Question

2. A student mixed 5.00 mL of 0.0120 M Pb(NO,), with 4.00 mL of 0.0300 M KI and 1.00 mL of 0.20 M KNO, and observed the formation of a yellow precipitate. Show all calculations for each of the following questions. a. What is the molecular formula of the precipitate? b. How many moles of Pb2+ are present initially? (moles = M.V) c. How many moles of I- are present initially? d. The concentration of I- at equilibrium is experimentally determined to be 6.5 x 10-M. How many moles of I- are present in the solution (10 mL)?

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Answer 1

+ 2KI PET - Pb(NO3), lag. + 2K I cag., PbIsot + 2k Norag) VY Pb I (moles - M. v) M = 0.020m und se V = 5 ml (0.005 L) = mo moles of Pb2+ = (0.0120 X 0.005) 120 xo.oos = 0.00006 moles 6x 10-3 moles. moles of pbt...c= present Initially c) How many moles of Initially ? I are re present (moles - Mou) M = V = 0.0300 M 4.00ml (0.004 e) moles of Ir present = (0.0300) (0.004) Initialle = 0.000/2 moles

moles of , 4.2x 10-moles. motes of of I I present. presenta Initially & The Concentrations of I- at equilibrium determined to be 6.5x 103 modes M in that oft are are present present How many moles of I solution (10 me) : moles - MV M= 6.5×16-3 M. = lome (0.010 l) I are presen moles of I are present in that folution h moles = (6.5x153) ( 0.010) (e) How many moles of I precipitated ? L moles of I precipitated =(initial numbero moles of I) - (number of moles. I remained in soll

- (1-2 X 16-4) - (6-5410-5) (12-605) x 165 - 315/365 (5,5x109) moles (f) How many moles of Pb2+ remain in solution? m As per chemical equation, a mole Pb2+ reacts with 2 moles of It to give precipitate moles of P627 precipitate = moles of I precipitate - (515x105 Janakes = 1 2.75 X16S moles → number of moles. of p62t remain in Solution =(initial number of moles al p627) (number of moles of pet preupitates , (6x10-5) - (2.1516-moles (6-2.15) x 105 moles 3.25 * 165 moles