# What actually happens to the equilibrium of solid and aqueous ions in a solution when concentrations are changed?

Taking as an example the equilibrium set up when $$\ce{BaSO4}$$ is added to water:

$$\ce{ BaSO4(s) <=> Ba^{2+}(aq) + SO^{2-}_4(aq) }$$

The solubility product constant is $$[\ce{Ba^{2+}}][\ce{SO^{2-}_4}]$$.

If I then add additional $$\ce{Ba^{2+}}$$ ions to the solution, what will happen to the concentrations of the products once equilibrium is restored, since the product of their concentrations is constant. Will the concentration $$[\ce{Ba^{2+}}]$$ remain larger than its original value before the addition of barium ions and the $$[\ce{SO^{2-}_4}]$$ become smaller than its original value to compensate and keep the solubility product constant?

• Any alternative idea? Oct 6, 2021 at 17:05
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• If somebody is going to vote the answer down, please kindly inform me why it is a bad question and where I’ve gone wrong. Oct 6, 2021 at 17:33
• (+1) The answer to your final question is simply yes: the sulfate ion concentration gets reduced and the barium ion concentration is increased. Adding sulfate ions gives the opposite situation: the barium ion concentration is reduced, etc. In barium medical “milk shakes”, given for fluoroscopic imaging of the throat and esophagus, the barium sulfate suspension is in a liquid with excess sulfate ions, as per the answer by Buttonwood. Barium is toxic, but it cannot do harm if it is tied up in barium sulfate and the equilibrium keeps it tied up.
– Ed V
Oct 6, 2021 at 21:44

On the microscopic scale, the thermodynamic equilibrium is dynamic, i.e. there are constant back- and forward reaction. Here: dissociation of $$\ce{BaSO4}$$ to yield $$\ce{Ba^{2+}}$$ and $$\ce{SO^{2-}_4}$$, and association to yield again $$\ce{BaSO4}$$. Once the thermodynamic equilibrium is installed, these ongoing microscopic reactions do not change anymore the concentration of $$\ce{Ba^{2+}}$$, nor the one of $$\ce{SO^{2-}_4}$$.
If disturbed at macroscopic level (e.g., by addition of further $$\ce{SO^{2-}_4}$$, an increase of $$[\ce{SO^{2-}_4}]$$), keeping the product of $$[\ce{Ba^{2+}}][\ce{SO^{2-}_4}]$$ constant is possible only if the other factor (i.e., $$[\ce{Ba^{2+}}]$$) decreases. Thus, in the case of gravimetric determination of barium in the form of $$\ce{BaSO4}$$, the addition of $$\ce{Na2SO4}$$, which is water-soluble, to obtain an exhaustive precipitation.