2. A precipitate of Pbl2 is formed when S mL of 0.012 M Pb(NO,)2 is added to 5 m
ID: 712713 • Letter: 2
Question
2. A precipitate of Pbl2 is formed when S mL of 0.012 M Pb(NO,)2 is added to 5 mL of 0.030 MKI. It was found experimentally that the final concentration of r in the 10 mL solution after equilibrium was 8 x 103M a) How many moles of Pb2 are originally in the solution? b) How many moles of I are originally present? c) How many moles of I remained in solution after equilibrium? d) How many moles of I precipitated? e) How many moles of PbI2 precipitated? f) How many moles of Pbl2 remain in solution at equilibrium? g) What is [Pb'] at equilibrium? h) What is Kp for the formation of Pbl,?Explanation / Answer
a) Pb(NO3)2 (s) ----> Pb2+(aq) + 2 NO3-(aq)
According the stoichiometry 1 mole of Pb(NO3)2 gives 1 mole of Pb+2
moles of Pb+2 originally in the solution = 0.012 mol/L * (5/1000)L = 6*10-5 mol
b) KI (s) ----> K+ (aq) + I- (aq)
one mole of KI gives 1 mole of I-
moles of I- originally in the solution = 0.030 mol/L * (5/1000)L = 1.5*10-4 mol
c) moles of I- remained in solution after equilibrium = 8*10-3mol/L * (10/1000)L = 8*10-5 mol
d) .Pb2+ (aq) + 2I- (aq) ------> PbI2 (s)
2 moles of I- react with 1 mole of Pb+2
6*10-5 mol of Pb+2 * ( 2 moles of I- / 1 mol of Pb+2 ) = 1.2*10-4 mol
e) Change in the moles of I- will be 2X according to the reaction stoichiometry,
here X is moles of PbI2 at equilibrium
moles of I- unreacted = 1.5*10-4 mol - 1.2*10-4 mol = 3*10-5 mol
At equilibrium I- moles will be = 3*10-5 + 2X
But there are 8*10-5 mol of I- after equilibrium so 3*10-5 + 2X = 8*10-5
X = 2.5*10-5 mol
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