Question

Addition of conc. $HCl$ to sturated $Ba{Cl}_2$ solution precipitate $Ba{Cl}_2$ because:

• A. at constant temperature the product $\left[{Ba}^{2+}\right]{\left[{Cl}^{-}\right]}^2$ remains constant in a saturated solution
• B. ionic product of $\left[{Ba}^{2+}\right]\left[{Cl}^{-}\right]$ remains constant in a saturated solution
• C. of common ion effect
• D. it follows Le-Chatelier's principle

Answer 1

Answer: (B)

ionic product of $\left[{Ba}^{2+}\right]\left[{Cl}^{-}\right]$ remains constant in a saturated solution

When you add a compound having a common ion, It decreases the solubility of the solute. This is because Le Chatelier's principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.

This is called common-ion effect. due to this effect, some solute becomes insoluble and precipitate.

In the above case $HCl$ and $Ba{Cl}_2$ have $C{l}^{-}$ common ions and thus $Ba{Cl}_2$ precipitates.