Question

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

Answer 1

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