Bila Ksp CaCO3 = 9 x 10^-8 mol L^-1 , maka berapa gram kelarutan CaCO3 dalam 250 ml air ? (Mr CaCO3 = 100)
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Pertanyaan
Bila Ksp CaCO3 = 9 x 10^-8 mol L^-1 , maka berapa gram kelarutan CaCO3 dalam 250 ml air ? (Mr CaCO3 = 100)
1 Jawaban
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1. Jawaban andrisaputra2
ksp CaCO3 = [Ca][CO3]
= s x s = s2
s = √ksp
= √9 x 10^-8
= 3 x 10^-4
s = M
M = gram/Mr x 1000/ml
3 x 10^-4 = gram/100 x 1000/250
gram x 4 = 3 x 10^-4 x 100
gram = 3 x 10^-2 : 4
= 7,5 x 10^-3