Chemistry 122
Mines, Spring 2012
Answer Key, Problem Set 6 (With explanations)
1. 15.56(a,d)*; 2. 15.58(a); 3. 15.64(a,c) Also find % ionization; 4. 15.68; 5. 15.80(a,b) For (a), assume 25°C, for
(b), assume 40°C (at which Kw 2.92 x 10-14); 6. 15.86; 7. 15.90 Also calculate [OH-]; 8. 15.94 & 15.98; 9. 15.100*;
10. 15.102*; 11. 15.104*; 12. 16.30*; 13. 16.42(a)*; 14. Conceptual Connection 16.1 (p. 720), except using the
following numbers in place of 4.82, 4.25, and 4.72: pKa = 5.58, pH = 6.13, and pH = 6.93; 15. 16.58, except switch
the word “masses” to “moles” and Add Part (b): Why can’t an H3PO4/NaH2PO4 buffer system be used to make the
buffer in this problem? Give specific reasoning
-----------------------------------1. 15.56(a,d).
Determine the pH of each solution.
(a) 0.048 M HI
Answer: 1.32
Reasoning:
HI is a strong acid, thus “% ionization” is 100%, and [I-]eq  [H3O+]eq  0.048 M
 pH  -log(0.048)  1.318..1.32
(d) A solution that is 1.09% HCl by mass (Assume a density of 1.01 g/mL for the solution.)
Answer: 0.520
Reasoning:
Assume 100 mL ( 0.1 L) of solution (exactly). Then:
mass of solution  1.01 g/mL x 100 mL  101 g
mass of HCl  101 g solution x
1.09 g HCl
 1.101 g HCl
100 g solution
MM (HCl) = 1.01 + 35.45 = 36.46 g/mol
 moles HCl  1.101 g HCl x
1mol HCl
 0.03019..  0.0302 mol
36.46 g
[HCl]0 = 0.0302 mol/0.100 L = 0.302 M
HCl is a strong acid, thus “% ionization” is 100%, and [Cl-]eq  [H3O+]eq  0.302 M
 pH  -log(0.302)  0.51999..0.520
2. 15.58(a).
What mass of HClO4 should be present in 0.500 L of solution to obtain a solution with each pH value?
(a) pH = 2.50
Answer: 0.159 g HClO4
Reasoning:
[H3O+]eq  10-pH  10-2.50 = 0.00316 M. Since HClO4 is a strong acid (100% ionized), the original
concentration of HClO4 must have been 0.00316 M.
0.00316 M x 0.500 L = 0.00158 mol HClO4
MM(HClO4) = 1.01 + 35.45 + 4(16.00) = 100.46 g/mol  0.00158 mol x
3. 15.64(a,c) Determine the pH (and % ionization) of an HF solution of each concentration.
make the simplifying assumption that x is small? [Assume 25°C.]
(a) 0.250 M;
(c) 0.0250 M
PS6-1
100.46 g
 0.1587 g HClO 4
mol
In which cases can you not
Answer Key, Problem Set 6
Answers: (a) pH = 2.03; % ionization = 3.7%; small x approximation is valid
(c) pH = 2.55; % ionization = 11%; small x approximation is not valid
Strategy:
1) Write out the acid ionization equation and Ka expression. Find Ka (25°C) from Table 15.5.
2) Set up an ICE table, letting x = [HF] that ionizes. Substitute into the Ka expression and solve
for x. Use the “small x” approximation if it looks reasonable. If not, use the quadratic
formula.
3) Check the small x approximation, if applicable. If good (or if you used the quadratic
formula), recognize that x = [H3O+]. (If the small x approximation is not good, use the
quadratic formula to get x.)
[H3 O  ]eq
4) pH = -log[H3O+]; % ionization =
x100
[HA]0
Execution of Strategy for Part (a):
F- + H3O+; K a 
HF + H2O
[F  ] eq [H3 O  ] eq
 3.5 x 10 -4
[HF] eq
[HF] (M)
[F-] (M)
[H3O+] (M)
I
0.250
0
~0
C
-x
+x
+x
E
0.250- x
x
x
Ka 
[F  ] eq [H3 O  ] eq
[HF] eq
 3.5 x 10 -4 
x x 
0.250  x 
 3.5 x 10 -4
Try the “small x” approximation:
x x 
0.250  x 
Assume 0.250 - x  0.250
 3.5 x 10 - 4 
    
x x 
0.250 
 3.5 x 10 - 4
 x 2  0.250(3.5 x 10 -4 )  8.75 x 10 -5
 x  8.75 x 10 -5   9.354 x 10 -3
A negative x here does not make sense (can’t have a negative [F-] or [H3O+]) so use the (+) one.

 Check assumption :



9.354 x 10 -3
x 100  3.74% (  5% so OK) 
0.250

[H3O+]  x  9.34 x 10-3  pH  -log(9.34 x 10-3)  2.029..  2.03 ; % ionization = 3.7% (see )
Execution of Strategy for Part (c):
The setup is the same as in (a), except for using 0.0250 M (new [HF]0) for 0.250 M:
Ka 
[F  ] eq [H 3 O  ] eq
[HF] eq
 3.5 x 10 - 4 
Try the “small x” approximation:
PS6-2
x x   3.5 x 10 -4
0.0250  x 
Answer Key, Problem Set 6
x x   3.5 x 10 -4
0.0250  x 
x x   3.5 x 10 -4
0.0250 
Assume 0.0250 - x  0.0250

     
 x 2  0.0250(3.5 x 10 -4 )  8.75 x 10 -6
 x  8.75 x 10 -6   2.958.. x 10 -3
A negative x here does not make sense so use the positive one.

 Check assumption :



2.958 x 10 -3
x 100  11.8% (  5% so NOT okay!
0.0250

Put in x  2.96 x 10-3 in (0.0250 – x) (successive approximations approach):
x x 
0.0250  x 
Assume 0.0250 - x  0.0250 -0.00296 0.0220
 3.5 x 10 -4 
         
x x 
 3.5 x 10 -4
0.0220 
 x 2  0.0220(3.5 x 10 -4 )  7.70 x 10 -6
 x  7.70 x 10 -6   2.77.. x 10 -3 (only (+) root meaningful here)

 Check assumption :


0.0220 - (0.0250 - 2.77 x 10 -3 )
x 100  -1.05% (  5% so OK 
0.0220

Same
OR, if you do the quadratic equation approach to solve for x:
( x )( x )
 3.5 x 10 - 4  x 2  3.5 x 10 - 4 0.0250  x 
(0.0250  x )
 x 2  8.75 x 10-6  3.5 x 10-4 x  x 2  3.5 x 10-4 x  8.75 x 10-6  0
 x
 3.5 x 10 -4 
3.5 x 10 
-4 2

 4(1)  8.75 x 10 -6
2(1)
   3.5 x 10
-4
 5.926..x 10 -3
 2.78..x 10 -3 (  )root
2
(Negative root is not meaningful for this physical system.)


(2.8 x 10 -3 )(2.8 x 10 -3 )
 Check (+) root :
 3.53 x 10 - 4 (matches K a ) 
-3
(0.0250  2.8 x 10 )


So, [H3O+]  2.78..x 10-3 M  pH -log(2.78.. x 10-3)  2.5546.. 2.55
% ionization =
4. 15.68.
2.78 x 10 -3
x100  11.1%
0.0250
Makes sense. More dilute HF should
be less acidic, and it is. (2.56 > 2.03)
Makes sense. More dilute solution
should have greater % ionization (Le
A 0.115 M solution of a weak acid (HA) has a pH of 3.29 [at some temperature]. Calculate the acid ionization
constant (Ka) for the acid [at this temperature].
Answer: Ka = 2.3 x 10-6
Strategy:
1) Analysis: It may not look like it at first, but this is fundamentally a “given the initial
concentrations and one equilibrium concentration, find K” kind of equilibrium problem. Although
[H3O+]eq is not technically “given”, you are just one step away since the pH is given! Thus:
2) Write out the acid ionization equation (the primary reaction that occurs here is acid ionization)
and its K expression.
3) Calculate [H3O+]eq from pH ([H3O+] = 10-pH)
PS6-3
Answer Key, Problem Set 6
4) Use stoichoimetry (ICE table is not necessary, but helpful here) to determine the other
equilibrium concentrations from [HA]0 (given) and [H3O+]eq (and recognizing that [A-]0 and
[H3O+]0 are both (effectively) zero, as assumed in all “normal” acid ionization problems).
5) Plug into Ka expression!
Execution of Strategy:
A- + H3O+;
HA + H2O
Ka 
[A - ]eq [H3 O  ]eq
[HA] eq
Since pH = 3.29, [H3O+]eq  10-3.29 = 5.128.. x 10-4 M  0.0005128 M
Since the stoichiometry is 1 : 1 : 1 and we assume there was essentially no [H3O+] to begin with,
this means that 0.0005128 M of H3O+ formed. Thus, the [A-] that formed and the [HA] that
reacted were also 0.0005128 M. If 0.0005128 M of the initial 0.115 M reacted, the concentration
remaining at equilibrium is just 0.115 – 0.0005128 M = 0.11448 M. If you wanted to use an ICE
table to help see these relationships, it would look like this:
I (initial)
[HA] (M)
[A-] (M)
[H3O+] (M)
0.115
0
~0
C (change in)
0.0005128
E (at equilibrium)
[HA] (M)
[A-] (M)
[H3O+] (M)
I (initial)
0.115
0
~0
C (change in)
- 0.0005128
+ 0.0005128

+ 0.0005128
0.0005128
E (at equilibrium)
[HA] (M)
[A-] (M)
[H3O+] (M)
I (initial)
0.115
0
~0
C (change in)
- 0.0005128
+ 0.0005128
+ 0.0005128
0.0005128
0.0005128
0.115 – 5.128.. x 10
M  0.11448
E (at equilibrium)
-4
NOTE: I suppose another way to use an ICE table for this is to use an “x”, but then recognize that you know x here
+
(because you know [H3O ]eq from the pH):
[HA] (M)
-
[A ] (M)
+
[H3O ] (M)
I (initial)
0.115
0
~0
C (change in)
-x
+x
+x
E (at equilibrium)
0.115 - x
x
x
x = 0.0005128
Substitute back into the equilibrium constant expression equation (Law of Mass Action):
Ka 
[A - ] eq [H3 O  ] eq
[HA] eq

(0.0005128)(0.0005128)
 2.297 x 10 -6  2.3 x 10 -6
(0.11448)
PS6-4
Answer Key, Problem Set 6
5. 15.80(a,b).
-
+
For each strong base solution, determine [OH ], [H3O ], pH, and pOH. For (a), assume 25°C, for (b),
-14
assume 40°C (at which Kw 2.92 x 10 )
-3
(a) 8.77 x 10 M LiOH (at 25°C)
(b) 0.0112 M Ba(OH)2 (at 40°C; Kw 2.92 x 10 )
-14
Answers: (a) [OH-] = 8.77 x 10-3 M; [H3O+] = 1.1 x 10-12 M; pH = 11.94; pOH = 2.057 (2.06 OK)
(b) [OH-] = 0.0224 M; [H3O+] = 1.3 x 10-12 M; pH = 11.88; pOH = 1.650 (1.65 OK)
Strategy:
1) Since these are identified as “strong base” solutions, you must assume that these (ionic)
compounds are strong electrolytes  “let the ionic compounds dissociate first!”
2) Being careful to note stoichiometry (1 OH- per formula unit in LiOH; 2 OH- per formula unit in
Ba(OH)2), calculate the [OH-] in each solution.
3) Use [H3O+][OH-]  Kw to calculate [H3O+]eq. (remembering to switch to Kw  2.92 x 10-14 in (b)!)
NOTE: If you calculate pOH first and then use pH + pOH  14.00, you will get (a) correct but (b) incorrect since
that equation only applies to a solution at 25°C!!). If you really want to go this route, you must use (the
general) relationship that pH + pOH  pKw
4) Calculate pH using pH = -log[H3O+].
Calculate pOH using pOH = -log[OH-].
Execution of Strategy, Part (a):
LiOH(aq)  Li+(aq) + OH-(aq)  [OH-]in solution  [LiOH]initial  8.77 x 10-3 M
[H3O+] 
1.0 x 10 -14
 1.14 x 10 -12 M  pH  -log(1.14 x 10-12)  11.94
-3
8.77 x 10
pOH  -log(8.77 x 10-3)  2.057 (but 2.06 is fine also, since usually pH’s and pOH’s are limited to
the hundredths place)
Execution of Strategy, Part (b):
Ba(OH)2(aq)  Ba2+(aq) + 2 OH-(aq)  [OH-]in solution  2 x[Ba(OH)2]initial  2(0.0112)  0.0224 M
[H3O+] 
2.92 x 10 -14
 1.303 x 10 -12 M  pH  -log(1.303 x 10-12)  11.88
0.0224
NOTE: Even though this solution has a lower pH than the
one in (a), it is clearly not “more acidic”!
Remember, “acidic” is not defined in terms of “pH”.
pOH  -log(0.0224)  1.6497.. 1.650 (or 1.65)
6. 15.86.
-
Write equations showing how each weak base ionizes [in?] water to form OH . Also write the corresponding
expressions for Kb.

2-
(a) CO3
CO32-(aq) + H2O(l)
HCO3-(aq) + OH-(aq) ;
Kb 
[HCO 3 ] eq [OH - ] eq
2
[CO 3 ] eq

+ OH (aq) ; K b 
-
(b) C6H5NH2
C6H5NH2(aq) + H2O(l)
C6H5NH3+(aq)
(c) C2H5NH2
C2H5NH2(aq) + H2O(l)
C2H5NH3+(aq) + OH-(aq) ; K b 
[C 6 H5 NH3 ] eq [OH- ] eq
[C 6H5 NH2 ] eq

PS6-5
[C 2H5NH3 ] eq [OH - ] eq
[C 2H5NH2 ] eq
Answer Key, Problem Set 6
7. 15.90.
Amphetamine (C9H13N) is a weak base with a pKb of 4.2 [at 25°C]. Calculate the pH of a solution containing an
amphetamine concentration of 225 mg/L. Also calculate [OH ]. [Assume 25°C.]
Answer: 10.5
Strategy:
1) Write out the base ionization equation and Kb expression.
2) Calculate Kb from pKb (the same way you calculate [H3O+] from pH: K b  10  pK ;
[because pKb = -log Kb])
b
3) Calculate the initial [C9H13N] (in moles/L) using the mg converted into grams, and then using
the (calculated) molar mass (9(12.01) + 13(1.01) + 14.00 = 135.2 g/mol) to get moles.
4) Set up an ICE table, letting x = [B] that ionizes. Substitute into the Kb expression and solve
for x. Use the “small x” approximation if it looks reasonable. If not, use the quadratic
formula.
5) Check the small x approximation, if applicable. If good (or if you used the quadratic
formula), recognize that x = [OH-]. To get pH, calculate [H3O+] from [OH-] using [H3O+][OH-]
= Kw = 1.0 x 10-14 (because we’re assuming T = 25°C).
6) pH = -log[H3O+]
Execution of Strategy:
C9H13NH+ + OH-; K b 
C9H13N + H2O
[C9H13N]0 
[C 9H13NH  ] eq [OH - ] eq
[C 9H13N] eq
 10  4.2  6.3 x 10 -5
225 mg
1g
1 mol
x
x
 1.664.. x 10-3 M
L
1000 mg 135.2 g
[C9H13N](M)
[ C9H13NH+] (M)
[OH-] (M)
I
0.001664
0
~0
C
-x
+x
+x
E
0.001664 - x
x
x
It does not look to me like the “small x” approximation is even worth trying here. K is not all that
small, and [B]0 is quite small. I will just go straight to the quadratic formula approach:
K b  6.3 x 10-5 


[C9H13NH ]eq [OH- ]eq
[C9H13N]eq


x 2  1.048.. x 10-7  6.3 x 10-5 x
x
6.3 x 10-5 
 6.3 x 10 
-5
2
2(1)

(x )(x )
(0.001664  x )


x 2  6.3 x 10-5  0.001664  x 


x 2  6.3 x 10-5 x  1.048.. x 10-7  0

 4(1) 1.048.. x 10-7

6.3 x 10-5  6.50..x 10-4
 2.93..x 10-4 (  ) root
2
Negative root is not meaningful for this physical system.



[C9H13NH ]eq [OH- ]eq (2.93..x 10-4 )(2.93..x 10-4 )

 6.26.. x 10-5 (close to K b ) 
 Check (+) root:
-4

[C9H13N]eq
(0.001664  2.93..x 10 )


So, [OH-]  2.93 x 10-4 M and thus [H3O+]  1.0 x 10-14/2.93 x 10-4  3.41.. x 10-11 M
pH -log(3.41.. x 10-11)  10.46.... 10.5
PS6-6
Answer Key, Problem Set 6
8. 15.94 & 15.98.
15.94. Determine whether each anion is [basic or neutral] a (non-negligible) base or a negligible base (“pH neutral”).
For those anions that are bases, write an equation that shows how the anion acts as a base.
-
is a weak base because its CA is HC7H5O2, a weak acid [not one of the six SA’s]
(a) C7H5O2
C7H5O2-(aq) + H2O(l)
(b) I
-
HC7H5O2(aq) + OH-(aq)
is a negligible base (i.e., pH neutral) because its CA, HI, is a strong acid
-
is a negligible base (i.e., pH neutral) because its CA, HNO3, is a strong acid
(c) NO3
-
is a weak base because its CA is HF, a weak acid [not one of the six SA’s]
(d) F
F-(aq) + H2O(l)
HF (aq) + OH-(aq)
15.98. Determine whether each cation is [acidic or pH neutral] a (non-negligible) acid or a negligible acid (“pH neutral”).
For those cations that are acids, write an equation that shows how the cation acts as an acid.
(a) Sr
2+
is a negligible acid because it is a Group II ion
3+
is an acid because it is a +3 ion (not a Gp I or Gp II ion)
(b) Mn
Mn(H2O)63+(aq) + H2O(l)
(c) C5H5NH
+
is an acid because it is the CA of the weak (amine) base C5H5N
C5H5NH+(aq) + H2O(l)
+
Mn(H2O)5(OH)2+(aq) + H3O+(aq)
C5H5N(aq) + H3O+(aq)
is a negligible acid because it is a Group I ion
(d) Li
9. 15.100*.
Determine whether each salt will form a solution that is acidic, basic, or pH-neutral.
(a) Al(NO3)3
Answers: (a) acidic
(b) C2H5HN3NO3
(b) acidic
(c) K2CO3
(c) basic
(d) RbI
(d) neutral
(e) NH4ClO
(e) basic
Strategy:
1) Let each salt ( “soluble ionic compound”) dissociate first!
2) Look at each ion (see outline at the bottom of the PS6 sheet) to assess whether it is an “acid”,
“base”, or a “negligible”
3) If:
a) one of the two ions is an acid and the other is a “negligible”, the solution will be acidic
b) one of the two ions is a base and the other is a “negligible”, the solution will be basic
c) if both of the ions are “negligibles”, the solution is neutral
d) if the cation is an acid and the anion is a base, then the solution will be:
i) basic, if Kb(anion) > Ka(cation)
ii) acidic, if Ka(cation) > Kb(anion)
NOTE: You will need K values in this case (not in the others).
Note: KaKb(for conjugates)  Kw, and you will need to use
this frequently to determine the Ka or Kb of ions from
their respective (molecular) conjugates.
Execution of Strategy:
(a) Al(NO3)3

Al3+
NO3-
+
acid (+3 ion)
negligible base (HNO3 is a strong acid)
 Solution is acidic
PS6-7
Answer Key, Problem Set 6
(b) C2H5NH3NO3

C2H5NH3+
NO3-
+
acid (CA to C2H5NH2)
negligible base (HNO3 is a strong acid)
 Solution is acidic
(c) K2CO3

K+
CO32-
+
negligible acid (Gp I ion)
non-negligible base (HCO3- is not a strong acid)
 Solution is basic
(d) RbI

Rb+
I-
+
negligible acid (Gp I ion)
negligible base (HI is a strong acid)
 Solution is neutral
(e) NH4ClO

NH4+
acid (CA to NH3)
Ka(NH4+) 
From Table 15.8:
-8
Kb(NH3) = 2.9 x 10
ClO-
+
non-negligible base (HClO is not a strong acid)
1.0 x 10 -14
1.0 x 10 -14
 5.6 x 10 -10 < Kb(ClO-) 
 3.4 x 10 -7
-5
1.8 x 10
2.9 x 10 -8
 Solution is basic
10. 15.102*.
-
From Table 15.5:
-8
Ka(HClO) = 2.9 x 10
+
(ClO ionizes “more” than NH4 )
Arrange the solutions in order of increasing basicity:
CH3NH3Br, KOH, KBr, KCN, C5H5NHNO2
Answer (at the level of this class):
C5H5NHNO2 or CH3NH3Br < KBr < KCN < KOH
NOTE: The precise determination of the relative placement of the two acidic solutions in this problem is really
beyond the scope of this course. I did not recognize this until I was writing up the key for the problem.
I have sent out an email regarding this and/or mentioned it in class. I apologize for this fundamental
“flaw” in the problem (I’m curious what the solution manual says about this problem…. ). If you’d like to
know how to determine the pH of the C5H5NHNO2 solution (0.1 M), see me in an office hour. It is
completely “accessible” to you given what we’ve done, but it is not something I would expect anyone
to figure out on his / her own.
Strategy:
Analyze each salt separately as in the prior problem. The first “sorting” would be by type—acid,
base, or (pH) neutral salt. If two salts end up in the “acid” or “base” category, then determining
relative strength would be the secondary sorting.
Clearly, any acid salt will make a solution that is less basic (more acidic) than a solution of any
neutral salt, and any neutral salt will make a solution less basic than a solution of any base salt. If
there are two salts with only a base ion, compare the Kb’s to determine which solution will be more
basic (bigger Kb), and if there are two salts with only an acid ion, compare the Ka’s.
At the level of this course, you are not responsible for determining the pH of a mixture containing a
weak acid and a weak base (e.g., a salt with an acid cation and a base anion)—that is beyond the
scope of this course (although it is a direct, reasonable, and accessible extension of principles
already covered). You are only responsible for determining if such a salt solution would be acidic
or basic (by comparing Ka(cation) vs. Kb(anion). So, as noted above, you will not be able to
completely answer this question as written because one of the two acidic solutions is from a salt
with an acid cation and base anion. If I were to put a similar problem on an exam, there would be
PS6-8
Answer Key, Problem Set 6
no such “problematic” pair of salts (the authors of this problem were just not careful enough in its
construction).
Execution of Strategy:
CH3NH3Br
 CH3NH3+ (a weak acid; CA to CH3NH2) + Br- (a negligible base; CB to SA HBr)
 (weakly) acidic
 K (a negligible acid; Gp I) + OH- (considered a strong base, although technically, it is KOH that is the
+
KOH
-
strong base [it completely ionizes to produce OH )
 (maximally) basic
KBr
 K+ (a negligible acid; Gp I) + Br- (a negligible base; CB to SA HBr)
KCN
 K (a negligible acid) + CN- (a weak base; CB to WA HCN)
 neutral
+
 (weakly) basic
C5H5NHNO2
From Table 15.8:
-9
Kb(C5H5N) = 1.7 x 10
 C5H5NH+ (a weak acid; CA to C5H5N) + NO2- (a weak base; CB to WA HNO2)
Ka(C5H5NH+) 
1.0 x 10-14
 5.88 x 10-6 <
1.7 x 10-9
 (weakly) acidic
Kb(NO2-) 
+
1.0 x 10-14
 2.17 x 10-11
4.6 x 10-4
From Table 15.5:
-4
Ka(HNO2) = 4.6 x 10
-
(C5H5NH ionizes “more” than NO2 )
Thus, at this point, you could certainly say:
Least basic: C5H5NHNO2 or CH3NH3Br < KBr < KCN < KOH
Most basic
Regarding the two acidic solutions, you need to look at more data: Namely, the Ka of CH3NH3+.
If the Ka of this acid were greater than the Ka for C5H5NH+, then you could clearly state that the
1.0 x 10-14
solution of CH3NH3Br would be more acidic. However, its Ka 
 2.27 x 10-11 , which is
4.4 x 10-4
considerably smaller. However, you cannot state based on this information alone that the
solution of CH3NH3Br must be less acidic (more basic). Although it is a poorer acid than
C5H5NH+, the solution of C5H5NH+ also has an equal concentration of a weak base present in the
solution (NO2-) which will make the solution more basic than if it were only C5H5NH+)! As such, it
is not technically possible to (correctly) conclude which solution is more basic (i.e., further
calculation would be needed). This is why the problem, as written, is not “solvable” at the level of
this course. As it turns out, the C5H5NHNO2 does end up more acidic, but not by all that much
(less than 2 pH units)! Again, come by my office if you want to see the “proof” of this.
11. 15.104*.
Determine the pH of each solution. [Assume 25°C.]
(a) 0.20 M KCHO2
Answers: (a) 8.52
(b) 5.67
(b) 0.20 M CH3NH3I
(c) 0.20 M KI
(c) 7.00
Strategy:
1) First “let each salt dissociate” to see what ions are present.
2) Analyze each ion as in the previous problems. If you are asked to calculate the pH, then at
most, one of the ions will be an acid or a base (not both).
3) If the cation and anion are both “negligibles”, the solution will be neutral (pH = 7.00 if you
assume 25°C).
PS6-9
Answer Key, Problem Set 6
4) If the cation is an acid (with the base negligible), determine the Ka of the acid* and treat the
problem as an “acid ionization” problem (because it is exactly that!). I.e., use the strategy
outlined in Problem 3 (15.64) of this problem set.
5) If the anion is a base (with the cation negligible), determine the Kb of the base and treat the
problem as a “base ionization” problem (because it is exactly that!). I.e., use the strategy
outlined in Problem 7 (15.90) [except that you’d get Kb likely by other means*].
6) *Note that you will likely get Ka or Kb by using KaKb = Kw as in the prior couple of problems.
Execution of Strategy, Part (a)
KCHO2 
K+ (a negligible acid) + CHO2- (a weak base; CB to WA HCHO2)
Thus, the solution is equivalent to being a solution of 0.20 M CHO2- (weakly basic)
1.0 x 10-14
 5.56 x 10-11
1.8 x 10-4
Ka(HCHO2) = 1.8 x 10-4 (Table 15.5)  Kb(CHO2-) 
HCHO2 + OH-; K b 
CHO2- + H2O
[CHO2  ]eq [OH- ]eq
[HCHO2 ]eq
[CHO2-](M)
[HCHO2] (M)
[OH-] (M)
I
0.20
0
~0
C
-x
+x
+x
E
0.20- x
x
x
 10 4.2  5.56 x 10-11
Try the “small x” approximation:
 x  x 
 0.20  x 
Assume 0.20 - x  0.20
 5.56 x 10-11 

 x  x 
 5.56 x 10 -11
 0.20 
 x 2  0.20(5.56 x 10-11 )  1.11 x 10-11
 x  1.11 x 10-11 
 3.33.. x 10-6
A negative x here does not make sense so use the positive one.

 Check assumption:


3.33 x 10-6
x 100  0.0016% ( << 5% so OK) 
0.20

So, [OH-]  3.33 x 10-6 M and thus [H3O+]  1.0 x 10-14/3.33 x 10-6  3.00.. x 10-9 M
pH -log(3.00.. x 10-9)  8.522.... 8.52
(weakly basic, as expected)
Execution of Strategy, Part (b)
CH3NH3I  CH3NH3+ (a weak acid; CA to CH3NH2) + I- (a negligible base; CB to SA HI)
Thus, the solution is equivalent to being a solution of 0.20 M CH3NH3+ (weakly acidic)
Kb(CH3NH2) = 4.4 x 10-4 (Table 15.8)  Ka(CH3NH3+) 
CH3NH3+ + H2O
CH3NH2 + H3O+; K a 
PS6-10
1.0 x 10-14
 2.27 x 10-11
4.4 x 10-4
[CH3NH2 ]eq [H3 O ]eq
[CH3NH3 + ]eq
 2.27 x 10-11
Answer Key, Problem Set 6
[CH3NH3+] (M)
[ CH3NH2] (M)
[H3O+] (M)
I
0.20
0
~0
C
-x
+x
+x
E
0.20- x
x
x
Ka 
[CH3NH2 ]eq [H3 O ]eq
+
[CH3NH3 ]eq
 2.27 x 10-11 
 x  x 
 0.20  x 
 2.27 x 10-11
Try the “small x” approximation:
 x  x 
 0.20  x 
 x  x 
 2.27 x 10-11
 0.20 
Assume 0.20 - x  0.20
 2.27 x 10-11 

 x 2  0.20(2.27 x 10-11 )  4.54 x 10-12
 x  4.54 x 10-12 
 2.13 x 10-6
A negative x here does not make sense (can’t have a negative [F-] or [H3O+]) so use the (+) one.

 Check assumption:


2.13 x 10-6
x 100  0.0011% (  5% so OK) 
0.20

[H3O+]  x  2.13 x 10-6  pH  -log(2.13 x 10-6)  5.671..  5.67 (weakly acidic, as expected)
Execution of Strategy, Part (c)
KI  K+ (a negligible acid; Gp I) + I- (a negligible base; CB to SA HI)
 neutral  at 25C, pH = 7.00
12. 16.30*.
Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution [at 25°C]:
Answers: (a) 4.55
(a) a solution that is 0.195 M in HC2H3O2 and 0.125 M in KC2H3O2
(b) 10.92
(b) a solution that is 0.255 M in CH3NH2 and 0.135 M in CH3NH3Br
Strategy and Execution:
Set up the problem as an acid ionization problem (regardless of whether or not the acid is of the
form HA or of the form BH+!) with some of the product ion present initially (“common ion”)
C2H3O2- + H3O+; K a 
(a) HC2H3O2 + H2O
[C2H3 O2  ]eq [H3 O ]eq
[HC2H3 O2 ]eq
[HC2H3O2] (M)
[C2H3O2-] (M)
[H3O+] (M)
I
0.195
0.125
~0
C
-x
+x
+x
E
0.195- x
0.125 + x
x
Ka 
[C2H3 O2  ]eq [H3 O ]eq
[HC2H3 O2 ]eq
 1.8 x 10-5 
PS6-11
 1.8 x 10-5 (Table 15.5)
 0.125 + x  x 
 1.8 x 10-5
 0.195  x 
Answer Key, Problem Set 6
Use the “small x” approximation:
 0.125 + x  x 
 1.8 x 10-5
 0.195  x 
Assume 0.197 - x  0.197



Assume 0.125 - x  0.125
 x  [H3 O  ]  1.8 x 10-5 x

 Check assumption:

 0.125  x 
 1.8 x 10-5
 0.195 
 0.195 
 2.80..x 10-5
 0.125 

2.80 x 10-5
x 100  0.022..% (  5% so OK) 
0.125

Answer is reasonable. pKa =
4.74 here, and there is more HA
than A . Thus pH should be a bit
more acidic (lower) than 4.74 & it
is!
 pH  -log(2.80..x 10 )  4.551..  4.55
-5
NOTE: Once you “get” this, you can go straight to this equation for buffer in the future if you wish:
[H3 O ]eq  K a x
[HA]0
[A - ]0
buffer 
Note how this equation matches this one
(b) The acid component of this buffer is CH3NH3+. Look up Kb for CH3NH2 in Table 15.8 and use
that to calculate Ka [2.27 x 10-11] (see prior problem, 11. 15.104(b)).
CH3NH2 + H3O+; K a 
CH3NH3+ + H2O
[CH3NH2 ]eq [H3 O ]eq
[CH3NH3 + ]eq
[CH3NH3+] (M)
[CH3NH2] (M)
[H3O+] (M)
I
0.135
0.255
~0
C
-x
+x
+x
E
0.135 - x
0.255 + x
x
Ka 
[CH3NH2 ]eq [H3 O  ]eq
+
[CH3NH3 ]eq
 2.27 x 10-11 
 2.27 x 10-11
 0.255 + x  x 
 2.27 x 10-11
 0.135  x 
With a Ka of ~10-11 the “small x” approximation is a “slam dunk”:
“Shortcut” once you
get familiar with buffer
situations. “Buffers
are WYSIWYG!”
 0.255 + x  x 
 2.27 x 10-11
 0.135  x 
Assume 0.255 - x  0.255



Assume 0.135 - x  0.135
 x  [H3 O  ]  2.27 x 10-11 x

[HA]0

 [H3 O ]eq  K a x
[A - ]0


 Check assumption (for kicks):

 0.255  x 
 2.27 x 10-11
 0.135 
 0.135 
 1.20..x 10-11
 0.255 

 buffer  


1.2 x 10-11
x 100 ~ 9 x 10-9 % (!)(  5%!) 
0.135

 pH  -log(1.20..x 10-11)  10.920..  10.92
13. 16.42(a)*.
Calculate the pH of the solution that results from each mixture:
(a) 150.0 mL of 0.25 M HF with 225.0 mL of 0.30 M NaF
PS6-12
Answer is reasonable. pKa = 10.64
+
here, and there is more B than BH .
Thus pH should be a bit more basic
(higher) than 10.64 & it is!
Answer Key, Problem Set 6
Answer: 3.71
Strategy:
1) BE CAREFUL! Although this is a buffer problem, it involves the mixing of two solutions. Thus,
you cannot use the “reagent bottle” concentrations (given in the problem). Recognize that
each reagent’s concentration will change upon mixing (because of dilution).
**Also remember the earlier guideline to “Let any soluble ionic compounds dissociate first!”
That is how you can see that NaF is effectively a solution of “F-“ in this problem.
2) You can either:
a) Calculate the initial concentrations after mixing (i.e., do a dilution calculation)
OR
b) Calculate initial moles and use those in place of initial concentrations.
NOTE: Why is it okay to use “moles” here (when moles is obviously not the same as concentration)? Because in
this specific instance, it is the ratio of concentrations that matters, not each individual value, and since the
volume is the same for both species in the buffer solution, the ratio of the moles equals the ratio of the
concentrations (see p. 721 in Tro).
3) Find the Ka for HF from Table 15.5 (Ka  3.5 x 10-4) (Recognize that you need the Ka for the
acid component of a buffer in order to determine buffer pH.)
4) Once (2) is completed, you can (if you are comfortable with the prior problem!!) substitute
directly into:
[H3 O ]eq  K a x
[HA]0
[A - ]0
buffer 
[H3 O ]eq  K a x
OR
moles HA 0
moles A 0-
buffer 
5) Calculate pH from [H3O+]
NOTE: You may use the Henderson-Hasselbalch in place of (4) and (5) if you like, but I have
chosen not to do that in my key since I have found that students tend to “lose touch” [and
make more sign errors] when trying to use the HH equation.
Execution of Strategy:
[HF]0  0.25 M x
[F- ]0  0.30 M x
150.0 mL
 0.10 M
(150.0 mL + 225.0 mL)

Vbefore 
 dilution calculation: Mafter  Mbefore x

Vafter 

225.0 mL
 0.18 M
(150.0 mL + 225.0 mL)
 [H3 O ]eq  3.5 x 10-4 x
0.10
 1.94 x 10-4
0.18
pH  -log(1.94 x 10-4)  3.711 = 3.71
Check answer. It makes sense! pKa = -log(3.5 x 10 ) = 3.46. [F ] > [HF]  pH should be a bit higher (more basic).
-4
-
14. Conceptual Connection 16.1 (p. 720), except using the following numbers in place of 4.82, 4.25, and 4.72:
pKa
= 5.58, pH = 6.13, and pH = 6.93.
-
A buffer contains the weak acid HA and its conjugate base A . The weak acid has a pKa of 5.58 and the buffer has a pH
of 6.13. (i) Which statement is true of the relative concentrations of the weak acid and conjugate base in the buffer?
-
(a) [HA] > [A ]
-
(b) [HA] < [A ]
-
(c) [HA] = [A ]
(ii) Which buffer component would you add to change the pH of the buffer to 6.93?
Answer: (i) (b) is correct;
(ii) Add more A-
Reasoning:
The pH of the buffer is greater (more basic) than the pKa. Thus, the buffer must contain more of
the base component than the acid component. (This is because the pH = pKa when the HA / A-
PS6-13
Answer Key, Problem Set 6
ratio is precisely 1. That’s why I call a pH of pKa “home” for a buffer. This is the pH around which
the buffer system will act as a buffer.)
Getting the pH from 6.13 to 6.93 means making the buffer more basic (higher pH). Thus, you
must add the base component of the buffer, which is A-.
15. 16.58.
(a)(i) Which buffer system is the best choice to create a buffer with pH = 9.00?
NH3/NH4Cl
HF/NaF
HNO2/KNO2
HClO/KClO
(a)(ii) For the best system, calculate the ratio of the moles of the buffer components required to make the buffer.
(b) Why can’t an H3PO4/NaH2PO4 buffer system be used to make the buffer in this problem? Give specific reasoning.
OR
[A - ]0

[HA]0
6
5
.
0
[HA]0

[A - ]0
8
.
1
Answers: (a)(i) The NH3/NH4Cl system; (ii)
(b) Ka1 for H3PO4 is 7.5 x 10-3 (Sorry, this was from Table 15.10). This is way too strong of an
acid to make a buffer at pH 9.00! Note that pKa = 2.12, and thus buffers with H3PO4
and NaH2PO4 would only be possible within a pH unit (or two) around 2.12 (i.e., the
maximum pH would be around 4.12, which is well below 9.00).
(Note: At pH 9.00, the HA / A- ratio would be 0.000000133 : 1 ! It will not act like a
buffer with that tiny ratio!)
Strategy:
1) Recognize that a buffer system tends to only work as a buffer at pH’s near to the pKa of its
acid component. This is because of the relationship:
[H3 O ]eq  K a x
[HA]0
[A - ]0
buffer 
If HA / A- is 1, [H3O+] = Ka and pH equals pKa. If the ratio is 10 or 1/10, the pH will be one unit
below or above pKa. If the ratio is 100 or 1/100, the pH will be two units below or above pKa.
Once the ratio gets this high (Tro says even lower than this), the system won’t act much like a
buffer since there will be too little of one component.
2) Find the Ka’s for all the acids of the potential buffer systems from either Table 15.5, or (in the
case of NH4+, from Kw/Kb, where Kb is from Table 15.8).
3) Calculate pKa for each (pKa = -logKa).
4) Pick the one that has the pKa closest to the desired pH (here 9.00).
5) For (a)(ii), use the relationship in (1) above to solve for the HA / A- ratio (in moles). You will
need to calculate [H3O+] from pH first.
Execution of Strategy:
Buffer System
Acid Component
HF/NaF
NH3/NH4Cl
HNO2/KNO2
HClO/KClO
HF
NH4+
HNO2
HClO
-4
Ka
3.5 x 10
pKa
3.46
10 /1.8 x 10 
-14
-5
5.6 x 10-10
9.25
-4
4.6 x 10
2.9 x 10-8
3.34
7.54
Thus, the NH3/NH4Cl buffer system will work best at pH 9.00 (9.25 is the closest to 9.00).
Part (a)(ii): [H3O+] = 10-9.00 M
PS6-14
Answer Key, Problem Set 6
[HA]0
[A - ]0

8
.
1
[H3 O ]eq  K a x

[HA]0 [H3 O ]eq
10 9.00


 1.79.. 
Ka
[A ]0
5.56 x 10-10
[A - ]0
Ka
5.56 x 10-10




[HA]0 [H3 O ]eq
10 9.00
did not specify which ratio it wanted.)
PS6-15
6
5
.
0
(You could also have solved for
since the problem