How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate? I have a hard time with selective precipitation problem.


  A mixture of $\ce{BaSO4}$ and $\ce{BaS2O3}$ is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is $[\ce{Ba^{2+}}]$ in the saturated solution? $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$ and $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$.


My solution: 

Because $K_\mathrm{sp}$ of $\ce{BaSO4} &lt; \ce{BaS2O3}$, $\ce{BaS2O3}$ will dissolved first. When $\ce{BaS_2O_3}$ dissolved, the concentration of $[\ce{Ba^{2+}}] = [\ce{S2O3^2-}] = \sqrt{K_\mathrm{sp}(\ce{BaS2O3})} = 2 \times 10^{-5}$.

That means, in the solution, there are:

\begin{align}
[\ce{Ba^2+}] &amp;= 2 \times 10^{-5}\\
[\ce{S2O3^2-}] &amp;= 2 \times 10^{-5}\\
[\ce{SO4^2-}] &amp;= \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} = \frac{9 \times 10^{-11}}{2 \times 10^{-5}} = 4.5 \times 10^{-6}
\end{align}

But the solution is $\ce{[Ba^2+]} = \sqrt{4.9 \times 10^{-10}}$.

What have I missed?

EDIT 1 : 
\begin{align}
\ce{BaSO4 &amp;&lt;=&gt; Ba^2+ + SO4^2- }\\
\ce{BaS2O3 &amp;&lt;=&gt; Ba^2+ + S2O3^2- }
\end{align}

The Solubility Product Equation : 

\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &amp;= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
K_\mathrm{sp}(\ce{BaS2O3}) &amp;= [\ce{Ba^2+}] \times [\ce{S2O3^2-}]
\end{align}

When the solids dissolves, it will produce $\ce{Ba^2+}$ ion, $\ce{SO4^2-}$ ion and $\ce{S2O3^2-}$ ion.

$\ce{BaSO4}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{SO4^2-}$ ion concentration is equal to solubility product of $\ce{BaSO4}$, which is $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$

And at the same time, 

$\ce{BaS2O3}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{S2O3^2-}$ ion concentration is equal to solubility product of $\ce{BaS2O3}$, which is $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$

Solubility of $\ce{BaSO4}$ in water is 
\begin{align}
 s &amp;= \sqrt{K_\mathrm{sp}(\ce{BaSO4})}\\
 s &amp;= \sqrt{9 \times 10^{-11}} \\
 s &amp;= 3 \times 10^{-5.5}
\end{align}

Solubility of $\ce{BaS2O3}$ in water is 
\begin{align}
 s &amp;= \sqrt{K_\mathrm{sp}(\ce{BaS2O3})}\\
 s &amp;= \sqrt{4 \times 10^{-10}}\\
 s &amp;= 2 \times 10^{-5}
\end{align}

Now, when the solids is mixed in water, There should be $[\ce{SO4^2-}] =3 \times 10^{-5.5} $ and $[\ce{S2O3^2-}] =2 \times 10^{-5} $.

Now I confuse what ion concentration should I use to determine $[\ce{Ba^2+}]$

EDIT 2:

\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &amp;= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} &amp;= [\ce{SO4^2-}] \\
\end{align}
and 
\begin{align}
K_\mathrm{sp}(\ce{BaS2O3}) &amp;= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} &amp;= [\ce{S2O3^2-}] \\
\end{align}

From @santimirandarp answer, I got that : 

\begin{align}
[\ce{Ba^2+}] &amp;= [\ce{S2O3^2-}] + [\ce{SO4^2-}]\\
[\ce{Ba^2+}] &amp;= \frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} + \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &amp;= \frac{4 \times 10^{-10}}{[\ce{Ba^2+}]} + \frac{9 \times 10^{-11}}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &amp;= \frac{4.9 \times 10^{-10}}{[\ce{Ba^2+}]} \\
[\ce{Ba^2+}]^2 &amp;= 4.9 \times 10^{-10} \\
[\ce{Ba^2+}] &amp;= \sqrt{4.9 \times 10^{-10}} \\
\end{align}

My problem solved. Thank You
This is just one possible approach:

First of all, the constants doesn't seem different enough to argue that $[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]$. So for now consider:

$$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}] \hspace{2em} (1)$$

and dividing both $K$s we can find: $$\frac{K_{\ce{BaS2O3}}}{K_{\ce{BaSO4}}}=\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}=4.44 \hspace{2em} (2)$$ 

Now we can solve for sulphate: $$K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}=[\ce{SO_4}^{2-}]^2(1+\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$

Using equation $(2)$ we get $[\ce{SO_4}^{2-}]=4.03\times 10^{-6}$

And for thiosulphate: $$K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}=[\ce{S2O_3}^{2-}]^{2}(1+\frac{[\ce{SO4}^{2-}]}{[\ce{S2O_3}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$

Using inverse of equation $(2)$ leads to $[\ce{S_2O_3}^{2-}]=1.8\times 10^{-5}$

The sum yields 
$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}]=0.00002215015=\sqrt(4.906295\times 10^{-10})$