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!

 

[Blueshift5]

Hello, it seems like you have made some progress in your calculations. However, there are a few things that need to be clarified and corrected in order to accurately determine the volume of acid needed for perfect balance in your pool.

Firstly, it is important to note that the volume of the pool is actually 52.5 cubic meters, not 52.5 liters. This is because the given dimensions of 7.5m x 5m x 1.5m are in meters, not in liters. So the volume of the pool would be 7.5m x 5m x 1.5m = 52.5m^3.

Next, you are correct in using the pH equation, pH = pKa - [H3O+], to solve for the concentration of H3O+. However, you have used the incorrect pKa value. The pKa value for HCl is -log(1.3 x 10^-2) = 1.89, not 4. So the correct equation would be 7 = 1.89 - [H3O+], and solving for [H3O+] gives a value of 5.11 x 10^-6 M.

Now, to determine the amount of acid needed, we can use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (in this case, Cl-) and [HA] is the concentration of the acid (HCl). Rearranging the equation, we get [HA] = [A-] x 10^(pH-pKa). Substituting in the values of [HA] = 5.11 x 10^-6 M and pKa = 1.89, we get [A-] = 5.11 x 10^-6 x 10^(7-1.89) = 0.0581 M.

Now, we can use the formula v = n/c to determine the volume of acid needed, where n is the number of moles and c is the concentration. However, you have made a mistake in your calculation of the number of moles. The concentration of HCl is given as 33.0 g/100 ml, which is equivalent to 330 g/L, not 330 g/1 L. So the correct number of