Question

At 25 degrees Celcius the Ksp for PbCl2 is 1.6 × 10–5.

Calculate Q  for the following: 125.0 mL of 0.0100 M Pb(NO3)2 is mixed with 75.0 mL of 0.0200 M NaCl at 25 degrees Celcius

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 = 0.01 M*125.0 mL/(125.0+75.0)mL

M(Pb(NO3)2) after mixing = 6.25*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 = 0.02 M*75.0 mL/(75.0+125.0)mL

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

So, we have now

[Pb2+] = 6.25*10^-3 M

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

The salt dissolves as:

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

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

Qsp = (6.25*10^-3)*(7.5*10^-3)^2

Qsp = 3.5*10^-7

we have,

Ksp = 1.6*10^-5

Since Qsp is less than ksp, precipitate will not form

Qsp = 3.5*10^-7

Since Qsp is less than ksp, precipitate will not form