Question

For the reaction: PbCl₂(s) ↔ Pb²⁺(aq)+2Cl^1-(aq), what is Q* when 1.5 mL of 0.035 M lead nitrate is added to 13 mL of 0.010 M sodium chloride?
K_sp of lead chloride is 1.6 x 10^-5 M³.
Hint given in general feedback

*Recall: Q is compared to K_sp to determine whether a precipitate forms.

Answer 1

Lets find the concentration after mixing for Pb(NO3)2

Concentration after mixing = mol of component / (total volume)

M(Pb(NO3)2) after mixing = M(Pb(NO3)2)*V(Pb(NO3)2)/(total volume)

M(Pb(NO3)2) after mixing = 0.035 M*1.5 mL/(1.5+13.0)mL

M(Pb(NO3)2) after mixing = 3.621*10^-3 M

Lets find the concentration after mixing for NaCl

Concentration after mixing = mol of component / (total volume)

M(NaCl) after mixing = M(NaCl)*V(NaCl)/(total volume)

M(NaCl) after mixing = 0.01 M*13.0 mL/(13.0+1.5)mL

M(NaCl) after mixing = 8.966*10^-3 M

So, we have now

[Pb2+] = 3.621*10^-3 M

[Cl-] = 8.966*10^-3 M

At equilibrium:

PbCl2 <----> Pb2+ + 2 Cl-

Qsp = [Pb2+][Cl-]^2

Qsp = (3.621*10^-3)*(8.966*10^-3)^2

Qsp = 2.91*10^-7

we have,

Ksp = 1.6*10^-5

Since Qsp is less than ksp, precipitate will not form

Answer: Qsp = 2.9*10^-7