250 mL of 4.25 x 10^-4 M Pb(NO₃)₂ is added to 275 mL of 6.25 x 10^-5 M NaI.

Will a precipitate form?

Data: K_sp [PbI₂] = 7.10 x 10^-9

Question

250 mL of 4.25 x 10^-4 M Pb(NO₃)₂ is added to 275 mL of 6.25 x 10^-5 M NaI.

Will a precipitate form?

Data: K_sp [PbI₂] = 7.10 x 10^-9

| General Chemistry 3

Accepted Answer

Dissolution reaction:

PbI₂ (s) → Pb²⁺ (aq) + 2 I^- (aq) [K_sp]

PbI₂ (s) forms if Q_sp > K_sp

Let’s determine Q_sp:

Q_sp = [Pb²⁺]₀ [I^-]₀²

Pb(NO₃)₂ (s) → Pb²⁺ (aq) + 2 NO₃^- (aq)

From the stoichiometry of this reaction: n_Pb2+,0 = n_Pb(NO3)2,0

⇒ n_Pb2+,0 = [Pb(NO₃)₂]₀ x V_Pb(NO3)2

⇒ n_Pb2+,0 = 0.250 x 4.25 x 10^-4 = 1.06 x 10^-4 mol

NaI (s) → Na⁺ (aq) + I^- (aq)

From the stoichiometry of this reaction: n_I-,0 = n_NaI,0

⇒ n_I-,0 = [NaI]₀ x V_NaI= 1.72 x 10^-5 mol

Q_sp = [Pb²⁺]₀ [I^-]₀²

Q_sp = $\frac{n_{{Pb}^{2 +}, 0} \times {n_{I^{-}, 0}}^{2}}{V^{3}}$

Q_sp = 2.17 x 10^-13 < K_sp

PbI₂ (s) does not form

Final - Question 7 | General Chemistry 3