$\text{The solubility of barium iodate in an aqueous solution prepared by mixing }200\mathrm{mL}\text{ of}\mathrm{ 0.010}M\text{ barium nitrate}$

$\text{with}\mathrm{ 100}\mathrm{mL}\text{ of}\mathrm{ 0.10}\mathrm{ M}\text{ sodium iodate is}\mathrm{ X}\times {10}^{-6}\mathrm{mol}{\mathrm{dm}}^{-3}.\text{The value of}\mathrm{ X}\text{ is}\text{\_\_\_\_\_}$
$\text{Use: Solubility product constant}\left(K_{\mathrm{sp}}\right)\text{of barium iodate}=1.58\times {10}^{-9}$

The correct answer is 3.95.

To find the solubility of barium iodate in the resulting mixture, we first need to determine the concentration of the excess ions remaining in the solution after the precipitation reaction goes to completion.

Step 1: Calculate the initial moles of barium and iodate ions.
Barium nitrate dissociates as follows:
$\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{NO}}_3^{-}$
The number of moles of ${\mathrm{Ba}}^{2+}$ ions is:
$n_{{\mathrm{Ba}}^{2+}}=200\text{mL}\times 0.010\text{M}=2.0\text{mmol}$
Sodium iodate dissociates as follows:
${\mathrm{NaIO}}_3\to {\mathrm{Na}}^{+}+{\mathrm{IO}}_3^{-}$
The number of moles of ${\mathrm{IO}}_3^{-}$ ions is:
$n_{{\mathrm{IO}}_3^{-}}=100\text{mL}\times 0.10\text{M}=10.0\text{mmol}$

Step 2: Determine the limiting reactant and excess ions.
Barium iodate precipitates according to the reaction:
${\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{IO}}_3^{-}\left(\mathrm{aq}\right)\to {\mathrm{Ba}({\mathrm{IO}}_3}_2\left(s\right)$
Since $2.0\text{mmol}$ of ${\mathrm{Ba}}^{2+}$ requires $2\times 2.0\text{mmol}=4.0\text{mmol}$ of ${\mathrm{IO}}_3^{-}$, and we have $10.0\text{mmol}$ of ${\mathrm{IO}}_3^{-}$ available, ${\mathrm{Ba}}^{2+}$ is the limiting reactant.
All of the ${\mathrm{Ba}}^{2+}$ will precipitate, leaving excess ${\mathrm{IO}}_3^{-}$ in the solution:
${\text{Remaining mmol of IO}}_3=10.0\text{mmol}-4.0\text{mmol}=6.0\text{mmol}$
The total volume of the mixture is: