How do I lower the pH of a water supply

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SUMMARY

The discussion focuses on lowering the pH of a water supply from 9.93 to 6.00 using a 30% HCl solution. The calculations indicate that approximately 0.6 liters of HCl would be required based on molarity assumptions. However, participants emphasize that this approach is flawed without understanding the water's buffering capacity and composition, particularly due to the presence of leachate from landfill waste. Accurate pH adjustment requires knowledge of the solution's alkalinity and other dissolved substances.

PREREQUISITES
  • Understanding of pH and its measurement
  • Knowledge of molarity and dilution calculations
  • Familiarity with acid-base reactions and buffering systems
  • Basic chemistry of hydrochloric acid (HCl) solutions
NEXT STEPS
  • Research the concept of buffering capacity in water chemistry
  • Learn about alkalinity testing methods for water samples
  • Explore the effects of landfill leachate on water quality
  • Study the principles of acid-base titration for precise pH adjustments
USEFUL FOR

Chemists, environmental scientists, water quality analysts, and anyone involved in water treatment processes will benefit from this discussion.

Biffinator87
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Moved from a technical forum, so homework template missing

Homework Statement


[/B]
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.

Homework 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

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.
 
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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).
 
Borek said:
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).
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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