2 on: January 16, 2013, 04:31:03 PM »
Oh, yes, that's true, my mistake, I meant to say MgCl2.
Well, considering that Mg(OH)2 + 2 HCl = MgCl2 + 2 H2O, and that we had 0,00625 mol of HCl, therefore, we had, initially, 0,003125 mol of Mg(OH)2. When it was titrated, it was transformed into 0,003125 mol of MgCl2 per 22,5 mL. Therefore, the concentration of the solution after being titrated would be of 0,139, approximately. The most proximate answer was 0,1, so that's what I marked. However, the "official" answer is 0,3, which is the concentration of the solution of Mg(OH)2 before being titrated. Well, since the volume does change, and if everything is right until here, I guess I'm going to consider asking for a revision of the question. I can't believe it's all happened because of a comma!  :o
Thanks!

Offline cezarrrr

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Re: Final volume of the titrated solution
« Reply #3 on: January 16, 2013, 04:34:07 PM »
Oh, I must point out that it was never clear or indicated that they wanted the concentration of Mg(OH)2 specifically. They only asked for "the concentration of a 10 mL sample of Mg(OH)2 that was titrated with 12,5 mL of HCl 0,5 mol/L", in a restrictive manner, although it must have gone unnoticed.

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