Sec 15'1 Solutions of Acids and Bases Containing a Common Ion:

1

• Chapter 15
• Sec 15.1 Solutions of Acids and Bases Containing
  a Common Ion
• In this section we will discuss equilibrium
  involving _______ acids or bases with their
  conjugate acids or bases. (Ex HF with NaF or
  NH3 with NH4NO3).
• Write the equilibrium for HF and NaF below
• HF (aq) ? H(aq) F- (aq)
• 2. Since, the salt of HF is present (NaF), by Le
  Chateliers principle, the equilibrium of the
  acid dissociation of HF will not shift as hard to
  the right. So there will be ______ of the acid
  present than if there were only HF in water (less
  H).
• 3. The shift in equilibrium due to this salt
  (containing what is called a common ion) is
  referred to as the common ion ________.
• We have already used common ion effect in Chapter
  14 with polyprotic acids

weak
more
effect
2

• First determine H contribution from strongest
  acid (H3PO4).

H3PO4 (aq) ? H (aq) H2PO 4 -
(aq) Initial 3.0 M 0 M 0
M Final 3.0 M x x
x Ka HH2PO4- / H3PO4 7.5.10-3 x2 /
3.0 x try neglecting x x
((7.5.10-3)3.0)1/2 x 0.150 M (5 of
3.0M) H H2PO4- 0.15 M H3PO4 3.0 M
0.150 M 2.9 M Now analyze the H2PO4-
dissociation
H2 PO4- (aq) ? H (aq) HPO 4 -2
(aq) Initial 0.150 M 0.150 M 0
M Final 0.150 x 0.150 x x
Ka HHPO4-2 / H2PO4-
6.2.10-8 0.150 x x
0.150 x neglect x x HPO4-2
6.2.10-8 M Now analyze HPO4- dissociation
H PO4-2 (aq) ? H (aq) PO 4 -3
(aq) Initial 6.2.10-8 M 0.150 M 0
M Final 6.2.10-8 x 0.150 x
x Ka HPO4-3 / HPO4-2 4.8.10-13
0.150 x x 6.2.10-8
x neglect x x 4.8.10-13(6.2.10-8) / (0.150) x
PO4-3 2.0.10-19 M
OH- Kw / H OH-
1.0.10-14 / 0.150 M OH- 6.7.10-14
M
3

• Sample 15.1
• HF (aq) ? H(aq) F- (aq)
• Initial 1.0 M 0 1.0 M
• Final 1.0 M x x 1.0M x
• Sec 15.2 Buffered Solutions
• The most important application of common ion
  effect is in the preparation of buffer solutions.
  These solutions are prepared by mixing a
  _____________ acid or base with its conjugate (a
  weak acid or base is required).
• A buffered solution is resistant to ______
  changes by addition of acid or base. Our blood
  is buffered so as to resist changes in ____ when
  acids or bases are absorbed by our body.
• Sample Exercise 15.2
• HC2H3O2 (aq) ? H(aq) C2H3O2 - (aq)
• Initial 0.50 M 0
  0.50 M
• Final 0.50 M x x 0.50
  M x

Ka HF- / HF 7.2.10-4 x 1.0 x /
1.0 - x x H 7.2.10-4 M
dissociation (x / HF) 100 dissociation
(7.20 .10-4 M / 1.0 M ) 100 dissociation
0.072 Much small H and dissociation than
when we explored 1.0 M HF (0.027 M H and 2.7 )
weak
pH
pH
Ka HC2H3O2- / HC2H3O2 1.8.10-5 x 0.50
x / 0.50 - x x H 1.8.10-5 M pH -
log H pH -log 1.8.10-5 pH 4.74
4

• Sample 15.3 When adding H or OH- perform the
  following 2 steps
• Step 1 NaOH will dissociate into Na and OH-
  ions.The OH- will react with the ___________ as
  follows K Ka/Kw or 1 / Kb of conjugate base
  (very large)
• OH- (aq) HC2H3O2 (aq) ? H2O (l)
  C2H3O2- (aq)
• Initial 0.010 mol 1.0 L (0.50 mol/L)
  1.0 L (0.50 mol/L)
• Final 0 mol 0.500 mol 0.010
  mol 0.500 mol 0.010 mol
• Final molarity 0.490 mol / 1.0L
  0.510 mol / 1.0 L
• Step 2
• Determine the pH of the buffer with the new
  adjusted concentration of acid and conjugate base
• HC2H3O2 (aq) ? H (aq) C2H3O2 - (aq)
• Initial 0.490 M 0 0.510 M
• Final 0.490 M x x 0.510 M
  x

Weak acid
When 0.010 mol of OH- added to 1.0 L of
water OH- 0.010 mol / 1.0 L OH- 0.0100
M pH 14.00 log 0.0100 pH 14.00
2.000 pH 12.00 (very basic) Good thing we have
buffers eh?
Ka HC2H3O2- / HC2H3O2 1.8.10-5 x
0.510 x / 0.490 - x x 1.8.10-5 0.490 /
0.510 x H 1.729 .10-5 M (less H than
15.2) pH - log H pH -log 1.729 .10-5 pH
4.76 (just slightly more basic, than the
original buffer before NaOH was added)
5

• Sample Exercise 15.4 (something will be Lac
  ing)
• HLac (aq) ? H (aq)
  Lac - (aq)
• Initial 0.75 M 0 0.25 M
• Final 0.75 M x x 025 M x

Ka HLac- / HLac 1.4.10-4 x 0.25 x
/ 0.75 - x x H 1.4.10-4 0.75 /
0.25 H 4.20.10-4 M pH - log H pH
-log 4.20 .10-4 pH 3.38
Ignore the Henderson Hasselbach equation It is
merely a restatement of the K equation with 2
steps of algebra H Ka acid /
base -log of both sides pH pKa log acid /
base pH pKa log base / acid If you
cant neglect x it doesnt work You always know
the K expression So dont waste your time
deriving or memorizing it!
6
Sample Exercise 15.5 2 ways you can go here (the
acid dissociation of ammonium)
NH4 (aq) ? H(aq) NH3 (aq) Initial
0.40 M 0 0.25 M Final
0.40 M x x 0.25 M x Or
you could look at the NH3 base removing ___ from
water NH3 (aq) H2O (l) ? OH-(aq) NH4
(aq) Initial 0.25 M 0 M
0.40 M Final 0.25 M x x 0.40 M
x
Ka HNH3 / HC2H3O2 5.556.10-10 x 0.25
x / 0.40 - x x H 5.556 .10-10 0.40
/ 0.25 H 8.889.10-10 M pH - log H pH
-log 8.889 .10-10 pH 9.05
H
Kb OH-NH4 / NH3 1.8.10-5 x 0.40 x
/ 0.25 - x x OH- 1.8 .10-5 0.25 /
0.40 OH- 1.125.10-5 M pH 14.00 log
OH- pH 14.00 log 1.125 .10-5 pH 14.00 -
4.9488 pH 9.05 Using NH4
dissociation was quicker
7
Sample Exercise 15.6 Added H will react with the
______________ Again, think of the 2 steps Step
1 Perform the stoichiometry on the acid/base
reaction (K Kb/Kw very large) NH3
(aq) H (aq) ? NH4
(aq) Initial 1.0L (0.25 M)
0.10 mol 1.0L (0.40 M) Final 0.250
mol 0.10mol 0 mol
0.400 mol 0.10 mol Final M 0.150 mol / 1.0 L
0 M 0.500 mol / 1.0
L Step 2 Determine the pH of the Buffer
obtained Now look at the acid dissociation
NH4 (aq) ? H(aq) NH3
(aq) Initial 0.500 M 0
0.150 M Final 0.500 M x x
0150 M x Side sample If OH- is added to a
weak base or H is added to a weak acid. There
is no reaction. The OH- is merely a common ion
of the base reacting with water (H would be a
common ion of the acid dissociation). The
contributions of the weak base or weak acid can
be ignored. Example 0.012 moles of NaOH is added
to 1.0 L of 1.0 M NH3. NH3 (aq) H2O (l) ?
NH4 (aq) OH-(aq) Initial 1.0 M 0 M
0.012 M Final 1.0 M
x x 0.012 M x
Weak base
Ka HNH3 / NH4 5.556.10-10 x 0.150
x / 0.500 - x x H 5.556 .10-10 0.500
/ 0.150 H 1.852.10-10 M pH - log H pH
-log 1.852 .10-10 pH 8.73
Can neglect the OH- from the NH3 pH 14.00 log
0.012 pH 12.08
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Sec 15.3 Buffer Capacity 1. The pH of a
buffer is determined by the __________ of
A-/HA. 2. The capacity of the buffer (or how
much acid or base can be added w/o altering the
pH to a great extent) is determined by the
magnitudes (amounts) of A- and HA. A buffer
will work best when you have ________ amounts of
A- and HA. 3. Therefore, when you need a
buffer of certain pH, pick a weak acid whose ____
is equal to the desired pH. (buffer range pKa
pH /- 1) Exercise 15.7 Adding 0.010 mol HCl to
Solution A C2H3O2- (aq)
H (aq) ? HC2H3O2 (aq) Initial 1.0 L (5.00
M) 0.010 mol 1.0 L (5.00 M) Final
4.99 mol 0 mol 5.01
mol Final 4.99 mol / 1.0 L
5.01 mol / 1.0 L HC2H3O2 (aq)
? H(aq) C2H3O2- (aq) Initial 5.01 M
0 4.99 M Final 5.01 M x
x 4.99 M x Adding 0.010
mol HCl to Solution B C2H3O2- (aq) H
(aq) ? HC2H3O2 (aq) Initial 1.0 L (0.050 M)
0.010 mol 1.0 L (0.050 M) Final
0.040 mol 0 mol 0.060
mol Final 0.040 mol / 1.0 L
0.060 mol / 1.0 L HC2H3O2
(aq) ? H(aq) C2H3O2 - (aq) Initial 0.060 M
0 0.040 M Final 0.060
M x x 0.040 M x

ratio
equal
pKa
Ka HC2H3O2- / HC2H3O2 1.8.10-5 x 4.99
x / 5.01 - x x H 1.8.10-5 M pH -
log H pH -log 1.807 .10-5 pH 4.74
Ka HC2H3O2- / HC2H3O2 1.8.10-5 x
0.040 x / 0.060 - x x H 2.70 .10-5
M pH - log H pH -log 2.70 .10-5 pH
4.56
9

• Exercise 15.8 For a quality buffering system the
  buffer range should be pKa pH /- 1
• pKa - log (1.35.10-3)
• pKa 2.87
• b. pKa -log (1.3.10-5)
• pKa 4.87
• c. pKa -log (6.4.10-5)
• pKa 4.19
• d. pKa -log (3.5.10-8)
• pKa 7.46
• Choice c is the choice with the pKa closest to
  the pH so a (benzoic acid / benzoate) system
  would
• make for the best buffer of the choices listed

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What volume of 1.00 M NaOH should be added to
200. ml of 2.00 M acetic acid to create a buffer
whose pH is 4.30? HC2H3O2 (aq) ? H (aq)
C2H3O2- (aq) Here, a buffer will be created by
the reaction of HC2H3O2 with OH-. In order to
solve this problem, the relative final amounts
of acetic acid and acetate will be
examined. 1.8.10-5 HC2H3O2- /
HC2H3O2 since we know the final pH we know the
final H C2H3O2- / HC2H3O2 1.8.10-5 /
10-4.30 C2H3O2- 0.3591 HC2H3O2 (since each
is in the same volume) moles C2H3O2- 0.3591
molesHC2H3O2 Ill call this equation 1 In
order produce the buffer, the OH- will be
reacting with acetic acid to create acetate
ion HC2H3O2 (aq) OH- (aq) ? C2H3O2-
(aq) H2O (l) Initial 0.200L (2.00M)
x 0 Final
0.4000 moles x 0 x
We see that the final moles of acetate are
equal to the moles of NaOH added Substituting
final values into equation 1 x 0.3591 (0.4000
moles x) x 0.1436 moles - 0.3591 x 1.3591 x
0.1436 moles x 0.1057 moles OH- originally
added Determine volume of NaOH ?L 0.1057 moles
NaOH (L / 1.00 mol NaOH) 0.11 L

Answer Check HC2H3O2 (aq) ?
H(aq) C2H3O2-(aq) (0.4000
0.1057)moles/0.3057 L x 0.1057 moles /
0.3057 L Ka x0.1057 moles / 0.3057 L /
0.2944 moles / 0.3057 L Notice when acid and
base are present, the volume cancels H
1.8.10-5 0.2944 / 0.1057 H 5.013
.10-5 M pH - log 5.013 .10-5 pH 4.30