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 $\ce{BaSO4} < \ce{BaS2O3}$, $\ce{BaS2O3}$ 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+}] &= 2 \times 10^{-5}\\
[\ce{S2O3^2-}] &= 2 \times 10^{-5}\\
[\ce{SO4^2-}] &= \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 &<=> Ba^2+ + SO4^2- }\\
\ce{BaS2O3 &<=> Ba^2+ + S2O3^2- }
\end{align}


The Solubility Product Equation : 


\begin{align}
K\_\mathrm{sp}(\ce{BaSO4}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
K\_\mathrm{sp}(\ce{BaS2O3}) &= [\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 &= \sqrt{K\_\mathrm{sp}(\ce{BaSO4})}\\
 s &= \sqrt{9 \times 10^{-11}} \\
 s &= 3 \times 10^{-5.5}
\end{align}


Solubility of $\ce{BaS2O3}$ in water is 
\begin{align}
 s &= \sqrt{K\_\mathrm{sp}(\ce{BaS2O3})}\\
 s &= \sqrt{4 \times 10^{-10}}\\
 s &= 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}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
\frac{K\_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} &= [\ce{SO4^2-}] \\
\end{align}
and 
\begin{align}
K\_\mathrm{sp}(\ce{BaS2O3}) &= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \\
\frac{K\_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} &= [\ce{S2O3^2-}] \\
\end{align}


From [@santimirandarp answer](https://chemistry.stackexchange.com/a/115750/79443), I got that : 


\begin{align}
[\ce{Ba^2+}] &= [\ce{S2O3^2-}] + [\ce{SO4^2-}]\\
[\ce{Ba^2+}] &= \frac{K\_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} + \frac{K\_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4 \times 10^{-10}}{[\ce{Ba^2+}]} + \frac{9 \times 10^{-11}}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4.9 \times 10^{-10}}{[\ce{Ba^2+}]} \\
[\ce{Ba^2+}]^2 &= 4.9 \times 10^{-10} \\
[\ce{Ba^2+}] &= \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 
$[\ce{Ba^{2+}}]=[\ce{S\_2O\_3}^{2-}]+[\ce{SO\_4}^{2-}]=0.00002215015=\sqrt(4.906295\times 10^{-10})$


 1