Question

1. If you started with 1.75g of lithium chloride and added that to 250. mL of a 0.150M solution of lead(II) acetate

   1. Write the reaction

   2. What would be the grams of precipitate produced

   3. Calculate the final molarity of the chloride ion

   4. Calculate the final molarity of the lead ion

Answer 1

Dete (0) 2 Lid (ag) + - (Lithium chloride) Pb (Q H₂O Lead acutate) lag, Dobcz og - Lead chloride +2 Li CC H₂O) Cag) - tithium acutate = 0.0412 mol. mole og Lid = 1.75 g / 42.4 & 1 mol / mod dj Pb (CH 3 ) = 0.150 mx 250x103 L / / = 0.03757 / c. Lice is liniting reagent. molex molar mall. mall of pricipitate (Pbcl) = mole of Psche = mole of Lid Consumed. - mole of Ph . 0.0375 -0.01875 "mals of Phcl, = 0.01875 ml x2 78:19 = 5.21 gram mole of Li (CLH₂ O 2 ) = 0.0375 nolarity = 0.0375 mol = 0.15M 0.250 L [- ] = 0; all casumed [lit] = 2x0.15 m = 0.3M