Question
Does any solid PbCl2 form when 3.1 mg NaCl is dissolved in 0.228 L of 0.12 M Pb(NO3)2.
Find the Qsp.
Find the Qsp.
Answers
DrBob222
PbCl2 ==> Pb^2+ + 2Cl^-
Ksp = )Pb^2+)(Cl^-)^2
NaCl ==> Na^+ + Cl^-
Convert 3.1 mg NaCl to moles and M NaCl = mole/L soln where L soln = 0.228.
Then Qsp = (Pb^2+)(Cl^-)^2
Substitute (Pb^2+) from Pb(NO3)2. Substitute (Cl^-) from NaCl M from above.
If Qsp > Ksp, a ppt will occur.
If Qsp < Ksp, a ppt will not occur.
Ksp = )Pb^2+)(Cl^-)^2
NaCl ==> Na^+ + Cl^-
Convert 3.1 mg NaCl to moles and M NaCl = mole/L soln where L soln = 0.228.
Then Qsp = (Pb^2+)(Cl^-)^2
Substitute (Pb^2+) from Pb(NO3)2. Substitute (Cl^-) from NaCl M from above.
If Qsp > Ksp, a ppt will occur.
If Qsp < Ksp, a ppt will not occur.