Question

At 25oC the Ksp for PbCl2 is 1.6 × 10–5

1) Calculate Q  for the following: 125.0 mL of 0.0700 M Pb(NO₃)₂ is mixed with 75.0 mL of 0.0200 M NaCl at 25^oC

2) Calculate Q  for the following: 125.0 mL of 0.0700 M Pb(NO₃)₂ is mixed with 75.0 mL of 0.0200 M NaCl at 25^oC

Answer 1

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 = 0.07 M*125.0 mL/(125.0+75.0)mL

M(Pb(NO3)2) after mixing = 4.375*10^-2 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 = 0.02 M*75.0 mL/(75.0+125.0)mL

M(NaCl) after mixing = 7.5*10^-3 M

So, we have now

[Pb2+] = 4.375*10^-2 M

[Cl-] = 7.5*10^-3 M

At equilibrium:

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

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

Qsp = (4.375*10^-2)*(7.5*10^-3)^2

Qsp = 2.461*10^-6

Answer: 2.46*10^-6

2)

we have,

Ksp = 1.6*10^-5

Since Qsp is less than ksp, precipitate will not form

Answer: No precipitate