1. Aqueous Equilibria, Part One
AP Chemistry
Aqueous Equilibria, Part One

2. Weak Acids
-- most acids are weak
(i.e., only partially ionized)
-- For a weak acid HX...
HX(aq) H+(aq) X–(aq)
[H+] [X–]
[HX]
-- acid-dissociation constant Ka =
large Ka:
“strong” weak acid
small Ka:
“weak” weak acid
The % of a weak acid
that is ionized is given
by the equation:
[H+] at eq.
x 100
% ionization =
[acid]orig.

3. For organic acids (containing only C, H, and O), the
“donated” H was connected
to...
Formic acid is the simplest carboxylic
acid. It is present in bee and ant stings.
O, NOT to C.
First isolated from apple juice,
malic acid is a dicarboxylic
acid. Because of its extreme sour
taste, it is used as a food additive,
notably in Mega Warheads ®.

4. A 0.020 M niacin solution has pH 3.26.
(a) What % of the acid is ionized?
(b) What is Ka?
[H+] at eq.
x 100
% ion. =
[acid]orig. a)
10–3.26.
x 100
% ion. =
0.020
Niacin (i.e., vitamin B3) is a water-soluble
vitamin and an essential nutrient. It is
found naturally in such foods as meat,
wheat germ, dairy products, and yeast.
A niacin deficiency leads to pellagra, a
condition whose symptoms include skin
lesions, mental confusion, and weakness.
= 2.7% ionized

5. (b) What is Ka?
niacin(aq) H+(aq) + conj. base–(aq)
init. [ ]
0.020
D [ ]
– x
+ x
+ x
at eq. [ ]
0.020 – x x x
But, from given info, x = [H+] = 10–3.26 = 5.5 x 10–4 M,
so “at eq.” line becomes…
at eq. [ ]
5.5 x 10–4
Thus, Ka =
(5.5 x 10–4)2
= x 10–5

6. Use a shortcut IF the power of K is –4, –5, –6, etc.
If Ka for niacin is 1.6 x 10–5, find the pH of a M
niacin solution.
niacin(aq) H+(aq) + conj. base–(aq)
init. [ ]
0.010
D [ ]
– x
+ x
+ x
at eq. [ ]
0.010 – x x x
Ka = P R
1.6 x 10–5 = x2
0.010 – x
Could use quadratic equation to solve for…
x = [H+] =
3.92 x 10–4 M
pH = 3.41
Use a shortcut IF the power of K is –4, –5, –6, etc.

7. Using the shortcut method for the previous problem…
niacin(aq) H+(aq) + conj. base–(aq)
init. [ ]
0.010
D [ ]
– x
+ x
at eq. [ ]
0.010 – x x
0.010 – x
1.6 x 10–5 = x2
0.010
1.6 x 10–5 = x2
x = [H+] =
4.00 x 10–4 M
pH = 3.40

8. % ionization of a weak acid at a given temperature...
% ionization / as acid [ ] /
Sometimes, the less concentrated the substance,
the more active is the amount that IS present.
In the same way, the less concentrated the weak
acid, the greater the % ionization.

9. X Calculate the % of HF molecules ionized in a
0.10 M HF solution. (Ka = 6.8 x 10–4)
HF(aq) H+(aq) + F–(aq)
init. [ ]
D [ ]
at eq. [ ]
0.10
– x
+ x
0.10 – x x
0.10 – x
6.8 x 10–4 = x2
x = [H+] = 8.25 x 10–3 M X
[acid]orig.
[H+] at eq.
x 100
% ion.=
0.10
8.25 x 10–3 =
(Without shortcut,
% ion. is 7.9%.)
8.3%

10. X Calculate the % of HF molecules ionized in a
0.010 M HF solution. (Ka = 6.8 x 10–4)
HF(aq) H+(aq) + F–(aq)
at eq. [ ]
0.010 – x x
0.010 – x
6.8 x 10–4 = x2
 x = [H+] = 2.61 x 10–3 M X
[acid]orig.
[H+] at eq.
x 100
% ion.=
0.010
2.61 x 10–3 =
(Without shortcut,
% ion. is 23%.)
26%

11. Polyprotic acids – like sulfurous acid, H2SO3 –
have more than one ionizable H.
H2SO3(aq) H+(aq) + HSO3–(aq) (Ka1 = 1.7 x 10–2)
HSO3–(aq) H+(aq) + SO32–(aq) (Ka2 = 6.4 x 10–8) --
It is harder to remove additional H+s,
so Ka values w/each H+ removed.
-- Usually, Ka2 is at least 1000X
smaller than Ka1. In such cases,
one can calculate [H+] and pH
based only on Ka1 (i.e., ignore
Ka2 and pretend you have a
monoprotic acid).

12. Find the pH of a 0.0037 M carbonic acid solution.
(Ka1 = 4.3 x 10–7, Ka2 = 5.6 x 10–11)
H2CO3(aq) H+(aq) + HCO3–(aq)
at eq. [ ]
– x x x
– x
4.3 x 10–7 = x2
shortcut
0.0037
4.3 x 10–7 = x2
x = [H+] = 3.97 x 10–5 M
x = [H+] = 3.99 x 10–5 M
pH = 4.40
pH = 4.40

13. ([H+] after 1st ionization = 3.97 x 10–5 M)
HCO3–(aq) H+(aq) CO32–(aq)
at eq. [ ]
3.97 x 10–5 – y
3.97 x 10–5 + y y
3.97 x 10–5 – y
5.6 x 10–11 =
(3.97 x 10–5 + y) (y)
3.97 x 10–5 – y
5.6 x 10–11 =
(3.97 x 10–5y + y2)
y = add’l [H+] = 5.6 x 10–11 M
totally
negligible
pH = 4.40

14. [ ] Weak Bases (i.e., proton acceptors)
weak base + H2O conjugate acid + OH–
[conj. acid] [OH–]
[weak base]
Kb =
Weak bases are often nitrogen-
containing molecules
(“amines”) or anions.
dimethylamine
(all of these have
lone e– pairs)
Example:
OH–(aq) +
H–N–CH3 (aq) H
H–N–CH3(aq) :
H2O(l)
[ ]

15. What is the conc. of ammonia
in a solution of pH 9.35?
Ammonia’s Kb = 1.8 x 10–5.
Since pOH = 4.65, [OH–] = 10–4.65 = 2.24 x 10–5 M.
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
x – 2.24 x 10–5
w.c?
2.24 x 10–5
2.24 x 10–5
1.8 x 10–5 =
(2.24 x 10–5)2
x – 2.24 x 10–5
x = [NH3]init. = 5.0 x 10–5 M

16. Anions related to Weak Acids
NaClO(aq) Na+(aq) ClO–(aq)
This is the conj. base of HClO,
so it acts as a weak base in H2O
(i.e., it accepts H+).
ClO–(aq) + H2O(l) HClO(aq) + OH–(aq)
Sodium hypochlorite is the active
ingredient in bleach, which is used
as a laundry additive and to disinfect
water for swimming pools.

17. What mass of NaClO is required to make 2.0 L of a
pH solution? The Kb for ClO– is 3.3 x 10–7.
Since pOH = 3.50, [OH–] = 10–3.50 = 3.16 x 10–4 M.
ClO–(aq) + H2O(l) HClO(aq) + OH–(aq)
x – 3.16 x 10–4
w.c?
3.16 x 10–4
3.16 x 10–4
Kb =
[HClO] [OH–]
[ClO–]
M NaClO =
mol NaClO
L soln M
2.0 L
3.3 x 10–7 =
(3.16 x 10–4)2
x – 3.16 x 10–4
mol NaClO
x = [ClO–]init.
= [NaClO]init. = M
45 g NaClO

18. Relationship Between Ka and Kb
NH4+(aq) NH3(aq) + H+(aq)
(1)
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
(2)
Ka =
[NH3] [H+]
[NH4+]
Kb =
[NH4+] [OH–]
[NH3]
[NH3] [H+] [NH4+] [OH–]
[NH4+] [NH3]
Then Ka x Kb =
= [H+] [OH–]
For an acid and its conj. base:
KaKb = Kw = 1.00 x 10–14

19. Because they are easily calculated from Ka values,
reference tables are often “short” on Kb values.
Just like pH = –log [H+],
pKa =
–log Ka
If [H+] [OH–] = 1.00 x 10–14
pKb =
–log Kb
and pH + pOH = 14
then since Ka Kb = 1.00 x 10–14
pKa + pKb = 14
25oC)

20. For hydrofluoric acid, Ka = 6.8 x 10–4.
a. Write the formula of the conjugate base. F–
b. Write the equation for which Kb applies.
F– + H2O HF + OH–
c. Write the equation for which Ka applies.
HF H+ + F–
d. Find pKa, pKb, and Kb.
pKa = –log Ka
= –log (6.8 x 10–4)
pKa = 3.17
3.17
pKa + pKb = 14
pKb = 10.83
(6.8 x 10–4)
Ka Kb = 1 x 10–14
Kb = 1.5 x 10–11