1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Amount of 6M H2SO4 to lower pH of pool from 7.0 to 4.0
  1. Finding ph from Ka (Replies: 2)

Loading...