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I've completed an acid base titration experiment of $\ce{HCl}$ and $\ce{NaOH}$. This forms salt + water

$$\ce{HCl + NaOH -> NaCl + H2O}$$

Indicator was used and we evaporated the water to get the salt.

I've been asked to design a test to show that water was the other product of this reaction. I'm not sure what to do here. I'm thinking maybe a distillation type test but not sure where to go with this question …

I'm a middle school student and would like an answer that is of a middle to high school level if possible. Could you provide some examples of tests I could do, and explain how they work?

EDIT -

I really appreciate the answers guys/gals. I'm not going to pretend like they did not go over my head though!

I think the issue might be the wording of the question in the textbook. What I've written above is exactly how it was asked in the textbook but I get the feeling that the expected answer is much simpler considering the level of chemistry I'm dealing with. Even though I don't expect you to read all of this I've gone ahead and scanned the book page anyway. Below is the experiment I completed, and question 4 in the discussion section which is highlighted is what I am trying to answer.

enter image description here

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  • $\begingroup$ I don't see how to do it as you described. By definition an "acid base titration" is done in water. You get just a tiny amount of extra water with the reaction. The only way that I could think to do it is to blow dry HCl gas over solid NaOH which is heated and trap the resulting H2O in a cold trap. Identification of H2O could be done via freezing point, melting point, and density (or instrumental techniques such as mass spec, IR, NMR). // The gist is that HCl would be blowing over NaOH in a "tube". $\endgroup$ – MaxW Oct 28 '15 at 6:18
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    $\begingroup$ The point is that you have a restriction "I've been asked to design a test to show that water was the other product of this reaction." In other words, the reaction that is happening in the solution. So doing the reaction with gaseous HCl leads to another problem. The "reactions" are not occurring under the same conditions. How do you show the reactions are the same? $\endgroup$ – MaxW Oct 28 '15 at 6:31
  • $\begingroup$ It appears you have created two accounts. Have a look at this link to obtain information what you can do. $\endgroup$ – Martin - マーチン Oct 28 '15 at 17:35
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As was written in comments, it is nigh impossible to detect the formation of more water in an aquaeous solution. Unless you coloured the oxygen atoms in $\ce{NaOH}$ pink or purple you would have no clue which water molecule is new and which isn’t.

Say you had a $1~\mathrm{M}$ solution of both acid and base and used $25~\mathrm{ml}$ each; you would create $25~\mathrm{mmol}~\ce{H2O}$ in the reaction or approximately $0.45~\mathrm{g}$. That’s just about $1~\%$ of the total water added, so you may be able to detect a slight increase in weight if you distilled away the water.

Do not try to get your hopes high, though.

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    $\begingroup$ I mostly agree, but there is a nasty wrinkle. How do you figure out the weight of the water in the HCl solution? For the NaOH solution I could at least start with distilled water and solid NaOH. $\endgroup$ – MaxW Oct 28 '15 at 8:40
  • $\begingroup$ @MaxW I assume a perfect solution containing $36.46~\mathrm{g}$ of $\ce{HCl}$ in $1~\mathrm{l}$ $\ce{H2O}$ and right there is another catch. $\endgroup$ – Jan Oct 28 '15 at 8:43

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