Question

A sample of 1.45 g of lead (II) nitrate is mixed with 129 mL of 0.100M sodium sulfate solution.

Part A: What is the concentration of SO₄^2- ion that remains in solution after the reaction is complete?

Part B: What is the concentration of Na⁺ ion that remains in solution after the reaction is complete?

Part C: What is the concentration of NO₃^- ion that remains in solution after the reaction is complete?

Answer 1

1 mole sodium sulfate gives one mole sulfate ion and 2 moles sodium ions in solution

1 mole lead nitrate gives one mole lead ion and 2 moles nitrate ions in solution.

2- Exus 4:38003 = 8:52x10? mor Ime Pb(NO3) (28) + Na, 504 (ac) – PbSO4 (5) + zaven@, (ae) NO. OF molesMASS OF Pb (403) SO4 = 12.9x10 - of Pb(NO3)2 19w of pb (103) = 10459 Molarity = 8.52x10 mol = 10.0669 331.2 gmoet 129 m2 x 0.0014 = 4.38x10 moe No.of males of Nay s4 = 0.100 mol x 125ml x 0.0011 ® No.of moles of nat = , 1 ml 2 x 12.9 x103 = 25.8x10 mee = 12.9 x 103 mar 3 Molarity = 25.8 x 103 moe Pb (NO3) Anup na, sog REACT IN 1:1 moli . 129mc x 0.001L = 10:20m RATIO. Pb(NO3). Is umiring Renaeos. C NOOF MOLES of Noz = 2x40384 so 4.38 x63 mor of pb (103) REACT WITH 4.38 x 10 mar (out of molarity = 8.76x10 moe. =10.068M 12.8 x 103 ma) of na, sog. Ime . = 8.76x10 mal 12 ml x 0.00ll ime