What weight of calcuim carbonate will dissolve in 100cm3 of 0.2M of HCL?

Answer 1

mole HCl =M*V =0.2*0.1 =0.02

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

1 mole CaCO3 reacts with 2 mole HCl
x mole CaCO3 react with 0.02 mole HCl

Thus x=0.01 mole CaCO3
and similarly 0.01 mole of CaCl2 are formed. Considering that there is no change in volume you would have 0.01/0.1= 0.1 M Ca+2 coming from CaCl2

For CaCO3 Ksp=3*10^-9 =>solubility is 5.5*10^-5 M so the strong common ion effect of 0.1 M CaCl2 will practically not allow any more CaCO3 to be dissolved and we can consider that the amount of CaCO3 dissolved is equal to that that reacted with HCl.

Thus 0.01 mole CaCO3 dissolved and since the molecular weight is 100, mass=mole*MW=0.01*100 =1g