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

Concentration of
`Pb^(2+)` decreases

Step-by-step 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)`