Answer to Question #68881 in Chemistry for Mhamad

Question

How much solid Pb(No3)2 must be added to 1L of 0.001M Na2So4 sloution for a pricipitation of PbSo4 (ksp=1.6*10^-8) to form ? assume no change in volume when solid is added

Answer 1

Pb(NO3)2 + Na2SO4 = 2NaNO3 + PbSO4;
Pb2+ + 2NO3- + 2Na+ + SO42- = 2Na+ + 2NO3- + PbSO4 (s);
Pb2+ + SO42- = PbSO4 (s).
Ksp = [Pb2+]*[SO42-] =1.6*10-8;
We write the dissociation equation for sodium sulfate, Na2SO4:
Na2SO4 = 2Na+ + SO42-;
By the reaction equation:
C (Na2SO4) = [SO42-] = 0.001 M;
Then,
[Pb2+] = Ksp / [SO42-] = 1.6*10-8 / 0.001 = 1.6*10-5 (M);
We write the dissociation equation for lead nitrate, Pb(NO3)2:
Pb(NO3)2 = Pb2+ + 2NO3-;
By the reaction equation:
C (Pb(NO3)2) = [Pb2+] = 1.6*10-5 M.
C = n/V;
n (Pb(NO3)2) = C*V = 1.6*10-5 M * 1L = 1.6*10-5 mol;
n = m/M;
M (Pb(NO3)2) = 331.2 g/mol;
m (Pb(NO3)2) = n*M = 1.6*10-5 mol * 331.2 g/mol = 0.0053 g;

Answer: 0.0053 g Pb(NO3)2 must be added.