CHEM 116-Dr. Babb’s Sections
Answer Key to Exam III Lecture Problem Sheet
146.
In general, any factor that weakens the bond to hydrogen, makes the substance more acidic.
For binary acids within the same group such as HCl, HF, HI, and HBr, the atom bonded to hydrogen (F, Cl, Br, I) becomes larger as proceed down the periodic table and this tends to weaken the bond to H. Thus, binary acids become more acidic as proceed from top to bottom with a column. Most acidic: HI > HBr > HCl > HF :Least acidic; Most acidic: H
2
Te
> H
2
S > H
2
O :Least acidic.
147.
For binary acids within the same row such as CH
4
, NH
3
, H
2
O, and HF, the atom the hydrogen is bonded to (C, N, O and F) becomes more electronegative as proceed across the row and this tends to weaken the bond to H. Thus these binary acids become more acidic as proceed from left to right within a row. Most acidic: HF > H
2
O > NH
3
> CH
4
:Least acidic.
148.
In oxyacids, the H is always bonded to one of the oxygens. Any factor that weakens the O-H bond, makes the substance more acidic. Within an oxyacid series such as HClO, HClO
3
,
HClO
2
and HClO
4
, as the number of oxygens bonded to the central atom increases, the oxidation number of the central atom increases causing a weakening of the O-H bond strength and an increase in the acidity. Most acidic: HClO
4
> HClO
3
> HClO
2
> HClO.
For a series of oxyacids with the same number of oxygens but with different central atoms such as HOBr, HOCl, HOI, as the electronegativity of the central atom increases, the O-H bond strength weakens and the acidity increases. Most acidic: HOCl > HOBr > HOI :Least acidic.
149.
Compare strengths of acids on reactant and product sides. The stronger acid has a greater tendency to give up its proton; therefore, the equilibrium will lie on the opposite side from the strongest acid.
A.
HBr is a stronger acid than HF. Eq lies on side opposite the HBr (i.e. on reactant
B.
side); thus, [Reactants] > [Products] and K < 1.
H
2
O is a stronger acid than NH
3
. Eq lies on side opposite the H
2
O (i.e. on product side); thus, [Products] > [Reactants] and K > 1.
C.
D.
HClO
2
is a stronger acid than HClO. Eq lies on side opposite the HClO
2
(i.e. on reactant side); thus, [Reactants] > [Products] and K < 1.
H
3
O + is a stronger acid than H
2
O. Eq. lies on side opposite the H
3
O + (i.e. on the reactant side); thus, [Reactants] > [Products] and K < 1.
E.
HBrO
3
is a stronger acid than HIO
3
. Eq. lies on side opposite the HBrO
3
(i.e. on the product side); thus, [Products] > [Reactants] and K > 1.
154.
As solid NaC
2
H
3
O
2
is added to 0.500 M HC
2
H
3
O
2
, the [H
3
O + ] decreases and the pH increases.
Why? Look at effect of addition of NaC
2
HC
2
NaC
H
2
3
H
O
2
(aq) + H
2
3
O
2
O(l) <–> C
2
H
3
(s) –> Na + (aq) + C
2
H
3
H
O
O
2
2
3
-
O
2
, a strong electrolyte, on the acetic acid eq.
(aq) + H
(aq)
3
O
+
(aq)
C
2
H
3
O
2
As NaC
2
H
3
O
2
is added to an eq. mixture of 0.500 M HC
2
H
3
O
2
, the common ion
is added which shifts the acetic acid eq. to the left decreasing the [H
3
O + ] and increasing the pH.
Addition of NaCl to an eq. mixture of 0.500 M HC
2
H
3
O
2
has no effect on the pH or acetic acid eq. because NaCl has no ion in common with the acetic acid eq.

155.
As solid NH
4
Cl is added to 0.500 M NH
3
, the [OH
-
] decreases, the [H
3
O
+
] increases and the pH decreases. Why? Look at effect of addition of NH
4
Cl, a strong electrolyte, on the ammonia eq.
NH
3
(aq) + H
2
NH
4
O(l) <–> NH
4
Cl(s) –> NH
4
+
(aq) + Cl
+ (aq) + OH (aq)
-
(aq)
As NH
4
Cl is added to an eq. mixture of 0.500 M NH which shifts the ammonia eq. to the left decreasing the [OH
3
-
, the common ion NH
4
+ is added
] and decreasing the pH.
Addition of KNO
3
to an eq. mixture of 0.500 M NH
3
has no effect on the pH or ammonia eq. because KNO
3
has no ion in common with the ammonia eq.
156.
For a 0.500 M HC
2
H
3
O
2
/0.250 M NaC
2
H
3
O
2
: pH = 4.44 and % Ionization = 0.0072%.
For a 0.500 M HC
2
H
3
O
2
: pH = 2.52 and % Ionization = 0.60%.
157.
Buffer solution: A solution consisting of a weak acid and the conjugate base (or salt) of the weak acid. OR A solution consisting of a weak base and the conjugate acid (or salt) of the weak base.
Function of Buffer: Acts to maintain the pH of the solution at a relatively constant value even with the addition of strong base or strong acid.
If the following substances are mixed in equimolar amounts, will a buffer solution be formed?
A.
B.
C.
HF and NaF; Yes
HC
2
H
3
O
2
and NaC
2
H
3
O
2
; Yes
HCN and NaCN; Yes
D.
E.
F.
G.
NH
3
and NH
4
Cl; Yes
HCl and KCl; No
NaOH and NaCl; No
HF and NaOH; No
C.
D.
E.
F.
H.
NH
3
and HCl; No
If equal volumes of the following solutions are mixed, will a buffer solution result?
A.
B.
0.2 M HF and 0.1 M NaOH; Yes
0.2 M HF and 0.3 M NaOH; No
0.2 M HF and 0.2 M NaOH; No
0.2 M NaF and 0.1 M HCl; Yes
0.5 M NH
3
and 1.0 M HCl; No
0.2 M HCN and 0.2 M KOH; No
G.
0.3 M HOCl and 0.15 M Ba(OH)
2
; No
H.
0.3 M HOCl and 0.1 M Ba(OH)
2
; Yes
159.
pH = 4.43
160.
pH = 4.56
161.
Henderson-Hasselbach equation: pH = pK
A
+ log[base]/[acid]
162.
When [Base] = [Acid] then ratio [base]/[acid] = 1 and log(1) = 0 so pH = pK
A
. Thus, the pK
A is the midpoint of the buffer range.
163.
pH = 3.13
164.
If a buffer with a pH = 4.0 is needed in an experiment, then a buffer system with a pK
A value around 4.0 is needed.
A.
B.
C.
D.
E.
F.
HNO
HC
C
6
H
HCN/NaCN
NH
2
3
5
H
2
3
/NaNO
O
COOH/NaC
/NH
2
4
Cl
2
/NaC
2
H
3
6
O
HCOOH/NaCOOH
2
K
A
=4.5×10 -4
K
A
=1.8×10 -5
K
A
=1.8×10 -4
=6.5×10
-5
H
5
COO K
A
K
A
=4.9×10 -10
K
B
=1.8×10 -5 pK
A
3.35
4.74
3.74
4.19
9.31
9.26

The best buffer system is that given in D because its pK
A
value is closest to 4.0. To obtain a pH=4.0 using this buffer system, a concentration ratio of base:acid of 0.65 should be used.
165.
Therefore, if the [C
6
H
5
[NH
3
]/[NH
4
+ ] = 1.7
COOH] is 1.0 M; the [NaC
6
H
5
COO] should be 0.65 M.
167.
pH of original buffer solution = 5.04
pH after addition of 0.0040 mol HCl = 4.80
pH after addition of 0.0060 mol NaOH = 5.6
168.
After addition of 0 mL of 0.25 M KOH: pH = 0.30
After addition of 25 mL of 0.25 M KOH: pH=0.9 (prior to eq. pt.; excess strong acid present)
After addition of 50 mL of 0.25 M KOH: pH = 7 (at eq. pt; only neutral salt present)
After addition of 60 mL of 0.25 M KOH: pH = 12.5 (after eq. pt; excess strong base present)
Volume of base needed to reach eq. pt = 50. mL
169.
After addition of 0 mL of 0.20 M KOH: pH = 2.23
After addition of 2.5 mL of 0.20 M KOH: pH =3.46 (½ way to eq. pt; buffer present)
After addition of 5.0 mL of 0.20 M KOH: pH = 8.14 (at eq. pt; only basic salt present)
After addition of 6.0 mL of 0.20 M KOH: pH =12.1 (after eq. pt; excess strong base present)
Volume of base needed to reach eq. pt. = 5.0 mL pH ½ way to eq. pt is equal to pK
A
170.
After addition of 0 mL of 0.20 M HCl: pH = 11.13
After addition of 5.0 mL of 0.20 M HCl: pH = 9.26 (½ way to eq. pt; buffer present)
After addition of 10.0 mL of 0.20 M HCl: pH = 5.22 (at eq. pt; only acidic salt present)
After addition of 15.0 mL of 0.20 M HCl: pH = 1.54 (after eq. pt; excess strong acid present)
Volume of acid needed to reach eq. pt. = 10. mL pH ½ way to eq. pt is equal to pK
A
for NH
3
172.
For phenolphthalein: color in acidic solution is colorless; color in basic solution is pink.
173.
Phenolphthalein will change color around its pK
A,In
value which is 9.10. Phenolphthalein could be used in titration of strong acid with strong base since pH around eq. pt. changes from 4-10 with addition of 1 drop of base. Phenolphathalein could also be used in titration of weak acid with strong base since at eq. pt a basic salt with a basic pH is present. However, phenolphthalein could NOT be using in titration of a weak base with a strong acid since at eq.
pt. an acidic salt with an acidic pH is present.
174.
To answer find pK
A
,
In
for each indicator which gives pH around which indicator changes color. Phenolphthalein, pK
A
= 9.10; methyl orange, pK
A
= 3.80; bromothymol blue, pK
A
=
6.80. Now decide the relative pH for each of the titrations in A-C.
Titration Eq. Pt. Salt pH at Eq. Pt. Indicator
A.
B.
HNO
HNO
3
3
with NH
3
with KOH
NH
4
NO
3
KNO
3 pH<7 pH = 7 methyl orange bromothymol orange
C.
HClO
2
with NaOH NaClO
2 pH > 7 phenolphthalein
NOTE: On the lecture problem sheet 174 C. should have had HClO
2
as the acid not HClO
3
.
177.
K sp
= 1.19×10 -5
178.
K sp
= 2.11×10 -33
179.
Solubility: 7.3×10
-7
M; 1.4×10
-4
g/L
180.
Solubility: 4.5×10 -5 M; 1.6×10 -2 g/L
181.
[Ag + ] = 2.9×10 -2 M; [SO
4
-2 ] = 1.4×10 -2 M

182.
Solubility in pure water: 3.6×10
-3
M
Solubility in 0.30 M KF: 2.0×10 -6 M
183.
Both K
2
SO
4
and BaCl
2
will decrease the solubility of BaSO
4 common ion. Solubility in 0.10 M K
M; Solubility in pure water: 1.0×10
2
-5
SO
4
M
: 1.1×10 -9
in water due to addition of
M; Solubility in 0.25 M BaCl
2
: 4.4×10 -10
184.
Any salt that contains the conjugate base of a weak acid will have its solubility increased by addition of a strong acid (i.e. the salt will be more soluble at lower or acidic solution pH).
Salts more soluble in acidic solution than in pure water are CaF
2
, MnS, and ZnCO
3
.
186.
Any salt that is a metal hydroxide will have its solubility increased by addition of a strong acid
(i.e. the salt will be more soluble at lower or acidic solution pH).
187.
A.
B.
pH = 10.35
Solubility in pure water: 1.12×10 -4 M; Solubility in soln buffered at pH 8.00: 5.6 M.
188.
Precipitate of MgCO
3
will form because IP > K sp
.
189.
If IP = K sp
; no precipitate will form and the solution is saturated.
If IP > K sp
; a precipitate will form and the solution is supersaturated.
If IP < K sp
; a precipitate will not form and the solution is unsaturated.
190.
Precipitate of Ag
2
SO
4
191.
[Cl
?
] = 1.2×10
-7
M
will not form because IP < K sp
.
192.
To precipitate as much Hg
2
+2 and Ag
0.089 M. At this [Cl ], the [Hg
2
+2
+ as possible without precipitating Pb
] left in solution is 1.8×10 -16
+2 need [Cl ] =
M. Therefore, most of the
Hg
2
+2 has been precipitated.
193.
A solution with a pH=0.52 will precipitate Pb
+2
and Hg
+2
.
A pH of slightly higher than 7.2 is needed to precipitate all of the Group III ions.
194.
Yes, these two ions can be separated by adjusting the solution pH. To precipitate as much of the Cr +3 as possible without precipitating the Ni +2 the solution pH should be adjusted to
6.8.
195.
Hg
196.
Sn
2
+2
Br
2
(ppts. first): AgBr: CuBr: PbBr
will precipitate first.
2
(ppts. last)