Optimizing Pool Water pH: Calculating Volume of Acid Needed for Perfect Balance

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SUMMARY

The discussion focuses on calculating the volume of hydrochloric acid (HCl) needed to lower the pH of a newly built swimming pool from 7 to 4. The pool dimensions are 7.5m x 5m x 1.5m, resulting in a total volume of 52.5m³ or 52,500 liters. The acid concentration is 33.0g/100ml, which translates to a molarity of approximately 9.052 moles per liter. The final calculation involves using the formula [H3O+] = Vacid(Macid)/(Vi + Vacid) to determine the required volume of acid.

PREREQUISITES
  • Understanding of pH and hydronium ion concentration
  • Basic knowledge of molarity and molar mass calculations
  • Familiarity with algebraic manipulation for solving equations
  • Knowledge of acid-base chemistry, specifically HCl behavior in water
NEXT STEPS
  • Learn about the relationship between pH and hydronium ion concentration
  • Study the concept of molarity and how to calculate it from mass and volume
  • Explore algebraic techniques for manipulating chemical equations
  • Research common methods for adjusting pool water chemistry effectively
USEFUL FOR

Pool owners, chemistry students, and anyone involved in maintaining swimming pool water quality will benefit from this discussion.

Taryn
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You have just built a swimming pool that has the dimensions of 7.5m in length and is 5 metres in width.
When filled with water the depth of the pool os exactly 1.5metres.
The lable on the 5 litre pool acid bottle indicates that this has a conc of 33.0g/100ml of HCl. Makin the assumption that the initial volume of pool water had a pH of 7 calculate the volume of pool acid you would need to add to the pool to bring the pH of the pool water to a value of 4.

This is wat I did, noting that I am a shocker at anything to do with pH, which is y I am trying to work out this qu b4 exams on mon!

Volume of the pool=52.5m^3=52500L

I sed that pH=pKa-[H30+]
7= 4-log[H30+]
[H30+]=0.001
then I sed that v=n/c

This is the dodgey thing that I did, I sed that in 1L there was 330g
Then n=330/36.453= 9.052mol

V= 9.052/0.001=9052
Then what? is that right? I feel as though I have done something wrong
If you have any idea of wat to do next or how to help agen I would be much appreciative.
And also just a general way or process of steps to solving pH problems I would also be appreciative
 
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Here's the general outline for solving this problem,

The total volume of the pool as the acid solution is added is Vi+Vacid.

The final desired pH of 4 has an hydronium acid concentration associated with it, use pH=-log[H3O+].

[H3O+]=moles of HCl/(Vi+Vacid), however moles of HCl is Vacid(the molarity of the acid solution)

Thus what you need to solve for finally is

[H3O+]=Vacid(Macid)/(Vi+Vacid), all are constants except Vacid, solve for Vacid through algebraic manipulation. Have fun!
 

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