chemistry

how do i find ratio of volumeHcl/volumeNaoh for my first trail of titration?

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asked by Riana
  1. It would be easier if I saw the numbers; however, usually, if it's just the ratio you want, its mL HCl/mL NaOH. Be careful with your caps (or no caps) for in chemistry there are so many different meanings. For example, Co is cobalt, CO is carbon monoxide, co is the abbreviation for company.

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  2. volume of HCl is 10.0ml. initail volume of NaOH is 1.0ml and final is 21.0ml. this is for the first trial only.

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    posted by Riana
  3. OK. Then it is 10.0/(21.0-1.0) = ?
    Do you know what you should calculate this? (other than because that's in the procedure)

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  4. i have to find concentration of NaOH and this is one of the requirement.the answer for the first trial is 0.5
    and should there be a unit with it?

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    posted by Riana
  5. No unit for the units cancel. mL/mL = 1/1.
    Yes, you need the ratio to calculate molarity; however, you can do that without doing the ratio first. One reason to do it now is that the second trial, third trial, etc. will give you a quick and dirty way to see how good your precision is. The ratio should be the same, within the limits of how careful you are in performing each titration.

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  6. 2nd and 3rd titration ratio are the same. our teacher said if you got two same your good. and the number of moles of NaOH used will be same for all trails ?

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    posted by Riana
  7. Your teacher is right on both accounts.

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  8. ok i have a question. if i take the average of NaOh concentration then that's the concentration i was looking for right and then i use the same concentration oh NaOH with unknown concentration of H2SO4 to find its concentration ?

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    posted by Riana
  9. All of that is true assuming that the trials you average are within the acceptable limits of the precision you need. Otherwise you practice more titrations until the precision is good enough, then make two or more (I always did triplicates but some call that overkill--I call it making sure) determinations for the M of the NaOH and proceed with the H2SO4.

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