Bila Ksp CaCO3 = 9 x 10^-8 mol L^-1 , maka berapa gram kelarutan CaCO3 dalam 250 ml air ? (Mr CaCO3 = 100)
monika1005
CaCO3 ---> Ca+ + CO3 - Ksp = s x s = s2 s= akar 9x10-8 = 3 x 10-4 M M = gr/Mr x 1000/v 3 X 10-4 = gr/100 x 1000/250 3 x 10-4 = 4 gr / 100 3 x 10-4/10+2 = 4gr 3 X 10-2 = 4 gr 75 x 10-4 = gr
1 votes
Thanks 6
Cindyy23
Perjelas lagi sih, bingung gua