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Homework Help: Amount of 6M H2SO4 to lower pH of pool from 7.0 to 4.0

  1. Feb 25, 2008 #1
    [SOLVED] Amount of 6M H2SO4 to lower pH of pool from 7.0 to 4.0

    1. The problem statement, all variables and given/known data

    A new swimming pool is filled with fresh, clean water with a pH of 7.0. The width is 25 ft, the length is 60 ft, and the depth linearly ranges from 3.5 to 9 ft. How much 6 M H2SO4 must be added to reduce the water pH to 4.0 so that the plaster liner in the pool can be cured?

    2. Relevant equations

    kw = ka*kb, pKa = -log(Ka) and for pOH and pH, any the usual concentration formulas.

    3. The attempt at a solution

    I'm honestly not entirely sure how to tackle this. I first started off by finding things I may need. I found the capacity of the pool to be 265,643 liters. The molecular weight of H2SO4 is 98.065 g/mol. Ka of H2SO4 is 10^3.

    If you guys could bump me in the right direction it'd be appreciated. Thanks!
     
    Last edited: Feb 25, 2008
  2. jcsd
  3. Feb 25, 2008 #2
    Nevermind, got it figured out :).
     
  4. Feb 25, 2008 #3
    Well since you figured it out, maybe you could tell me. I have not got to that section of the book yet in school, so I just read ahead. The way I saw it being done was this: Since you want a pH of 4, I computed Molar by using wanted pH 10^-4 = .0001 Molar. Then I took the equation M1V1=M2V2 solved for V1 so I got (.0001M * 265643L)/6M =4.43L
    I have a feeling that is not correct because H2SO4 is diprotic. What answer did you come up with?
     
  5. Feb 25, 2008 #4
    Well, if you want a pH of 4 then you will need a [H+] = 1 x 10^-4 M ([H+] = antilog (-pH)

    I assumed that the moles of H+ ion that will be present in the swimming pool after adding the acid will be the same as the number of moles of H+ provided by the acid. So I came up with an equation:

    moles of H+ from acid = moles of H+ in swimming pool after the acid is added

    H2SO4 is for all practical purposes completely ionized into the H+ ions and SO4-2 ions according to the equation:
    H2SO4 --> 2 H+ + SO4-2

    so the moles of H+ for acid = volume of acid added x molarity of the acid x 2 H+/acid molecule
    = x L (6mol/L)(2 H+/molecule)

    the moles of H+ in the pool = molarity of H+ in the pool x volume of the pool
    = (1 x 10^-4 M)( 2.66 x 105) L

    Plugging the values in for H+ yields:
    x L(6mol/L)(2 H+/molecule) = (1x 10^-4M)(265643L)

    I just solved for X and got 2.2L. I'm not 100% this is correct but it makes sense to me. It's basically a molarity balance thing. Moles of [H+] for the ace = moles of [H+] after it's in the water. I think what you did is correct for one stage of the dissociation, but not both, since the only difference is that I multiplied the 6M by two from the chemical equation above.

    Good luck and godspeed. Engineering majors like me need it! haha
     
    Last edited: Feb 25, 2008
  6. Feb 26, 2008 #5
    Yeah that makes sense. I think I messed up by not accounting for the H2, hence my number was 2x. Thanks for clarification. Funny you ask for help and end up giving it :)
     
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