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Acid-Base Titrations - Soda Lime and Hydrochloric Acid

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    "Soda lime is 85.0% NaOH and 15.0% CaO. What volume of 0.500M Hydrochloric Acid is required to completely neutralise 2.50g of the Soda Lime?"
    Hint: Consider the reactions separately


    2. Relevant equations

    Formation of Soda Lime:
    CO2 + H2O >> H2CO3
    H2CO3 + 2NaOH >> Na2CO3 + heat
    Na2CO3 + Ca(OH)2 >> CaCO3 + 2NaOH

    number of moles = mass / relative formula mass
    volume = number of moles divided by concentration


    3. The attempt at a solution

    1)Find the number of moles of Soda Lime (from the equations describing its formation)
    2)Deduce the number of moles of HCL from the equation describing the titration
    3)Work out volume of HCL using number of moles/0.500

    4. Main problems encountered

    *I don't know the chemical formula for soda lime/I cannot deduce its relative formula mass
    *Because of the above I cannot write a balanced equation for the titration

    Guidance is greatly appreciated.

    Regards,
    Adam
     
  2. jcsd
  3. Oct 24, 2009 #2

    Borek

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    Staff: Mentor

    Forget about soda lime and its production, it doesn't matter. You have 2.5 g of a mixture of NaOH and CaO. How much NaOH in the mixture? How much acid required for titration of NaOH? What about CaO?
     
    Last edited by a moderator: Aug 13, 2013
  4. Oct 24, 2009 #3
    Since there 85% of NaOH, the mass of NaOH in the 2.5g mixture is:
    (85/100)*2.5 = 2.12500

    Therefore the mass of CaO in the mixture is:
    2.5 - 2.12500 = 0.375g

    The titration of NaOH:
    NaOH + HCL >> NaCl + H20
    number of moles of NaOH = m/Mr = 2.12500/40 = 0.053125 mol
    From equation number of moles of HCL = 0.053125 mol
    Therefore volume(HCL) = n/c = (0.053125/0.500)*1000 = 106.25 cm3

    Is this first value for volume correct?
     
  5. Oct 24, 2009 #4

    Borek

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    Staff: Mentor

    Looks OK.
     
    Last edited by a moderator: Aug 13, 2013
  6. Oct 24, 2009 #5
    Continuing with the titration of CaO

    CaO + 2HCL >> CaCl2 + H2O

    number of moles = m/mr = 0.375/56 = 0.00669642857
    From equation, number of moles (HCL) = 2*0.00669642857 = 0.013938571
    Therefore:
    V = n/c = (0.0134/0.500)*1000 = 26.8cm3
     
    Last edited: Oct 24, 2009
  7. Oct 24, 2009 #6
    Adding the two calculated volumes gives
    total volume of HCL required = 26.8 + 106.25 = 133.05cm3 or 133cm3 (to 3 sig fig0

    Borek: Please confirm :)
     
    Last edited: Oct 24, 2009
  8. Oct 24, 2009 #7

    Borek

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    Staff: Mentor

    Approach is correct, but you have eaten one digit, so the final result is off.
     
    Last edited by a moderator: Aug 13, 2013
  9. Oct 24, 2009 #8
    It should be: 0.0133928571
    Which when rounded to 3 sig fig is 0.0134

    NB: See adjusted Calculation and answer
     
    Last edited by a moderator: Aug 13, 2013
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