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

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[SOLVED] Amount of 6M H2SO4 to lower pH of pool from 7.0 to 4.0

Homework Statement



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?

Homework Equations



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

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!
 
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Nevermind, got it figured out :).
 
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?
 
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
 
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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 :)