Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.

In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+IO3- 
some00 
less+x+2x 
lessx2x 

Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³

K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
x = molar solubility = 5.6 x 10⁻³ M

The answer is 5.6 x 10 ^ 3 M. (molar solubility)

5.62 x 10⁻³ M

Explanation:


1) Equilibrium equation:


Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³

2) By stoichimetry ratio, the concentrtion of IO⁻³ ions is the double than the concentraion of Ca⁺² , so call them x and 2x.


3) Solubility product constant,Ksp:


Ksp = [Ca⁺²] [IO⁻³]² = x(2x)² = 4x³


=> 4x³ = 7.1 x 10⁻⁷


=> x^3 = 1.775 x 10⁻⁷


=> x = 5.62 x 10⁻³ M

The answer would be

It shows that the experiment is designed properly.