# (1)425ml of a saturated solution of lanthanum iodate, La(IO_3)_3, has 2.93 xx 10^-4 mole of La^(3+) (a)what is the concentration of IO_3^-? (b)calculate the solubility product of lantham iodate?

Feb 9, 2016

Here's what I got.

#### Explanation:

So, you know that you're dealing with a $\text{425-mL}$ sample of a saturated solution of lanthanum(III) iodate, "La"("IO"_3)_3.

You also know that this solution contains $2.93 \cdot {10}^{- 4}$ moles of lanthanum(III) cations, ${\text{La}}^{3 +}$.

Use this information to find the molarity of the lanthanum(III) cations - do not forget that molarity uses liters of solution!

$\textcolor{b l u e}{c = \frac{n}{V}}$

["La"^(3+)] = (2.93 * 10^(-4)"moles")/(425 * 10^(-3)"L") = 6.89 * 10^(-4)"M"

Now, the equilibrium reaction that describes the partial dissociation of lanthanum iodate in aqueous solution looks like this

${\text{La"("IO"_3)_text(3(s]) rightleftharpoons "La"_text((aq])^(3+) + color(red)(3)"IO}}_{\textrm{3 \left(a q\right]}}^{-}$

Notice that you have a $1 : 1$ mole ratio between lanthanum(III) iodate and the lanthanum(III) cations. This tells you that every mole of the solid that dissolves in solution will produce $1$ mole of lanthanum(III) cations.

Likewise, the $1 : \textcolor{red}{3}$ mole ratio that exists between the solid and the iodate anions, ${\text{IO}}_{3}^{-}$, tells you that every mole of the solid that dissolves in solution will produce $\textcolor{red}{3}$ moles of iodate anions.

So, using this mole ratio, you can say that the molarity of the iodate anions will be $\textcolor{red}{3}$ times higher than that of the lanthanum(III) cations.

$\left[{\text{IO"_3^(-)] = color(red)(3) xx ["La}}^{3 +}\right]$

["IO"_3^(-)] = color(red)(3) xx 6.89 * 10^(-4)"M" = color(green)(2.07 * 10^(-3)"M"

The expression of the solubility product constant, ${K}_{s p}$, for lanthanum(III) iodate will look like this

${K}_{s p} = {\left[{\text{La"^(3+)] * ["IO}}_{3}^{-}\right]}^{\textcolor{red}{3}}$

Plug in your values to get

${K}_{s p} = 6.89 \cdot {10}^{- 4} \cdot {\left(2.07 \cdot {10}^{- 3}\right)}^{\textcolor{red}{3}}$

${K}_{s p} = \textcolor{g r e e n}{6.12 \cdot {10}^{- 12}}$

The answers are rounded to three sig figs.