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How do I lower the pH of a water supply

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  1. Mar 24, 2016 #1
    • Moved from a technical forum, so homework template missing
    1. The problem statement, all variables and given/known data

    Just a general question to see if i'm doing this right. The volume just doesn't seem right. Anyways we have a water supply of 1.6 million gallons with a pH of 9.93. We want to get the water to a pH of 6 using a 30% HCl solution.


    2. Relevant equations

    M1V1+M2V2=M3V3

    M1=9.93pH or 10-9.93
    V1=1.6 million gallons or 6056656 liters
    M2=pH of 30% HCl or molarity of 10.2
    V2=Volume of HCl (What I want to find)
    M3=6.00pH or 10-6
    V3=V1+v2

    3. The attempt at a solution

    I also converted into liters. I assumed a molarity of 30% HCl to be around 10.2 and I assumed that V3=V1+V2

    (1.17486x10-10)(6056656 liters)+(10.2)(V2)=(1.0x10-6)(6056656+V2)

    Manipulating the equation to solve I ended up with a volume of approximately 0.6 liters of HCl needed to reduce the pH. It just seems like a very small amount of acid in order to handle that much water.


    ANY help at all is greatly appreciated.
     
    Last edited: Mar 24, 2016
  2. jcsd
  3. Mar 24, 2016 #2

    Borek

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    Staff: Mentor

    Looking just at dilutions - your final volume is 6×107 L, expected H+ concentration is of the 10-7 order, so the volume of around 1 L doesn't sound off.

    But I am afraid it is completely off. It is not just about dilution. You can't calculate how much acid is needed not knowing WHY the pH is not neutral. Acid added has to react with the base present, and most likely it has to overcome a buffering effect of whatever is dissolved in the water. No way to deal with the problem without knowing the exact composition of the solution (or some combined parameter like alkalinity of the solution).
     
  4. Mar 24, 2016 #3
    I was afraid of that. This is a leachate pond example so there is all kinds of stuff flowing into the pond from landfill waste. So there could be any number of things going in there. Thank you for the help!
     
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