Question

If I mix 1000 mg of CaCO3 in 25 ml of water with 25 ml of 0.1 N HCl, what will the resultant pH be and how is it calculated?

Answer 1

I would like to know it

Explanation:

Since MM `CaCO_3` is 100 g/mol, 1000 mg are equal to 10 mmol.
since for HCl N=M for solutions `n= M xx V= 0,1 (mol)/L xx 25 mL = 2,5 mmol`

the reaction is
`CaCO_3 + 2HCl = CaCl_2 + H_2CO_3`
but since you have more`CaCO_3` it remain 10-(2,5/2)= 8,75 mmol of `CaCO_3`

that will react with 1,25 mol of `H_2CO_3` obtained
`CaCO_3 + H_2CO_3 = Ca(HCO_3)_2`

and it will remain
7,5 mmol of `CaCO_3` insoluble,
1,25 mmol of `Ca(HCO_3)_2`
1,25 mmol of `CaCl_2`

the concentartion of `HCO_3^-` is 2,5 mmol in 50 mL = 0,05 mol/L. Its pH is given by the formula:`pH = 1/2(pK_1+pK_2) =1/2( 8,5+10,3)= 8,4` very unstable
Instead `CaCl_2` is a soluble salt with a weak acid idrolisis that is difficult to calculate
pH will be near at the neutrality