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