Question

Jika ksp CaCO3 = 2,5 x 10^-9, berapa gram CaCO3 (Mr = 100 ) yang terkandung dalam 500 mL larutan jenuh ?

Answer 1

Ksp = 2,5.10^-9
M = n/v = 2,5 . 10^-9 / 5.10^-10 = 5. 10^-9 M
mol = M.V= 5.10^-9 . 5.10^-1 = 25. 10 ^-10 mol
mol = massa / Mr 
massa = mol . Mr = 25 . 10 ^-10 x  100 = 25 . 10^-8 gram

Answer 2

Ksp = 2,5.10^-9
M = n/v = 2,5 . 10^-9 / 5.10^-10 = 5. 10^-9 M
mol = M.V= 5.10^-9 . 5.10^-1 = 25. 10 ^-10 mol
mol = massa / Mr 
massa = mol . Mr = 25 . 10 ^-10 x  100 = 25 . 10^-8 gram