Question

Question #2d18a

Answer 1

The balanced equation of reaction of `CaCO_3` with `HCl` is as follows

`CaCO_3+2HCl->CO_2+H_2O+CaCl_2`

Molar mass of `CaCO_3=(40+12+3*16)g"/mol"=100g"/mol"`

So 1g `CaCO_3=1/100mol=0.01mol`
Let 2000cc or `2L" "HCl` solution contains `xmol" "HCl`

As per equation the reacting ratio of `CaCO_3` and `HCl` is `1:2`
Hence `0.01/x=1/2`

So `x=0.02mol` (Assuming the acid has been completely nutralized by 1g `CaCO_3`)

Hence the strength of the `HCl` solution is `c=(0.02mol)/(2L)=0.01M`

So pH if the solution is

`=-logc=-log(0.01)=2`