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Homework Help: Cocentration and molarity

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    What volume of water must be added to 45.0 mL of a 1.0 M solution of H2SO4 in order to create a 0.33 M H2SO4 solution?
    answer:91.4 mL

    vat contains 2.24 M hydrochloric acid solution. How many kg of Ca(OH)2 will be required to react completely (neutralize) 796 L of the solution?

    2. Relevant equations


    3. The attempt at a solution

    for the first one, i tried using mv=mv, but noticed that it wouldnt make sense jsut plugging in the numbers directly b/c of the problem's wording

    for the second one, i'm not sure how to start it
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    In regards to your first question: what did you determine to be the final volume of the 0.33 M H2SO4 solution?

    A good starting point for the second question is this: figure out how many moles of HCl you have in the vat then determine how many moles of Ca(OH)2 will be required to neutralize them.
  4. Mar 9, 2008 #3
    for the first one:
    ahh! i see now, subtract 45 from it

    for the second one

    M = moles/volume
    moles = 2.24*796=1783.04mol
    mols*(g/mol) = g = 1783.04/74 = 24g
    dont get the right answer..not sure what i did wrong there
    Last edited: Mar 9, 2008
  5. Mar 9, 2008 #4
    I agree with your math used to determine the total mass of 1783.04 moles of Ca(OH)2:

    moles Ca(OH)2 * g/moles = g of Ca(OH)2

    but your calculations shown are off.

    But this mass of base will not lead you to the right answer. Go back to the neutralization reaction (this should have been the first thing you did) and see if this gives you some ideas. I believe that the problem assumes something that may not be obvious. As a first hint, look at the acid dissociation constants (pKa values) of each proton in H2SO4. This may point you to a simplifying assumption (accurate or not) that will lead you to the given answer.
    Last edited: Mar 9, 2008
  6. Mar 9, 2008 #5
    hmm, i'm not too sure what the acid dissociation constants are (as in: we havent studied that yet(?)) but something i noticed between what i came up with and the answer is is that 66/24 = 2.75. hopefully thats significant...not sure where that comes from though
    Last edited: Mar 9, 2008
  7. Mar 9, 2008 #6
    The first mistake your a making is only a computational one. Try going back to your equation mol * g/mol = g and running your numbers again. I agree that there are 1783.04 moles of HCl in the vat and that the molecular weight of Ca(OH)2 is 74 g/mol.

    The second mistake will be easier to find if you write out a balanced chemical reaction for the neutralization. This will take the form of aA + bB → cS + dH20 where a, b, c, and d are integer values, A is the acid, B is the base, and S is the salt (byproduct).
  8. Mar 9, 2008 #7
    ah ok
    first mistake was a dumb one
    1783.04*74=131944.96 g Ca(OH)2

    then for the second part, 131944.96 g is 131.944 kg
    from the equation, 1 Ca(OH)2 : 2 HCl (...i guess i also could have done this part earlier when i had mols HCl)
    132/2 = 66kg

    thanks for your help!
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