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Chem lab - titration

  1. Oct 31, 2007 #1
    We recently did an experiment in chem lab with titrations and we are supposed to find out the molarities of some unknown solutions. I thought that I was doing my calculations right, but the checking program that I am using keeps telling me that I am wrong.
    We made a standard solution that was 0.09769 M and was 250 mL. We used 25mL aliquots of this solution to titrate against an unknown concentration of NaOH. The mean volume of the base that was used was 23.10 mL. I thought, since our acid was KHP with one titratable proton, that I could just use the equation M1V1=M2V2 like so: (0.09769M)(25.00mL)=M2(23.10mL)...can you please explain where my thinking where wrong?
  2. jcsd
  3. Oct 31, 2007 #2


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    You did nothing wrong. Perhaps the problem is with the checking program.
  4. Nov 3, 2007 #3
    I think the correct/complete formula is :

    (M1)(V1)/N1 = (M2)(V2)/N2

    M1, M2 = Molarity of the acid and the base
    V1, V2 = Volume of the acid used and the mean volume of the base (Na OH)
    N1, N2 = No. of moles of acid and base

    Your equation is perfectly correct, but only N1 and N2 are missing.

    N1 and N2 can be found out by writing the balanced equation of the Titration.

    I hope i have helped. If i am wrong please correct me.
  5. Nov 3, 2007 #4


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    How many moles of acid = (0.09769 moles/liter)*(0.025 liter)

    How many moles of NaOH = (M)*(0.02310 liter)

    The moles of acid equals the moles of NaOH. You have the information to find M in moles/liter. Very simple algebra.
  6. Nov 8, 2007 #5


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    Actually, (M1)(V1) = N1 if M1 is in moles/liter and V1 is in liters. Thus the expression (M1)(V1)/N1 is equal to 1. The same goes for M2, V2 and N2. The resulting expression (my favorite, 1=1) is always true.
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