Respuesta :
In order to find the answer, use an ICE chart:
Ca(IO3)2...Ca2+......IO3-
some.......0..........0
less.......+x......+2x
less........x.........2x
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)
Ca(IO3)2...Ca2+......IO3-
some.......0..........0
less.......+x......+2x
less........x.........2x
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)
Answer: 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
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