SOLUBILITY PRODUCT

Question

唔該各位大佬幫幫忙..

Q1: if 100cm3 of a 0.040M AgNO3 solution is added to 50cm3 of a 0.050M NaCl solution,will a precipitate form?justify ur ans by showing ur calculations.

Q2: will a precipitate exist at equilibrium if 30cm3 of 0.004M Pb(NO3)2 solution is mixed 70cm3 of 0.01M FeCl3 solution?The Ksp for PbCl2 is 1.7*10^-5 mol^3 dm^-9.

THX A LOT!~~~~

Answer 1

1.
Ksp of AgCl = 1.76 x 10-10
Total final volume = 100 + 50 = 150 cm^3
[Ag+] = 0.04X(100/1000) / (150/1000) = 0.02667 M
[Cl-] = 0.05X(50/1000) / (150/1000) = 0.01667 M
[Ag+][Cl-] = (0.02667)(0.01667) = 4.446 x 10^-4 > Ksp of AgCl ; ppt of AgCl will be formed.
2. Total final volume = 30 + 70 = 100 cm^3
[Pb2+] = 0.004X(30/1000) / (100/1000) = M
[Cl-] = 3 X 0.01 X (70/1000) / (100/1000) = 0.021 M
[Pb2+][Cl-]^2 = 0.0012 X (0.021)^2 = 5.292 x 10^-7 < Ksp of PbCl2 ; No ppt will be formed.
5 分好似小左D！

Answer 2

Q1. no. of mole of Ag+(aq) present in AgNO3
=100/1000x0.04
=0.004mol
concentration of Ag+
=0.04x1000/(100+50)
=0.267M

no.of mole of Cl-(aq) present in NaCl
=50/1000x0.05
=0.0025mol
concentration of Cl-
=0.0025x1000/(50+100)
=0.0167M

(Ag+)(Cl-)=0.267x0.0167
=0.00446M^2

then you can compare the value of Ksp, if the value you calculate is greater than Ksp, the precipitate is expexted to be formed.

Q2. no.of mole of Pb2+ present in Pb(NO3)2
=30/1000x0.004
=0.00012mol
concentration of Pb2+
=0.00012x1000/(30+70)
=0.0012M

no of mole of Cl- present in FeCl3
=70/100x0.01
=0.0007mol
concentration of Cl-
=0.0007x1000/(30+70)
=0.007M

(Pb2+)(Cl-)^2=0.0012x0.007^2
=5.88x10^ -10M^3

the precipitate will not exist. as the value is smaller than the Ksp