Question

Worksheet 17c (Gen Chem) Other Equilibria 37- Will a precipitate form if 10.00 mL. of 0.010 M sodium iodide is added to a solution that was previously prepared by mixing 16.0 ml. of thiosulfate ions (1.75 M) to 250. ml. of lead(II) ions (0.020 M)? 23

Answer 1

Ans :- PbI₂ (s) will not form precipitates.

Explanation :-

Overall Reaction is :

2 NaI (aq) + PbS₂O₃ (aq) ----------------> PbI₂ (s) + Na₂S₂O₃ (aq)

From the definition of Molarity as :

Molarity = Number of moles / Volume of solution in L

So,

Number of moles of NaI = Molarity of NaI x Volume of solution

= 0.010 M x 0.010 L

= 0.0001 mol

Similarly,

Number of moles of Pb²⁺ = Molarity of Pb²⁺ x Volume of solution

= 0.020 M x 0.250 L

= 0.005 mol

Now,

Total volume becomes = 10.0 ml + 16.0 ml + 250.0 ml = 276 ml = 0.276 L

So,

Molarity of NaI or [I^-] in complete solution = 0.0001 mol / 0.276 L = 0.000362 M

Also,

Molarity of Pb²⁺ or [Pb²⁺] in complete solution = 0.005 mol / 0.276 L = 0.0181 M

Partial dissociation of PbI₂ in solution is :

PbI₂ (s) <-------------------> Pb²⁺ + 2 I^-

Expression of reaction quotient (Q_sp) for PbI₂ is :

Reaction Quotient (Qsp) is the product of the molar concentration of products raise to the power of stoichiometric coefficient with respect to each products species at any stage of the reaction. Solubility product (Ksp) is the product of the molar concentration of products raise to the power of stoichiometric coefficient with respect to each products species at equilibrium stage of the reaction.

Q_sp = [Pb²⁺].[I^-]²  

Q_sp = (0.0181 M).(0.000362 M)²

Q_sp = 2.37 x 10^-9 M³

Which is smaller than Solubility product (K_sp) value of PbI₂ i.e. 7.1 x 10^-9 M³

Therefore, PbI2 (s) will not form precipitates.