What would be the solubility in grams of calcite in 1L of water? The Ksp of CaCO3 is 3.36 x 10^-9.?

Answer 1

If the molar solutbilty is s mole/L then

.. ... .. .. .. CaCO3 <=> Ca(+2) + CO3(-2)
Dissolve.. .. s
Produce .. .. .. .. .. .. .. .. .. s .. .. .. .. s

Ksp= [Ca+2][CO3(-2)] =s^2 =>
s= squareroot (Ksp) =SQRT(3.36*10^-9) = 5.80*10^-5 M

the molecular weight is 100 and since mass= mole*MW we have
5.80*10^-5 mole/L =5.80*100*10^-5 g/L= 5.80 *10^-3 g/L