The Kp of calcium iodate is 7.1 x 107 What is the concentration of Ca2 + in equilibrium with Ca 3.3 x 10-2 M? 2.3.3 * 10M b. 8.1 x 10 SM c. 2.2 * 10-SM d. 6.5 x 10 M e. 7.1 x 10-3 M

Hey there!:

Given Ksp of Ca (IO₃) ₂ = 7.1 x 10⁻⁷

At the equilibrium cation will be:

Ca (IO₃) ₂ (s) ⇌ Ca²⁺ (aq) + 2 IO₃⁻ (aq):

Ksp = [ Ca²⁺ ] [ IO₃⁻ ]² given [ IO₃⁻ ]² = 3.3 x 10⁻² M

therefore:

Ksp = [ Ca²⁺ ] [ IO₃⁻ ]²

7.1 x 10⁻⁷ = [ Ca²⁺ ] [ 3.3 x 10⁻² ] ²

[ Ca²⁺ ] = [ 7.1 x 10⁻⁷ ] / [ 3.3 x 10⁻² ] ²

[ Ca²⁺ ] = [ 7.1 x 10⁻⁷ ] / [ 10.89 x 10⁻⁴ ]

[ Ca²⁺ ] = 6.5 x 10⁻⁴ M