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Titration Calculations

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    2.0g of pure undiluted HsSO4 is cautiously added to water and the solution diluted to 500 cm3. What volume of 0.1M potassium hydroxide, KOH, would exactly neutralise 25cm3 of the acid solution?

    2. Relevant equations
    concentration = number of moles/volume
    mass in grams = number of moles*relative formula mass



    3. The attempt at a solution

    H2SO4
    2.0g/500cm3
    Divide by 20 to get values for 25cm3
    0.1g/25cm3

    Conversion of grams to M
    mass in grams = number of moles*relative formula mass
    0.1g = number of moles*98
    0.1g/98 = number of moles = 1.02moles

    We now have

    1.02M/25cm3

    Let us work out the concentration...

    25/100 * 1.02 = 0.0255molar

    Now let us work out the volume

    Concentration = number of moles/volume
    Concentration*volume = number of moles
    volume = number of moles/concentration
    volume = 1.02/0.0255
    volume = 40cm3

    Is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 2, 2009 #2
    After you calculate the moles of H2SO4, I think you have problems. 1.02M/25cm3 makes no sense with molarity over volume, but it should be 1.02 moles/25 cm3. You also don't use the molarity of the KOH.
    Write a balanced equation first.

    You should start this way, which has the moles you calculated.

    [tex]25 cm^3 solution \left(\frac{1.02 mol H_2SO_4} {500 cm^3 solution}\right)[/tex]

    You have moles of the acid, then multiply it by conversion factors using the balanced equation and the molarity of the base, and you will get the volume of the base.
     
  4. Jun 2, 2009 #3
    Can you show me the complete process?
     
  5. Jun 2, 2009 #4
    Write the balanced equation first. :smile:
     
  6. Jun 2, 2009 #5
    2koh + h2so4 --> k2so4 + 2h2o
     
  7. Jun 2, 2009 #6
    It is a 2:1 ratio.
     
  8. Jun 2, 2009 #7
    If H2SO4 is 1.02 then KOH is 2.04moles
     
  9. Jun 2, 2009 #8
    volume = concentration/number of moles
    volume = 0.1M/2.04 (KOH) * 1000
    volume = 49?
     
  10. Jun 2, 2009 #9
    So you know you need twice as much KOH as H2SO4 given from
    [tex]
    25 cm^3 solution \left(\frac{1.02 mol H_2SO_4} {500 cm^3 solution}\right)
    [/tex]

    which gives you moles of KOH. The molarity of the base is 0.1M KOH or [tex]\frac {0.1 mol KOH} {1 L KOH}[/tex]. How can you use this fraction with your moles of KOH from above to give you the volume of KOH solution?

    Edit: Looks like you don't need these steps. Read the next post.
     
  11. Jun 2, 2009 #10
    1.02 is the moles of H2SO4 in the 500 cm3 of acid solution, but you need to know how many moles are in 25 cm3 of solution since that is how much is being neutralized by the base. You need to calculate this from what I wrote above. That will give you the actual number of moles of acid you're working with in 25 cm3 that's being neutralized. Then the 2:1 ratio you said is correct and you should get the correct answer.
     
  12. Jun 2, 2009 #11
    0.051 moles in 25cm^3. Therefore 0.102 moles for KOH.

    Revised volume calculation

    V = 0.1/0.102
     
  13. Jun 2, 2009 #12
    Yes, that looks right.
     
  14. Jun 2, 2009 #13
    I need to multiply the answer by 1000 right?
     
  15. Jun 2, 2009 #14
    That would give you mL of solution, but the question didn't say in what units to give the volume. You should be okay with leaving it in L unless you're expected to use a certain unit when the question doesn't specify.
     
  16. Jun 2, 2009 #15

    Borek

    User Avatar

    Staff: Mentor

  17. Jun 2, 2009 #16
    You do have a mistake here. 0.1/98 is .00102, so that is the actual number of moles of H2SO4. This is also wrong

    Concentration = moles/volume, so volume = moles/concentration

    I hadn't checked your work thoroughly before, but I think these are all the mistakes you had.
     
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