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
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Expert's answer
2017-06-20T09:20:10-0400
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;
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