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r-ruslan [8.4K]
3 years ago
6

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

Chemistry
2 answers:
e-lub [12.9K]3 years ago
5 0
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)
goldenfox [79]3 years ago
5 0
<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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Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

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As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

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