Chemistry lab question on dilutions in titration

Hi everyone,

We had a lab this week where we learned how to identify an acid by titration. However, I'm having a little trouble with some calculations.
So basically we had 0.1g of an unknown solid that we dissolved in 100ml. Then we took 25ml of this solution to which we added NaOH for the titration.
So after finding the equivalence point and all, I had to find the concentration of the solution at that point. Since at equivalent point conccnetration of [ NaOH ] = [Unknown] then we can do C1V1=C2V2 in order to find the concentration of NaOH after adding it to the unknown (which is also the concentration of the unknown).
However, I'm not sure what volumes I must take...
V1 = volume of NaOH that was added (in this case 7.5ml) and the total volume is (25ml +7.5).
Afterwards, in order to determine the number of moles in the unknown, we used M = n/V. But here is my problem, would M be the concentration that was determined previously ( concentration of NaOH), or would it be something different? Also, would the V be 25+7.5ml, or would it be 100ml??

Thanks in advance for any help.

Reading through your question, it seems the only information you have about the titration is …

mass of acid consumed = 0.025 g
volume of acid solution = 25 mL
volume of basic solution = 7.5 mL

I’m afraid, if that’s all you have, it’s futile to continue. You can’t possibly find the identity of the unknown - or, indeed, anything useful - from that information .

Are you sure you weren’t told the concentration of the NaOH solution? If you knew that, you could proceed.
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WHAT FOLLOWS IS JUST AN ILLUSTRATION TO SHOW HOW IT WOULD BE DONE IF YOU KNEW THE NaOH CONCENTRATION.

Let’s say, for the sake of argument, that the concentration of the NaOH solution was 0.0500 mol/L. And let’s also assume that the acid was a monoprotic acid, which we can write as HX. The reaction would then be …