[NaOH]

On a quiz I did today we had this question

If a pool has a volume of 1000 litters of water how many grams of HCl is needed to lower the pH from 7 to 4?I got it wrong, the teacher didn´t explain well to me nor gave me the correct answer for the question so I thought you guys at the forum could help me with the question.

[Borek]

Show how you tried and we will tell you what went wrong.

[willian921]

This is what i thoguht, not sure it is right though

HCl --> H3O+ + Cl-

in the pH of 7.0 pool, you have [H3O+] of 10^-7 mol/L, you multiply this by volume and molar mass, you have mass which is 0.00365 g

in the pH of 4.0 pool, you have [H3O+] of 10^-4 mol/L, you multiply this by volume and molar mass, you have mass which is 3.65g

the difference between the two is 3.645 g, in order to bring a pool with pH of 7 (with 10^-7mol/L of hydronium ions) down to pH of 4 (with 10^-4 mol/L of hydronium ions), you need 3.645 more grams.

this probably right, but i am not sure! sorry if it is wrong.

[AWK]

in the pH of 7.0 pool, you have [H3O+] of 10^-7 mol/L, you multiply this by volume and molar mass, you have mass which is 0.00365 g

Pure water shows pH=7. There is no need to add some acid in this case.

[Borek]

There is a small error in your thinking. Initial pH is not due to the presence of the diluted HCl, but due to the water autodissociation, so it is not as simple as just subtracting initial concentration of [H⁺]. When you add strong acid, concentration of [H⁺] from water autodissociation goes down, which means you have to add slightly more HCl than you calculated. At the same time difference between your result and more correct approach would be neglectable.

OTOH - what you did is completely wrong, but for the reasons that you probably have not learnt yet, and amount of information given in the question doesn't allow to calculate better answer, so you can safely ignore what I wrote in this paragraph

Edit: AWK posted while I was composing my post, it took a lot of time because I got two phone calls in the meantime.