Molar mass of Al(OH)3,
MM = 1*MM(Al) + 3*MM(O) + 3*MM(H)
= 1*26.98 + 3*16.0 + 3*1.008
= 78.004 g/mol
Molar mass of Al(OH)3= 78.004 g/mol
s = 1.127*10^-3 g/L
To covert it to mol/L, divide it by molar mass
s = 1.127*10^-3 g/L / 78.004 g/mol
s = 1.445*10^-5 mol/L
At equilibrium:
Al(OH)3 <----> Al3+ + 3 OH-
s 3s
Ksp = [Al3+][OH-]^3
Ksp = (s)*(3s)^3
Ksp = 27(s)^4
Ksp = 27(1.445*10^-5)^4
Ksp = 1.177*10^-18
Answer: 1.177*10^-18
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