Question

A student mixes 41.0 mL of 2.84 M Pb(NO3)2(aq) with 20.0 mL of 0.00235 M Nal(aq). How many moles of Pbla(s) precipitate from the resulting solution? Number K,,[Pb12()] – 9810 mol What are the values of [Pb2+), [1]. [NO3-), and [Na*) after the solution has reached equilibrium at 25 °C? Number Number [Pb?*] = Number Number [no; ]- 0 M [nat)-

Answer 1

Polz (s) + 2 NANO Pb(NO3)2 (aq) 2.84M 410 me + 2 NaI (aq) 0.00235 M 200 ml 116.44 mmol o.ou 7 mmol limiting Regent 0.0235 116.41 mmol - -0.047 Total volume = 61 me Now, soll contain Pb(NO3)2, Polz and Nanog Pb(NO3)2 Pb2+ + 2NO3 116.41 mnol 116.41 monol 232.82 mmal Nat & Nog NaNO₃ - oroua mmol 0.047 mnol 07047 mmol Total moles of No = 232.82 +0.047 = 232.867 mmol Pros) = 232.867 = 3.817 m 1 (woz) = 3.817M 6! Total moles of Nat = 0.047 mmol nat) = 0.047 - - 47 = 0.00077 M [Nat) = 0.00077 M mola of Pb²+ = 116.41 mmol (ignoring P624 from POI) (Pb) = 116,41 - = .908M [Pb²4) = 1.908M - Pb 2+ +27- (springly soluble salt) ksp = 9.8 %70-9 Ksp = 1864] [1-12 [3-] = ksp pbla - 716X10-5 J (Por] 1908 I-= 7.16 8185 m