The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $$X \times 10^{-6}$$ mol dm$$^{-3}$$. The value of $$X$$ is ______.

Use: Solubility product constant ($$K_{sp}$$) of barium iodate = $$1.58 \times 10^{-9}$$

The dissolution equilibrium for barium iodate is
$$Ba(IO_3)_2(s) \rightleftharpoons Ba^{2+}+2\,IO_3^-$$
and its solubility-product constant is
$$K_{sp}=1.58\times10^{-9}= [Ba^{2+}][IO_3^-]^2$$

Step 1: Initial ion concentrations after mixing
Moles of $$Ba^{2+}$$ added: $$0.200\text{ L}\times0.010\text{ M}=2.0\times10^{-3}\,\text{mol}$$
Moles of $$IO_3^-$$ added: $$0.100\text{ L}\times0.10\text{ M}=1.0\times10^{-2}\,\text{mol}$$
Total volume after mixing: $$0.200+0.100=0.300\text{ L}$$

Initial concentrations
$$[Ba^{2+}]_0=\frac{2.0\times10^{-3}}{0.300}=6.67\times10^{-3}\,\text{M}$$
$$[IO_3^-]_0=\frac{1.0\times10^{-2}}{0.300}=3.33\times10^{-2}\,\text{M}$$

Step 2: Check for precipitation
Ionic product $$Q=[Ba^{2+}]_0[IO_3^-]_0^{\,2}$$
$$Q=6.67\times10^{-3}\times(3.33\times10^{-2})^{2}=6.67\times10^{-3}\times1.11\times10^{-3}=7.41\times10^{-6}$$
Since $$Q\;(7.41\times10^{-6}) \gg K_{sp}\;(1.58\times10^{-9})$$, the solution is supersaturated and $$Ba(IO_3)_2$$ will precipitate until equilibrium is attained.

Step 3: Let $$x$$ M of $$Ba(IO_3)_2$$ precipitate
Every mole of precipitate removes $$1\,Ba^{2+}$$ and $$2\,IO_3^-$$.
Final concentrations:
$$[Ba^{2+}]=6.67\times10^{-3}-x$$
$$[IO_3^-]=3.33\times10^{-2}-2x$$

Step 4: Apply $$K_{sp}$$ at equilibrium
$$(6.67\times10^{-3}-x)\,(3.33\times10^{-2}-2x)^{2}=1.58\times10^{-9}\;-(1)$$

Step 5: Simplify using the small-quantity approximation
Because $$K_{sp}$$ is extremely small, the final $$[Ba^{2+}]$$ will be much smaller than the initial $$6.67\times10^{-3}$$ M. Let $$z=[Ba^{2+}]_{eq}=6.67\times10^{-3}-x\ll6.67\times10^{-3}\;,$$ so $$x\approx6.67\times10^{-3}$$.

Then $$[IO_3^-]_{eq}=3.33\times10^{-2}-2x=3.33\times10^{-2}-2(6.67\times10^{-3}-z)=0.020+2z$$ Since $$z\ll0.020$$, take $$[IO_3^-]_{eq}\approx0.020\;\text{M}$$.

Insert these approximations into $$(1)$$:

$$z\,(0.020)^{2}=1.58\times10^{-9}$$
$$\Rightarrow\;z=\frac{1.58\times10^{-9}}{4.00\times10^{-4}}=3.95\times10^{-6}\,\text{mol dm}^{-3}$$

Step 6: Interpret the result
The equilibrium concentration of dissolved $$Ba^{2+}$$ equals the molar solubility of $$Ba(IO_3)_2$$ in the given medium. Therefore, the solubility is $$3.95\times10^{-6}\,\text{mol dm}^{-3}$$.

Expressed as $$X\times10^{-6}$$ mol dm$$^{-3}$$, $$X=3.95\;(\text{any value in }3.85-4.15\text{ is accepted}).$$