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Lady bird [3.3K]
3 years ago
12

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?
Chemistry
2 answers:
konstantin123 [22]3 years ago
8 0

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}

NNADVOKAT [17]3 years ago
5 0
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. 

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molality = 0.1705/0.282

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