Data given - PbI 2 (s) <===> Pb 2+ (aq) + 2I - (aq) ( Ksp = [Pb 2+ ][I - ] 2 ) 5

Question

Data given -
PbI2(s) <===> Pb2+(aq) + 2I- (aq) ( Ksp = [Pb2+][I-]2 )

5.0mL 0.0120 M Pb(NO3)2
2.0 mL 0.0300 M KI
3.0 mL 0.20 M KNO3
Total volume = 10 mL
Absorbance of solution = 0.280
[ I- ] in moles/liter at equilibrium = 4.2

CALCULATE -
a) The initial no. of moles Pb2+ Answer
b) The initial no. of moles I-
c) No. of moles I- at equilibrium
d) No. moles I- precipitated
e) No. moles Pb2+ precipitated
f) No. moles Pb2+ at equilibrium
g) [Pb2+] at equilibrium
h) Ksp PbI2

Answers of a to f are all raised to 10-5 (show calculations)

Explanation / Answer

a) The initial no. of moles Pb2+

mol of Pb+2 = M1*V1 = 0.012 * 5*10^-3 = 0.00006 mol of Pb+2

b) The initial no. of moles I-

mol of I- = M2*V2= 0.03 *2*10^-3 = 0.00006 mol of I-

c) No. of moles I- at equilibrium

in equilibrium:

[ I- ] in moles/liter at equilibrium = 4.2

V =10 mL = 10*10^-3 = 0.01 L

mol = MV = 4.2*10^-5 *0.01 = 4.2*10^-7 mol of I-

d) No. moles I- precipitated

difference --> (0.00006 -  4.2*10^-7) =0.00005958 mol of I-

e)

ratio is 2 mol of I- = 1 mol of PB+2 --< 1/2*0.00005958 = 0.00002979 mol of Pb+2

f)

Pb+2 in eq =0.00006 - 0.00002979 = 0.00003021 mol of Pb+2

g)

[Pb+2] = mol/V = (0.00003021)/(10*10^-3) = 0.003021 M

h)

Ksp = [Pb+2][I-]^2 = (0.003021)(4.2*10^-5)^2

Ksp = 5.329*10^-12

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