Question

soluble blz dissociah. - A student mixed 5.00 mL of 0.0120 M Pb(NO), with 5.00 mL of 0.0300 M KI and observed a yellow precipitate. Show all calculations for each of the following questions. a. What is the molecular formula of the precipitate? 0.000mlm Co.0120) (-005) b. How many moles of Pbare present initially? (moles = M.V) c. How many moles of l' are present initially? 0.00016 moleso.03)00-05) . The concentration of l' at equilibrium is experimentally determined to be 8.0 X 10-'M. How many moles of I are present in the solution (10 mL)? 0.oooos mold Pain +7X1N03)

Answer 1

a) Pb(NO3) + KI Coco (aq) Pb I V + 2 KNO₂ 5 (aq) molecular formula of precipitate: PbI, b) 0.0120 m x5.00ml x 10²2 imt 0.00006 mol. - mo X S.00 0.000 a LX 0.0300 mol - 9 0.0300 mel x5.00 hlx mol = 0.0015 mol d) 8.0 x103 mol - in 1 litre x - in 10ml x 1032 8.0 x 103 mol x 10RY = 8x105 mol = 0.00008 mol. moles of I precipitated = initial number of moles of I - number of i moles of I remained in solution = 0.00015 mol - 0.00008 mol = 0.00007 mol.

f) As per chemical equation, Imol phat reacts with 2 moles of I to give precipitate moles of I precipitated moles of ph2+ precipitated = ? 0.00007 = = 0.000035 mol number of moles of pb2+ remain in solution ini member of moles of p624 - number of moles, of phat precipitated = 0.00006 - 0.000035 = 0.000025 mol

g) concentration /molarity = I number of moles volume (L) number of moles of pb2t present in solution= 0.000025 mol volume = 10ml = 10 ml x 1032 - 0²L molarity = 2.5x 10 mol = 2.5 x 103 m. 1o L h) solubility, 5= 25 x 103 m Ksp = 48² = 4 (2.5 x103) 3 = 6.25 X108 m