Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.
2 answers:
In order to find the answer, use an ICE chart: Ca(IO3)2...Ca2+......IO3- <span>some.......0..........0 </span> <span>less.......+x......+2x </span> <span>less........x.........2x </span> <span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³ </span> K sp = [Ca⁺²][IO₃⁻]² K sp = (x) (2 x)² = 4 x³ 7.1 x 10⁻⁷ = 4 x³ <span>x = molar solubility = 5.6 x 10 </span>⁻³ M The answer is 5.6 x 10 ^ 3 M. (molar solubility)
<span>Answer: 5.62 x 10⁻³ M </span><span>Explanation : </span> <span /><span /><span>1) Equilibrium equation: </span><span /> <span>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. </span><span /> <span>3) Solubility product constant,Ksp : </span> <span /><span /><span> Ksp = [Ca⁺²] [IO⁻³]² = x(2x)² = 4x³ </span><span /> <span>=> 4x³ = 7.1 x 10⁻⁷ </span><span /> <span>=> x^3 = 1.775 x 10⁻⁷ </span><span /> <span>=> x = 5.62 x 10⁻³ M</span>
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Find the NUMBEr of half lives...then multiply by 54.3
2.27 = 6.5 (1/2)^n
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