Molar mass of AgBrO3 = 236 g/mol
Molar mass of AgBrO3= 235.8 g/mol
s = 7.2*10^-3 g/L
To covert it to mol/L, divide it by molar mass
s = 7.2*10^-3 g/L / 236 g/mol
s = 3.05*10^-5 mol/L
At equilibrium:
AgBrO3 <----> Ag+ + BrO3-
s s
Ksp = [Ag+][BrO3-]
Ksp = (s)*(s)
Ksp = 1(s)^2
Ksp = 1(3.05*10^-5)^2
Ksp = 9.3*10^-10
Answer: 9.3*10^-10
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