Study Guide
Buffer Solution Solubility
-
University:
Kentucky State University -
Course:
CHE 101 | General Chemistry I Academic year:
2020
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Views:
275
Pages:
33
Author:
embohescay9l
Buffer Solution
Solubility
CHEM-102
1 The common ion effect is the shift in equilibrium caused
by the addition of a compound having an ion in common
with the dissolved substance.
The presence of a common ion suppresses the ionization of
a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CHEM-102
CH3COONa (s)
Na+ (aq) + CH3COO- (aq)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
common
ion
2 Buffer Solution
A buffer solution is a solution of:
[weak conjugate acid-base pair]
1. A weak acid or a weak base and
2. The salt of the weak acid or weak
base
Both must be present!
A buffer solution has the ability to resist changes in pH
upon the addition of small amounts of either acid or base.
https://thebiologynotes.com/buffer-solutions/
CHEM-102
3 Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COO- (aq) + H2O (l)
Added strong acid/base react completely with
buffer component
CHEM-102
4 To have a good buffer (Higher Buffer capacity):
➢ weak conjugate acid-base pair
➢ Relatively high Conc.
➢ and ≈ equal [ conc. ] of the two components
A good buffer solution → The one has the best ability to
resist changes in pH
CHEM-102
5 Example 1:
Which of the following solutions can be classified as
buffer systems?
(a) H3PO4/KH2PO4 (b) HClO4/NaClO4 (c) HNO2/NaNO2
(d) C5H5N/C5H5NHCl [C5H5N is pyridine (base)]
Solution:
(a) H3PO4 is a weak acid, and its conjugate base, H2 PO-4 (weak
base)→ It is a buffer system.
(b) HClO4 is a strong acid, its conjugate base, ClO-4 (weak base)
→ The system cannot act as a buffer system.
(C) HNO2 is a week acid, its conjugate base (weak base)
→ The system It is a buffer system.
(d) C5H5N is a weak base and its conjugate acid, C5H5N+H
(weak acid) → It is a buffer system.
6 Example 1:
Describe how you would prepare a “phosphate buffer” with a
pH of about 7.40.
Solution:
For a buffer to function effectively, the concentrations of the acid
component must be roughly equal to the conjugate base
component. According to Equation (17.3), when the desired pH
is close to the pKa of the acid, that is, when
pH ≈ pKa,
[conjugate base]
log
0
[acid]
or
CHEM-102
[conjugate base]
1
acid
7 Triprotic acid:
H 3 PO 4 (aq ) ƒ H + (aq ) + H 2 PO -4 (aq)
K a1 = 7.5 × 10 -3 ; pK a1 = 2.12
H 2 PO -4 (aq ) ƒ H + (aq ) + HPO 42- (aq)
K a 2 = 6.2 × 10 -8 ; pK a 2 = 7.21
+
3HPO 2(
aq
)
ƒ
H
(
aq
)
+
PO
4
4 (aq )
K a 3 = 4.8 × 10 -13 ; pK a 3 = 12.32
The most suitable is HPO2/H
PO
4
2
4 , because the pKa of the acid
is closest to the desired pH. H2 PO-4
pH = pK a + log
[conjugate base]
[acid]
[HPO 24 ]
7.40 = 7.21 + log
[H 2 PO -4 ]
[HPO20.19
4 ]
=
10
= 1.5
[H 2 PO4 ]
[HPO 24 ]
log
= 0.19
[H 2 PO 4 ]
we could dissolve 1.5 moles of Na2HPO4 and 1.0 mole of
NaH2PO4 in enough water to make up a 1-L solution.
CHEM-102
8 Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
HA (aq)
H+ (aq) + A- (aq)
[H+]
Ka [HA]
=
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[A-]
pH = pKa + log
[HA]
CHEM-102
[H+][A-]
Ka =
[HA]
Henderson-Hasselbalch
equation
[conjugate base]
pH = pKa + log
[acid]
pKa = -log Ka
9 Essential features of a buffer:
➢ Relatively high and ≈ equal [ conc. ] of the two components
HF/NaF
HF F
NaOH
HCl
F-
HF
HF
NaOH reacts completely with HF
CHEM-102
-
F-
HCl reacts completely with F 10 NH3/NH4Cl
NH3 NH4+
0.1 M NaOH
NH3
→
H2O + NH3
NH4+
NH4+ + OH-
NaOH reacts completely with NH4+
CHEM-102
0.1 M HCl
NH3
NH4+
H+ + NH3 → NH4+
HCl reacts completely with NH3
11 1.0 M
1.0 M
1.0 M
1.0 M
HF F HCl
0.1 M
NaOH
0.1 M
HF
F-
HF
NaOH reacts completely with HF
1.0-0.1= 1.0+0.1=
0.9 M
HF
1.1 -M
F
F-
HCl reacts completely with F -
1.0+0.1=
1.1 M
HF
1.0-0.1= 1.0+0.1= 1.0-0.1=
0.9 -M
F
pH = pKa+ log
pH = pKa+ log
1.1
1.0-0.1= 1.0+0.1=
1.1 M
0.9 M
0.9 M
1.1 M
NH3
NH4+
NH3
NH4+
0.9
1.1
0.9
-log(6.8 x 10-4)
12 Example 1:
Calculate the pH of a buffer system containing 1.0 M
CH3COOH and 1.0 M CH3COONa.
Then, what is the pH of the buffer system after the addition of
0.10 mole of gaseous HCl to 1.0 L of the solution? Assume
that the volume of the solution does not change when the
HCl is added.
Ka of CH3COOH is 1.8 x 10-5
Solution:
CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
CHEM-102
1.0
-x
1.0-x
H+(aq) + CH3COO-(aq)
0
+x
x
1.0
+x
1.0+x
13 CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
1.0
-x
1.0-x
H+(aq) + CH3COO-(aq)
0
+x
x
1.0
+x
1.0+x
[H + ][CH 3COO - ]
Ka =
[CH 3COOH]
1.8 × 10-5 =
(x )(1.0+x)
(1.0-x)
Assuming 1.0 + x ≈ 1.0 and 1.0 - x ≈ 1.0, we obtain
(x)(1.0+x)
x(1.0)
1.8 × 10 =
(1.0-x)
1.0
-5
x = [H+] = 1.8 x 10-5 M
pH = -log (1.8 x 10-5 ) = 4.74
CHEM-102
14 (b)
Initial (mol):
Change (mol):
Final (mol):
HCl(aq)
→ H+(aq) +
0.10
-0.10
0
0
+0.10
0.10
Cl-(aq)
0
+0.10
0.10
The neutralization reaction is summarized as:
CH3COO-(aq) + H+(aq) → CH3COOH(aq)
Initial (mol):
Change (mol):
1.0
-0.10
0.10
-0.10
1.0
+0.10
Final (mol):
0.90
0
1.1
CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
CHEM-102
H+(aq) + CH3COO-(aq)
1.1
0
0.90
-x
1.1-x
+x
x
+x
0.90+x
15 CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
H+(aq) + CH3COO-(aq)
1.1
-x
0
+x
0.90
+x
1.1-x
x
0.90+x
[H + ][CH 3COO - ]
Ka =
[CH 3COOH]
(x )(0.90+x)
1.8 × 10 =
(1.1-x)
-5
0.90 + x ≈ 0.90
(x)(0.90+x)
x(0.90)
1.8 × 10 =
(1.1-x)
1.1
-5
x = [H+] = 2.2 x 10-5 M
CHEM-102
pH = -log (2.2 x 10-5 ) = 4.66
16 Example 1:
Calculate the pH of a buffer solution containing 0.10 M NH3
and 0.20 M NH4Cl. Kb for NH3 (= 1.8 × 10-5).
NH3/NH4Cl
The conjugate acid is NH4+, Ka can be calculated from Kb for
NH3 →Ka for is 5.6 × 10-10.
pKa = -log(5.6 ×10-10) = 9.25
pH = 9.25+ log )0.10 /0.20(= 8.95
CHEM-102
17 Examples: Calculating the pH of a buffer
1) Calculate the pH of a buffer consisting of 0.50 M
HF and 0.45 M F -. (Ka HF = 6.8 x 10-4).
2) What is the pH of this HF / F - buffer after
adding 0.10 moles of NaOH to 1.0 L of the
buffer? (Ka HF = 6.8 x 10-4 ).
3) What is the pH of this HF / F - buffer after
adding 0.10 moles of HCl to 1.0 L of the buffer?
(Ka HF = 6.8 x 10-4 ).
TRY THEM!
CHEM-102
18 The question we asked earlier was how to prepare a buffer of a given
pH—for example, pH 4.90. You can see that you need to fnd a
conjugate acid–base pair in which the pKa of the acid is close to the
desired pH. Ka for acetic acid is 1.7 ×10-5, and its pKa is -log(1.7 × 105) = 4.77. You can get a pH somewhat higher by increasing the ratio
[base]/[acid].
Consider the calculation of the pH of a buffer containing 0.10 M NH3
and
0.20 M NH4Cl. The conjugate acid is NH4+, whose Ka you can
calculate from Kb for NH3 (= 1.8 × 10-5). Ka for NH4+ is 5.6 × 10-10,
and the pKa is -log(5.6 ×10-10) = 9.25. Hence,
pH = 9.25+ log (0.10 /0.20)= 8.95
CHEM-102
19 Solubility Equilibria
AgCl (s)
Ksp = [Ag+][Cl-]
MgF2 (s)
Ag2CO3 (s)
Ca3(PO4)2 (s)
Ag+ (aq) + Cl- (aq)
Ksp is the solubility product constant
Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
2Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
No precipitate
Precipitate will form
20 21 Molar solubility (mol/L) is the number of moles of solute
dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in
1 L of a saturated solution.
22 17.6
The solubility of calcium sulfate (CaSO4) is found to be
0.67 g/L. Calculate the value of Ksp for calcium sulfate.
We are given the solubility of CaSO4 and asked to calculate its
Ksp. The sequence of conversion steps, according to Figure
17.9(a), is
solubility of
CaSO4 in g/L
molar solubility
of CaSO4
[Ca2+] and
[ SO2−]
4
Ksp of
CaSO4
23 17.6
Solution Consider the dissociation of CaSO4 in water. Let s
be the molar solubility (in mol/L) of CaSO4.
CaSO4(s)
Initial (M):
Change (M):
Equilibrium (M):
ƒ
2−
Ca2+(aq) + SO4 (aq)
-s
0
+s
s
0
+s
s
The solubility product for CaSO4 is
Ksp = [Ca2+][ SO24 − ] = s2
24 17.6
First, we calculate the number of moles of CaSO4 dissolved in
1 L of solution:
0.67 g CaSO 4
1 mol CaSO 4
×
= 4.9 × 10-3 mol/L = s
1 L soln
136.2 g CaSO 4
From the solubility equilibrium we see that for every mole of
CaSO4 that dissolves, 1 mole of Ca2+ and 1 mole of SO24 − are
produced. Thus, at equilibrium,
[Ca2+] = 4.9 x 10-3 M and [ SO24 − ] = 4.9 x 10-3 M
Ksp = [Ca2+] ][ SO24 −]
= (4.9 x 10-3 )(4.9 x 10-3 ) = 2.4 x 10-5
25 17.7
Using the data in Table 17.2, calculate the solubility of
copper(II) hydroxide, Cu(OH)2, in g/L.
Strategy
Ksp of
Cu(OH)2
[Cu2+] and
[OH-]
Initial (M):
Change (M):
Equilibrium (M):
molar solubilty
of Cu(OH)2
solubility of
Cu(OH)2 in g/L
Cu(OH)2(s)
Cu2+(aq) +
2OH-(aq)
-s
0
+s
s
0
+2s
2s
Ksp = [Cu2+][OH-]2
= (s)(2s)2 = 4s3
26 17.7
From the Ksp value in Table 17.2, we solve for the molar
solubility of Cu(OH)2 as follows:
2.2 × 10-20 = 4s 3
Hence
-20
2.2
×
10
s3 =
= 5.5 × 10-21
4
s = 1.8 × 10-7 M
Finally, from the molar mass of Cu(OH)2 and its molar
solubility, we calculate the solubility in g/L:
1.8 10−7 mol Cu(OH) 2
97.57 g Cu(OH) 2
solubility of Cu(OH) 2 =
×
1 L soln
1 mol Cu(OH)2
= 1.8 ×10-5 g / L
27 28 17.8
Exactly 200 mL of 0.0040 M BaCl2 are mixed with exactly
600 mL of 0.0080 M K2SO4. Will a precipitate form?
29 17.8
Solution
The number of moles of Ba2+ present in the original 200 mL of
solution is
0.0040 mol Ba 2+
1L
200 mL ×
×
= 8.0 × 10-4 mol Ba 2+
1 L soln
1000 mL
The total volume after combining the two solutions is
800 mL. The concentration of Ba2+ in the 800 mL volume is
-4
8.0
×
10
mol 1000 mL
[Ba 2+ ] =
×
800 mL
1 L soln
= 1.0 × 10-3 M
30 17.8
The number of moles of SO24 −in the original 600 mL solution is
0.0080 mol SO24−
1L
600 mL ×
×
= 4.8 × 10-3 mol SO24−
1 L soln
1000 mL
2−
SO
The concentration of
4 in the 800 mL of the combined
solution is
-3
4.8
×
10
mol 1000 mL
2−
[SO 4 ] =
×
800 mL
1 L soln
= 6.0 × 10-3 M
31 17.8
Now we must compare Q and Ksp. From Table 17.2,
BaSO4(s)
Ba2+(aq) + SO24 − (aq)
Ksp = 1.1 x 10-10
As for Q,
Q = [Ba2+]0[ SO24 − ]0 = (1.0 x 10-3)(6.0 x 10-3)
= 6.0 x 10-6
Therefore, Q > Ksp
The solution is supersaturated because the value of Q indicates
that the concentrations of the ions are too large. Thus, some of
the BaSO4 will precipitate out of solution until
2−
[Ba2+][ SO4 ] = 1.1 x 10-10
32 The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the
salt.
33
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of 33
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