Question

If 100. mL of 0.025 M NaCl is mixed with 150. mL of 0.0015 M Pb(NO3)2, will a precipitate form? (Ksp PbCl2 = 1.6 × 10−5) - Answer: No- (Justify this answer) please show work

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 = 0.0015 M*150.0 mL/(150.0+100.0)mL

M(Pb(NO3)2) after mixing = 9*10^-4 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 = 0.025 M*100.0 mL/(100.0+150.0)mL

M(NaCl) after mixing = 1*10^-2 M

So,

[Pb2+] = 9*10^-4 M

[Cl-] = 1*10^-2 M

The salt dissolves as:

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

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

Qsp = (9*10^-4)*(1*10^-2)^2

Qsp = 9*10^-8

we have,

Ksp = 1.6*10^-5

Since Qsp is less than ksp, precipitate will not form