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

[Taryn]

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

 

[GCT]

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!