# How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = 7.1 x 10^-7

Jun 9, 2016

$\text{Solubility of calcium iodate } \cong 0.25 \cdot g \cdot {10}^{-} 1 {L}^{-} 1$

#### Explanation:

We write out the dissolution reaction:

$C a {\left(I {O}_{3}\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s C {a}^{2 +} + 2 I {O}_{3}^{-}$

And then we write out the equilibrium expression, in which $C a {\left(I {O}_{3}\right)}_{2}$, as a solid, does not appear:

${K}_{s p} = \left[C {a}^{2 +}\right] {\left[I {O}_{3}^{-}\right]}^{2} = 7.1 \times {10}^{-} 7$

If we dub the solubility of calcium iodate under these conditions as $S$, then ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2}$ $=$ $4 {S}^{3}$.

This expression follows the equilibrium equation: each equiv of salt that dissolves gives 1 equiv of $C {a}^{2 +}$, but 2 equiv iodate ion.

Thus ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2}$ $=$ $4 {S}^{3}$ $=$ $7.1 \times {10}^{-} 7$

S=""^3sqrt{(7.1xx10^-7}/4)

This gives an answer in $m o l \cdot {L}^{-} 1$. You will have to use the formula mass of calcium iodate, $389.88 \cdot g \cdot m o {l}^{-} 1$, to give an answer in the units required.