Fullerton College Office of Special Programs
SI Review 111B
Handout #8 Solutions
1. The balanced equations and expressions for Ksp are:
a.
BaSO4(s)
b.
PbBr2(s)
c.
Ag2CrO4(s)
Ba2+(aq) + SO42-(aq)
Pb2+(aq) + 2Br-(aq)
2Ag+(aq) + CrO42-(aq)
2-
Ksp  [Ba2 ][SO4 ]
Ksp  [P b2 ][Br- ]2
2-
Ksp  [A g ]2[C rO4 ]
2. To determine Ksp, we write the solubility product expression in terms of a variable “x”, and then
substitute the molar solubility for the variable in the expression:
a.
BaCrO4(s)
“x”
I
C
E
Ba2+(aq)
0
+x
x
+ CrO42-(aq)
0
+x
x
x = molar solubility of barium sulfate = 1.08 x 10-5 mol / L
2-
Ksp  [Ba2 ][C rO4 ]  (x)(x)  x2  (1 .0 8x 1 0-5)2  1 .1 7x 1 0-10
b.
Ag2SO3(s)
“x”
I
C
E
2Ag+(aq)
0
+2x
2x
+ SO32-(aq)
0
+x
x
4 .5 9x 1 0-3 g 1 mol A g2SO3
1 .551 x 1 0-5 mol
x
x
L
2 9 5 .8 7g
L
x = molar solubility of silver(I) sulfite = 1.551 x 10-5 mol / L
2-
Ksp  [A g ]2[SO3 ]  (2 x)2(x)  4 x3  4 (1 .551 x 1 0-5)3  1 .4 9x 1 0-14
c.
CuCl(s)
“x”
I
C
E
Cu+(aq)
0
+x
x
+ Cl-(aq)
0
+x
x
3 .9 1mg
1 0 0 0mL
1g
1 mol C uC l
3 .949 x 1 0-4 mol
x
x
x
x
1 0 0 .0mL
1L
1 0 0 0mg
9 9 .0 0g
L
x = molar solubility of copper(I) chloride = 3.949 x 10-4 mol / L
Ksp  [C u ][C l-]  (x)(x)  x2  (3 .949 x 1 0-4)2  1 .5 6x 1 0-7
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3. For most solubility problems, it is best to begin by constructing an ICE table. From this table, it is
straightforward to write out the solubility product expression.
Ag2C2O4(s)
“x”
2Ag+(aq)
0
+2x
2x
I
C
E
+ C2O42-(aq)
0
+x
x
2-
Ksp  [A g ]2[C2O4 ]  (2 x)2(x)
To determine the value of “x”, we rely on the information given in the problem. The equilibrium
concentration of Ag+ is 2.2 x 10-4 M. So, from the ICE table…
[A g ]  2 x

x
[A g ] 2 .2 x 1 0 - 4 M

 1 .1 x 1 0-4 M
2
2
And,
Ksp  (2 x)2(x)  4 x3  4 (1 .1x 1 0-4)3  5 .3 x 1 0-12
4. To calculate the molar solubility, we begin by constructing an ICE table:
a.
In pure water, the initial concentration of barium and fluoride ions will be 0 M.
BaF2(s)
“x”
Ba2+(aq)
0
+x
x
I
C
E
+ 2F-(aq)
0
+2x
2x
x = molar solubility of BaF2
Ksp  [Ba2 ][F- ]2  (x)(2 x)2  4 x3  2 .4 x 1 0-5

b.
x  1 .8 x 1 0-2 M
(x is the molar s olubility
!)
In 0.10 M Ba(NO3)2, the initial concentration of barium and fluoride ions are 0.10 M and 0 M,
respectively.
BaF2(s)
“x”
I
C
E
Ba2+(aq)
0.10
+x
x + 0.10
+ 2F-(aq)
0
+2x
2x
x = molar solubility of BaF2
2
Ksp  [Ba2 ][F-]2  (x  0 .1 0 )(2 x)
 2 .4 x 1 0-5
To solve this cubic equation, you will want to use the “Small K” approximation. In using the
“Small K” approximation, we ignore any change (x) when added to, or subtracted from, a nonzero
concentration.
2
Ksp  [Ba2 ][F- ]2  (x  0 .1 0 )(2 x)
 2 .4 x 1 0-5

x  7 .74 x 1 0-3 M

2
(0 .1 0 )(2 x)
 2 .4 x 1 0-5
Before we can trust this value, we must check our approximation using the “5% Rule”.
% c hanged
amount c hanged
7 .74 x 1 0-3
(1 0 0 % )
(1 0 0 % ) 7 .7 % 5 % !
originalamount
0 .1 0
Because the percent change (dissociated) is greater than 5%, our initial approximation is not
valid. In this case, we will want to continue the problem with “successive approximations”.
We plug the value that we just solved for back into the original expression, and then solve for “x”
again. This process is repeated until we obtain two successive calculations provide the same
value (within the proper number of significant figures).
2
(7 .74 x 1 0-3  0 .1 0 )(2 x)
 2 .4 x 1 0-5

c.
0 .430 x2  2 .4 x 1 0-5

0 .429 x2  2 .4 x 1 0-5
x2  7 .46 x 1 0-3 M
2
(7 .46 x 1 0-3  0 .1 0 )(2 x)
 2 .4 x 1 0-5


x3  7 .47 x 1 0-3 M  7 .5 x 1 0-3 M
(x is the molar s olubility
!)
In 0.15 M NaF, the initial concentration of barium and fluoride ions are 0 M and 0.15 M,
respectively.
BaF2(s)
“x”
I
C
E
Ba2+(aq)
0
+x
x
+ 2F-(aq)
0.15
+2x
2x + 0.15
x = molar solubility of BaF2
Ksp  [Ba2 ][F-]2  (x)(2 x  0 .1 5 )2  2 .4 x 1 0-5
To solve this cubic equation, you will want to use the “Small K” approximation. In using the
“Small K” approximation, we ignore any change (x) when added to, or subtracted from, a nonzero
concentration.
2
Ksp  [Ba2 ][F- ]2  (x  0 .1 0 )(2 x)
 2 .4 x 1 0-5


(x)(0 .1 5 2)  2 .4 x 1 0-5
x  1 .06 x 1 0-3 M
Before we can trust this value, we must check our approximation using the “5% Rule”.
% c hanged
amount c hanged
2 (1 .0 6x 1 0-3)
(1 0 0 % )
(1 0 0 % ) 1 .4 % 5 % !
originalamount
0 .1 5
Because the percent change (dissociated) is less than 5%, our initial approximation is valid. We
can now report the molar solubility…
x  1 .1 x 1 0-3 M
(x is the molar s olubility
!)
5. To begin this precipitation problem, we first calculate the new concentrations of all ions as a result
of mixing the two solutions; the new volume is 175.0 mL + 145.0 mL = 320.0 mL. We will do this
using the dilution equation…
Mf Vf  MiVi

Mf 
MiVi
Vf
(1 7 5 .0mL )(0 .0 0 5 5M KC l)
 0 .0 0 300 M KC l
3 2 0 .0mL

(1 4 5 .0mL )(0 .0 0 1 5M A gN O3 )
 0 .0 0 0 6
79 M A gN O3
3 2 0 .0mL
0 .0 0 300 M C l-

0 .0 0 0 6
79 M A g
To determine whether or not a precipitate will form, we must calculate the ion product, Q.
Qsp  [A g ][C l-]  (0 .0 0 0 6
79 )(0 .0 0 3
00 )  2 .0 x 1 0-6  1 .6 x 1 0-10  Ksp
Because the ion product is larger than the solubility product, the solution is said to be
supersaturated. The precipitate of silver chloride will form, while the remaining ions will remain in
solution as spectator ions.
Next, we will assume complete reaction and calculate the concentration of the excess ion. This is
easily accomplished with an “amounts table”. The form of the table will be similar to the ICE
table, only now we will assume complete reaction.
(Note: We can only use molarities if we have taken the time to calculate the new concentrations.)
Ag+(aq)
0.000679
-0.000679
0
+ Cl-(aq)
0.00300
-0.000679
0.002321

I
C
F
AgCl(s)
Once the concentration of the excess ion is known, we can determine the concentration of the
silver ion by considering the shift back towards equilibrium:
AgCl(s)
I
C
E
Ag+(aq)
0
+x
x
+ Cl-(aq)
0.002321
+x
x + 0.002321
Ksp  [A g ][C l-]  (x)(x  0 .0 0 232 1 ) 1 .6 x 1 0-10
To solve this quadratic equation, we can use the “Small K” approximation:
(x)(x  0 .0 0 232 1 ) 1 .6 x 1 0-10

(x)(0 .0 0 2
32 1 ) 1 .6 x 1 0-10

x  6 .89 x 1 0-8
Before we can trust this value, we must check our approximation using the “5% Rule”.
% c hanged
amount c hanged
6 .89 x 1 0-8
(1 0 0 % )
(1 0 0 % ) 0 .0 0 3 0 % 5 % !
originalamount
0 .0 0 232 1
Because the percent change (dissociated) is less than 5%, our initial approximation is valid. We
can now report the concentration of the silver ion…
[A g ]  x  6 .9 x 1 0-8 M
[C l- ]  x  0 .0 0 232 1  0 .0 0 2 3M
6.
a.
To determine which cation will precipitate first, we will want to calculate the maximum
concentration of sulfate that can be found before precipitation occurs:
2-
2-
Ksp  [Ba2 ][SO4 ]  (0 .0 1 0M )[SO4 ]  1 .5 x 1 0-9
2-
2-
Ksp  [A g ]2[SO4 ]  (0 .0 2 0M )2[SO4 ]  1 .4 x 1 0-5


2-
[SO4 ]max  1 .5 x 1 0-7 M
2-
[SO4 ]max  3 .5 x 1 0-2 M
Given that barium has a lower “maximum sulfate concentration”, we would expect barium
sulfate to precipitate first.
Barium sulfate should be expected to precipitate when the concentration of sodium sulfate
exceeds 1.5 x 10-7 M.
b.
The second cation, silver, will begin to precipitate when the concentration of sulfate begins to
exceed 3.5 x 10-2 M. Given this concentration of sulfate, we can calculate the concentration of
barium remaining…
2-
Ksp  [Ba2 ][SO4 ]  [Ba2 ] (3 .5 x 1 0-2)  1 .5 x 1 0-9

[Ba2 ]  4 .3 x 1 0-8 M