# PH fish tank water problem

• marcelo
In summary, the conversation discusses the need for a fish breeder to adjust the water's pH from 5.0 to 6.5 using only calcium carbonate. The calculation for the amount of calcium carbonate needed is complicated due to the presence of other buffers and the type of neutralization required.

#### marcelo

Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5

B)5,5

C)6,5

D)7,5

E)9,5

What do YOU think?

I think if I have 10-5 of pH value we need 10-9 of pOH value to make Kw=-14
the reaction of water with CaCo3 is 2 per 1
(H+) + (OH-) + CO3(2-) + Ca(2+)----> HCO3(-) + Ca2(+) OH(-)
(H+) + (OH-) +HCO3(-) + (Ca(2+) + OH(-)-----> H2CO3 + Ca(OH)2
since I need only 10^-2 concentration of OH- and we have only 5L of water:

5.10(-2) mols of OH-. But 1 mol of CaCO3 gives 2 mols OH, we need only 2.5x10(-2) but this answer is wrong,

As asked this question has no answer, as we don't know what is responsible for the initial pH - typically water contains some buffers and the amount of base/acid required to change the pH depends of their concentration. Then, adding carbonates creates another buffer set, making the calculation a bit more convoluted than just the neutralization would do.

marcelo said:
I think if I have 10-5 of pH value we need 10-9 of pOH value to make Kw=-14
This is way too cryptic for me to guess what you mean. Or, if I guess right, you are mixing pH with [H+] and so on as if they were the same thing - they are not.

Spinnor