[bingobongo]

Starting with a pH of 4.5 how much CaCo3 does one need to change the pH of 2,400,000,000 L of water from pH 4.5 to 6?

ok so

CaCO3 + H20 --> CO2 + Ca(OH)2

I calculated my starting [OH] and the final [OH] that would correspond to pH of 6.  I took the difference and came up with about 2000 kg of CaCo3.... hmmmmmmmmm.

pH=4.5  [H+]=3.16E-05 and from [OH][H]=1E-14 i got [OH]=3.15E-10

ph=6   [H]=1E-06 and [OH]=1E-08

change in [OH] needed is 9.684E-09 M so since its a 1/1 molar ratio I got 23,240 moles of CaCO3 and 2325kg....

I feel I really screwed this one up I am confused because we are not provided with an molaritys.


help?

[Borek]

Won't you be dissolving CaCO₃ in acid?

And if so, won't it be easier to write reaction between CaCO₃ and H⁺?

[bingobongo]

There is no other information given in the problem..

But I think writing it out as you said would make more sense..