Question

For AgBrO3 in water at 25 degrees celsius and 1 bar, K(sp)=5.38x10^-5 mol^2/kg^2. Calculate the solubility of AgBrO3 in water at 25 degrees C, neglecting ion pairing.

Answer 1

AgBrO3 is relatively insoluble in water
the Ksp tells how much of the compound actually dissociates to ions

Ksp = [Ag⁺][BrO₃^-] = equilibrium equation

first, let's look at an ICE chart
AgBrO3.....[Ag+].....[BrO3 -]
some............0...............0
less............+x............x
less............x.............x

Ksp = [Ag+][BrO3 -]. we know that [Ag+] = x and [CO3 2-] = x
so, Ksp = x²
5.38 * 10^-5 = x²
x = Sq.rt ( 5.38 * 10^-5 )
molar solubility of AgBrO3 = 7.3 * 10^-3M
7.3 * 10^-3M x235.7704g/mole =1.721g/L