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 $[\ce{Ba^2+}]$


\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, 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

  • 2
    $\begingroup$ There is no "dissolved first". The salts don't have the central command to tell them when to start dissolving, nor the intelligence to find out what other salts are sitting nearby. They all start dissolving at once. $\endgroup$ May 22 '19 at 11:27
  • $\begingroup$ @IvanNeretin okay, what do you think happened ? What is the $[\ce{Ba^{2+}}]$ ? $\endgroup$
    – valkedin
    May 22 '19 at 11:36
  • 1
    $\begingroup$ I think that both salts have dissolved a little, and you have all three ions in the solution. $\endgroup$ May 22 '19 at 11:44
  • 2
    $\begingroup$ Salts don't have military insignia either. Any individual $\ce{Ba^2+}$ does not know whether it is from BaSO4 or from BaS2O3. Write down all unknowns, then all equations, then see what you can do. $\endgroup$ May 22 '19 at 12:16
  • 1
    $\begingroup$ $[\ce{SO4^2-}] =3 \times 10^{-5.5} $ is a very odd way to write a concentration. Integer powers of 10 are expected. $\endgroup$
    – MaxW
    May 22 '19 at 13:25

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


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