4.
14.
(E) Let s = solubility of each compound in moles of compound per liter of solution.
bgbg
Br = bg
sb
2 sg= 4 s = 4.0 × 10
F = bg
s bg
3s = 27 s = 8 × 10
AsO
= bgb
3s 2 sg= 108s = 2.1 × 10
(a)
Ksp = Ba 2+ CrO 4
(b)
Ksp = Pb 2+
(c)
Ksp = Ce 3+
(d)
Ksp = Mg 2+
2−
− 2
2
− 3
3
3
3− 2
3
s = 1.1 × 10−5 M
= s s = s2 = 1.2 × 10−10
−5
4
3
s = 2.2 × 10−2 M
−16
2
5
s = 7 × 10−5 M
s = 4.5 × 10−5 M
−20
4
(M) The titration reaction is Ca ( OH )2 ( aq ) + 2HCl ( aq ) ⎯⎯
→ CaCl2 ( aq ) + 2H2O(l)
1L
0.1032 mol HCl 1 mol OH −
×
×
1000 mL
1L
1 mol HCl
= 0.0221 M
1L
50.00 mL Ca(OH) 2 soln ×
1000 mL
10.7 mL HCl ×
⎡⎣OH − ⎤⎦ =
In a saturated solution of Ca ( OH )2 , ⎡⎣Ca 2+ ⎤⎦ = ⎡⎣OH− ⎤⎦ ÷ 2
2
2
Ksp = ⎡⎣Ca 2+ ⎤⎦ ⎡⎣OH − ⎤⎦ = ( 0.0221 ÷ 2 )( 0.0221) = 5.40 ×10 −6 ( 5.5 × 10−6 in Appendix D).
15.
(E) We let s = molar solubility of Mg (OH )2 in moles solute per liter of solution.
(a)
(b)
2
2
K sp = !"Mg 2+ #$!"OH − #$ = ( s ) ( 2s ) = 4s 3 = 1.8 ×10 −11
2+

Equation :
Mg OH s 

 Mg aq
s = 1.7 × 10−4 M
( )()
( )
−
−
−
0.0862 M
+s M
(0.0862 + s ) M
2
Initial :
Changes :
Equil :
( )
+
2OH − aq
≈ 0M
+2s M
2s M
2
2
Ksp = ( 0.0862 + s ) × ( 2s ) = 1.8 ×10−11 ≈ (0.0862 ) × ( 2s ) = 0.34 s 2 s = 7.3 ×10−6 M
(s << 0.0802 M, thus, the approximation was valid.)
(c)
OH − = KOH = 0.0355 M
Equation :
( ) (s )
2+



 Mg aq
−
−
−
0M
+ sM
sM
Mg OH
Initial :
Changes :
Equil :
2
( )
2
+
( )
2OH − aq
0.0355M
+ 2s M
(0.0355 + 2s ) M
2
Ksp = ( s )(0.0355 + 2s ) =1.8 ×10−11 ≈ ( s )(0.0355) = 0.0013 s
20.
()
(M) Equation: CaSO 4 s
Soln:
—




Ca 2+ aq +
( )
SO 2−
aq .
4
0M
0.0025 M
890
( )
s =1.4 ×10−8 M
Chapter 18: Solubility and Complex-Ion Equilibria
Add CaSO4 (s )
—
+x M
Equil:
—
xM
+x M
(0.0025 + x ) M
If we use the approximation that x << 0.0025, we find x = 0.0036. Clearly, x is larger than
0.0025 and thus the approximation is not valid. The quadratic equation must be solved.
2−
Ksp = ⎡⎣Ca 2+ ⎤⎦ ⎡⎣SO4 ⎤⎦ = 9.1×10−6 = x ( 0.0025 + x ) = 0.0025 x + x 2
x 2 + 0.0025 x − 9.1×10−6 = 0
−b ± b2 − 4ac −0.0025 ± 6.3 × 10−6 + 3.6 × 10−5
x=
=
= 2.0 × 10−3 M = CaSO4
2a
2
−3
2.0 × 10 mol CaSO 4 136.1 g CaSO 4
mass CaSO4 = 0.1000 L ×
×
= 0.027 g CaSO 4
1 L soln
1 mol CaSO 4
28.
(E) The solutions mutually dilute each other.
155 mL
⎡⎣Cl− ⎤⎦ = 0.016 M ×
= 6.2 ×10−3 M
155 mL + 245 mL
245 mL
⎡⎣ Pb 2+ ⎤⎦ = 0.175 M ×
= 0.107 M
245 mL +155 mL
Then we compute the value of the ion product and compare it to the solubility product constant
value.
2
2
Qsp = ⎡⎣ Pb 2+ ⎤⎦ ⎡⎣Cl− ⎤⎦ = ( 0.107 ) ( 6.2 ×10−3 ) = 4.1×10−6 < 1.6 ×10−5 = K sp
Thus, precipitation of PbCl2 (s) will not occur from these mixed solutions.
36.
(M) [Ag+]diluted = 0.0208 M × 175 mL = 0.008565 M
425 mL
250
mL = 0.02235 M
[CrO4 ]diluted = 0.0380 M ×
425 mL
2−
2
2−
Qsp = ⎡⎣Ag + ⎤⎦ ⎡⎣CrO4 ⎤⎦ = 1.6 ×10−6 > Ksp
Because Qsp > Ksp, then more Ag2Cr2O4 precipitates out. Assume that the limiting reagent is
used up (100% reaction in the reverse direction) and re-establish the equilibrium in the reverse
direction. Here Ag+ is the limiting reagent.
Ag2CrO4(s)
2 Ag+(aq) + CrO42 (aq)
K
=1.1 × 10-12
sp
(Ag CrO )
2
4





Initial
0.008565 M
0.02235 M
−
Change
x = 0.00428 M
−
−2x
−x
100% rxn
0
0.0181
−
Change
re-establish equil
+2y
+y
−
Equil
(assume y ~ 0)
2y
0.0181 + y
−
−
1.1 × 10 12 = (2y)2(0.0181+ y) ≈ (2y)2(0.0181) y = 3.9 × 10 6 M
−
−
891
Chapter 18: Solubility and Complex-Ion Equilibria
y << 0.0181, so this assumption is valid.
2y = [Ag+] = 7.8 × 10 6 M after precipitation is complete.
7.8 × 10−6
% [Ag+]unprecipitated =
× 100% = 0.091% (precipitation is essentially quantitative)
0.00856
−
42.
(D) Normally we would worry about the mutual dilution of the two solutions, but the values
of the solubility product constants are so small that only a very small volume of 0.50 M
Pb(NO3)2 solution needs to be added, as we shall see.
(a)
Since the two anions are present at the same concentration and they have the same type
of formula (one anion per cation), the one forming the compound with the smallest Ksp
value will precipitate first. Thus, CrO 4 2− is the first anion to precipitate.
(b)
At the point where SO 4 2− begins to precipitate, we have
1.6 ×10−8
2−
2+
−8
2+
2+
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤
Ksp = ⎣ Pb ⎦ ⎣SO4 ⎦ = 1.6 ×10 = ⎣Pb ⎦ ( 0.010M ) ; ⎣Pb ⎦ =
= 1.6 ×10 −6 M
0.010
Now we can test our original assumption, that only a very small volume of 0.50 M
Pb(NO3)2 solution has been added. We assume that we have 1.00 L of the original
solution, the one with the two anions dissolved in it, and compute the volume of 0.50 M
Pb(NO3)2 that has to be added to achieve Pb2+ = 1.6 × 10−6 M.
Vadded = 1.00 L ×
1.6 ×10−6 mol Pb2+ 1 mol Pb ( NO3 )2
1 L Pb2+soln
×
×
1 L soln
1 mol Pb 2+
0.50 mol Pb ( NO3 )2
Vadded = 3.2 ×10−5 L Pb 2+soln = 0.0032mL Pb 2+soln
This is less than one drop (0.05 mL) of the Pb2+ solution, clearly a very small volume.
(b)
The two anions are effectively separated if Pb2+ has not reached 1.6 × 10−6 M when
CrO 4
2−
is reduced to 0.1% of its original value, that is, to
2−
0.010 ×10−3 M = 1.0 ×10−5 M = ⎡⎣CrO 4 ⎤⎦
Ksp = Pb 2+ CrO 4
2−
c
= 2.8 × 10−13 = Pb 2+ 1.0 × 10−5
h
2.8 ×10−13
⎡⎣ Pb2+ ⎤⎦ =
= 2.8 ×10−8 M
−5
1.0 ×10
Thus, the two anions can be effectively separated by fractional precipitation.
46.
(E) In each case we indicate whether the compound is more soluble in base than in water. We
write the net ionic equation for the reaction in which the solid dissolves in base. Substances are
892
Chapter 18: Solubility and Complex-Ion Equilibria
soluble in base if either (1) acid-base reaction occurs [as in (b)] or (2) a gas is produced, since
escape of the gas from the reaction mixture causes the reaction to shift to the right.
Water:
BaSO4 is less soluble in base, hydrolysis of SO 4 2− will be repressed.
Base: H 2C2O4 (s ) + 2OH − ( aq ) ⎯⎯
→ C2O 4
2−
(aq ) + 2H 2O(l)
Water: Fe(OH)3 is less soluble in base because of the OH − common ion.
Same: NaNO 3 (neither Na+ nor NO3- react with H2O to a measurable extent).
Water: MnS is less soluble in base because hydrolysis of S2− will be repressed.
57.
(M) We first compute the free Ag + in the original solution. The size of the complex ion
formation equilibrium constant indicates that the reaction lies far to the right, so we form as
much complex ion as possible stoichiometrically.
Equation:
Ag+ (aq) +
In soln:
Form complex:
0.10 M
−0.10 M
0M
+x M
x M
Changes:
Equil:
b g bg
à
2NH3 (aq)
+
Ag NH 3 2 aq
0M
+0.10 M
0.10 M
−x M
0.10 − x M
1.00 M
−0.20 M
0.80 M
+2x M
0.80 + 2x M
b
b
g
g
+
⎡ ⎡ Ag NH ⎤ ⎤
3 )2 ⎦ ⎥
⎢⎣ ⎣ (
0.10 − x
0.10
⎦ =
Kf = 1.6 ×10 =
≈
2
2
2
+ ⎡
⎤
⎡ Ag ⎤ ⎣⎢ NH3 ⎦⎥
x ( 0.80 + 2 x )
x ( 0.80)
⎣
⎦
7
x=
0.10
7
2
1.6 ×10 ( 0.80)
= 9.8 ×10−9 M .
(x << 0.80 M, thus the approximation was valid.)
Thus, ⎡⎣Ag+ ⎤⎦ = 9.8 ×10−9 M. We next determine the I − that can coexist in this solution
without precipitation.
8.5 ×10−17
⎡⎣I ⎤⎦ =
Ksp = ⎡⎣ Ag ⎤⎦ ⎡⎣ I ⎤⎦ = 8.5 ×10 = (9.8 ×10 ) ⎡⎣I ⎤⎦ ;
= 8.7 ×10−9 M
−9
9.8 ×10
Finally, we determine the mass of KI needed to produce this I −
+
−
−17
−9
−
−
8.7 ×10−9 mol I− 1mol KI 166.0 g KI
mass KI = 1.00 L soln ×
×
×
= 1.4 ×10−6 g KI
−
1 L soln
1mol I
1mol KI
60.
(M) Since the cation concentrations are identical, the value of Qspa is the same for each one. It
is this value of Qspa that we compare with Kspa to determine if precipitation occurs.
Qspa
⎡⎣M2+ ⎤⎦ [H2S] 0.05M × 0.10 M
=
=
= 5 ×101
2
2
( 0.010 M )
⎡⎣ H3O+ ⎤⎦
893
Chapter 18: Solubility and Complex-Ion Equilibria
If Qspa > Kspa , precipitation of the metal sulfide should occur. But, if Qspa < Kspa , precipitation
will not occur.
For CuS, Kspa = 6 × 10−16 < Qspa = 5 × 101
Precipitation of CuS(s) should occur.
For HgS, Kspa = 2 × 10−32 < Qspa = 5 × 101
Precipitation of HgS(s) should occur.
For MnS, Kspa = 3 × 107 > Qspa = 5 × 101
Precipitation of MnS(s) will not occur.
894