Animated Solution for Chemistry - Ionic Equilibrium: The solubility of barium iodate in an aqueous solution prepared by mixing $200\text{ mL}$ of $0.010\text{ M}$ barium nitrate with $100\text{ mL}$ of $0.10\text{ M}$ sodium iodate is $X\times 10^{-6}{\text{ mol dm}}^{-3}$. The value of $X$ is _______. Use: Solubility product constant ($K_{sp}$) of barium iodate = $1.58\times 10^{-9}$

Mixing the Potions

Imagine you are in a laboratory, holding two beakers. In the first beaker, you have $200\text{ mL}$ of a $0.010\text{ M}$ solution of Barium Nitrate, ${\text{Ba(NO}}_3{)}_2$. In the second, you hold $100\text{ mL}$ of a $0.10\text{ M}$ solution of Sodium Iodate, ${\text{NaIO}}_3$. When you pour them together, a chemical dance begins.

To understand this dance, we must first count the dancers. We calculate the initial millimoles of each reactant by multiplying their volume by their molarity:

Who Runs Out First?

As the solutions mix, the barium ions ($B{a}^{2+}$) and iodate ions ($I{O}_3^{-}$) find each other and form a solid precipitate of Barium Iodate, ${\text{Ba(IO}}_3{)}_2$. The balanced chemical equation for this precipitation is:

Notice the stoichiometry: one mole of barium nitrate requires exactly two moles of sodium iodate. Since we only have $2\text{ mmol}$ of barium nitrate, it will completely react with $4\text{ mmol}$ of sodium iodate. Barium nitrate is our limiting reagent. It dictates when the precipitation stops.

A Crowded Room

Because the barium nitrate ran out, we have leftover iodate ions floating in the solution. Let's calculate exactly how much:

These $6\text{ mmol}$ of iodate are now swimming in a new, combined volume of $300\text{ mL}$ ($200\text{ mL}+100\text{ mL}$). This gives us a new concentration for the iodate ions:

Now, the solid ${\text{Ba(IO}}_3{)}_2$ tries to dissolve back into the water to establish an equilibrium. If its solubility is $s$, it will produce $s$ amount of $B{a}^{2+}$ and $2s$ amount of $I{O}_3^{-}$. However, the solution is already crowded with the $2\times 10^{-2}\text{ M}$ of iodate we just calculated! This is the Common Ion Effect. The massive amount of pre-existing iodate suppresses the solubility, making the extra $2s$ completely negligible. Thus, we approximate the total iodate concentration to simply be $2\times 10^{-2}\text{ M}$.

Unlocking $K_{sp}$

We are now ready to use the master equation of solubility, the $K_{sp}$ expression:

We substitute our known values into the equation. Remember, the concentration of barium ions is exactly equal to our unknown solubility, $s$:

Squaring the iodate concentration gives $4\times 10^{-4}$. Dividing the $K_{sp}$ by this value reveals the solubility:

The question asks for the answer in the format $X\times 10^{-6}$. By comparing our result, it is beautifully clear that $X=3.95$.