Solutions:
2001 A
By:
I Schnizzle
2001 A Part A (i)
A.) At 10 degrees Celsius, 8.9 e-5 g. of
AgCl(s) will dissolve in 100. mL of water.
(i) Write the equation for the dissociation
of AgCl(s) in water.
Since AgCl is a solid the dissociation will
be fairly simple. Basically just breaking the
compound down into its ions.
AgCl(s) yields Ag+(aq) + Cl-(aq)
2001 A Part A (ii)
A.) At 10 degrees Celsius, 8.9 e-5 g. of
AgCl(s) will dissolve in 100. mL of water.
(ii) Calculate the solubility, in mol L-1, of
AgCl(s) in water at 10 degrees Celsius.
Convert the number of g. into moles and
than divide by the number of liters and you
will find the solubility.
2001 A Part A (ii) Cont’
8.9e-5 x 1mol/ 143.32 (molar mass)=
6.20988e-7
Than take that number of moles and divide
by .1 since you have 100 mL and need the
number in liters.
6.20988e-7/.1= 6.20988e-6 mol/L
2001 A Part A (iii)
A.) At 10 degrees Celsius, 8.9 e-5 g. of AgCl(s)
will dissolve in 100. mL of water.
(iii) Calculate the value of the solubility product
constant, Ksp, for AgCl(s) at 10 degrees Celsius.
Since the Ksp of AgCl is the products over the
reactants and no pure solids or liquids are
included the Ksp value is obtained simply by
multiplication of the solubility of AgCl found in
part ii.
2001 A Part A (iii) Cont’
Ksp= [Ag+][Cl-]
Ksp=[6.20988e-6]^2=3.9e-11
You square the value since the solubility of
the AgCl applies to both the Ag+ and Clions.
And that gets you your Ksp.
2001 A Part B (i)
B.) At 25 degrees Celsius the value of Ksp for
PbCl2(s) is 1.6e-5 and the value for Ksp for
AgCl(s) is 1.8e-10.
(i) If 60.0 mL of 0.0400 M NaCl(aq) is added to
60.0 mL of 0.0300 M Pb(NO3)2(aq), will a
precipitate form? Assume that volumes are
additive. Show calculations to support your
answer.
To answer this question we need to compare the
given Ksp to the one we find.
2001 A Part B (i) Cont’
First we need to find the molarity of the Pb+ and
Cl60mL/(120mL) x .0400mol/1000mL=.0200 M[Cl-]
60mL/(120mL) x
.0300mol/1000mL=.0150M[Pb2+]
Than we use these molarities to find the trial
Ksp.
[Pb2+][Cl-]=(.0150)^2(.0200)=6.0e-6
Since the trial Ksp is less than the actual Ksp a
precipitate will not form.
6.0e-6 is less than 1.6e-5
2001 A Part B (ii)
B.) At 25 degrees Celsius the value of Ksp for
PbCl2(s) is 1.6e-5 and the value for Ksp for
AgCl(s) is 1.8e-10.
(ii) Calculate the equilibrium value of [Pb2+(aq)]
in 1.00 L of saturate PbCl2 solution to which
0.250 mole of NaCl(s) has been added. Assume
that no volume change occurs.
To solve this problem all that is required is some
manipulation of the information you already
have.
2001 A Part B (ii) Cont’
Since the Ksp is equal to [Pb2+][Cl-]^2 the
only step is to divde the Ksp by the [Cl-]^2
concentration and you have your Ksp for
[Pb2+].
[Pb2+]=Ksp/[Cl-]^2
[Pb2+]=1.6e-5/(.250M)^2
Since the .250 mol is in 1L it can be used
as the molarity
[Pb2+]= 2.6e-4 M
2001 A Part B (iii)
B.) At 25 degrees Celsius the value of Ksp for
PbCl2(s) is 1.6e-5 and the value for Ksp for
AgCl(s) is 1.8e-10.
(iii) If 0.100 M NaCl(aq) is added slowly to a
beaker containing both 0.120 M AgNO3(aq) and
0.150 M Pb(NO3)2(aq) at 25 degrees Celsius,
which will precipitate first, AgCl(s) or PbCl2(s)?
Show calculations to support your answer.
The first thing that is necessary to complete this
question is to find the concentrations of Cl- in
each of the PbCl2 and AgCl compounds.
2001 A Part B (iii) Cont’
For AgCl the Cl- concentration is Ksp/[Ag+] so
1.8e-10/.120= 1.5e-9M
You get the .120 concentration of Ag+ from the
amount in the beaker from AgNO3(aq) since it
dissociates completely.
Next is the PbCl2, the concentration is the sqrt.
of the Ksp/[Pb2+] since there are two Cl’s for
every one Pb2+ that’s why it is sqrt.
So it is sqrt. of 1.6e-5/.150=.0103M
The .150 is the concentration of Pb2+ from the
Pb(NO3)2 since the NO3 dissociates completely
like in AgNO3.
2001 A Part B (iii) Cont’
Since the concentration of the AgCl
needed to form a precipitate is so much
less than that of the PbCl2 it will form the
precipitate before the PbCl2 does.
Gangluff rocks that CS
Baby!!!
Shout out to my CS boys Danny,
Andy, Mikey, and Bhavih…
Hi to you to Dave.