[Jenmf43]

We did an ion exchange experiment with EDTA and Mg2+.  I understand this part just fine.  My confusion is in our standardization we used NaOH with our unknown water sample to find the total acid in the unknown.  We did it on a 10-mL aliquot so I have the data calculated for that.  We need to then take that and put it on a 25-mL basis and that's where I'm lost!  I'm not sure how to do that.  Not sure if this is needed but for my first sample I  used 7.47 mL of 0.03145 M NaOH solution and found out there were 235 milimoles (mmol) H+ per 10-mL.  I need to find it per 25-mL but have no clue how. 

[Borek]

You mean - you know how much of something is in 10 mL, but you have no idea how to calculate amount in 25 mL of identical solution?

[Jenmf43]

Yes I think that's what we have to do.

[Borek]

Do you know what concentration is? Do you know what is relation between concentration, volume and amount of substance?

[Jenmf43]

I know what the concentration is of the NaOH solution we used, we had to find that first.  Then I said that the molarity of NaOH times the volume NaOH used to titrate was equal to the moles NaOH which equals the moles of acid.  That's how I found it for when we did it on based on 10 mL.  I'm not sure if that's what you are asking.  When we originally did the experiment we did it using 10mL samples of our unknown.  I don't see how we can say how much acid there would be if we had used 25 mL of the unknown because we don't have the data collected for that, I'm not sure how it relates.

[DrCMS]

Here's a clue 10x2.5 = 25

[Borek]

C = n/V

From what you write you have already calculated C, why can't you solve the equation for n and put known concentration and volume into it?

Or differently - you know how much substance is in 10 mL. You take two times 10 mL. How much substance do you have?

You know there is one cookie per box. You take two boxes. How many cookies do you have?

Now take 2.5 boxes (DrCMS posted while I was editing my answer)

[Jenmf43]

I did the C=n/V for the first step when I found the amount based on 10 mL of the sample.  I didn't divide by 10 though for volume, I divided by the volume it took to reach the equivalence point.  So I don't know how much it would take to reach the equivalence point if I had used 25 mL of the sample.  I think the idea to multiply by 2.5 will work though that makes sense to me!  Thank you for your *delete me*!