Question

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2 formed. How will this affect the concentration of Pb+2?

Answer 1

PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Pb²⁺(aq) + 2Cl⁻(aq) ⇄ PbCl₂(s) 

At increase the concentration of chloride ions - concentration of lead ions decreases, the lead chloride is formed. 

Answer 2

Answer:

Concentration of $Pb^{2+}$ decreases

Explanation:

In a saturated solution of $PbCl_{2}$, the following solubility equilibrium exist-

                         $PbCl_{2}\rightleftharpoons Pb^{2+}+2Cl^{-}$

Now, NaCl contains common ion $Cl^{-}$. Therefore, concentration of $Cl^{-}$ increases in solution.

But, at a particular temperature, equilibrium constant remain constant. Therefore to keep equilibrium constant same $Pb^{2+}$ will reacts with excess $Cl^{-}$ to decrease concentration of $Cl^{-}$.

Therefore concentration of $Pb^{2+}$ decreases.

Note: Equilibrium constant for the above equilibrium=$\left [ Pb^{2+} \right ]\left [ Cl^{-} \right ]^{2}$