Ch 16 Acid-Base Equilibria
Finding [H3O+1] or [OH−1] for weak acids and bases involves equilibrium constants.
This is because they do not dissociate fully and they have a reversible hydrolysis reaction.
[H3 O+1 ][A−1 ]
is the expression for HA + H2O ⇌ A-1 + H3O+1
-
Acid Ionization Constant Ka =
-
For acetic acid (HOAc)
Ka =
-
Correspondingly
For acetic acid
This is similar to
pKa = − log(Ka) and Ka = 10−pKa
pKa = − log(1.7 × 10−5) = 4.75 and Ka = 10−(4.75) = 1.7 × 10−5
pH = − log[H3O+1] and [H3O+1] = 10−pH
-
[HA]
[H3 O+1 ][OAc−1 ]
[HOAc]
= 1.7 × 10−5
Ka must be determined experimentally (two general methods)
- Degree of ionization (or % ionization) is determined by measuring electrical conductivity or
another colligative property (that depends only on total moles of all solutes).
- Measure pH or [H3O+1] directly, then calculate all other concentrations with an eqm table.
Ex 16.1 0.012 M HNic has pH = 3.39, Find Ka and degree of ionization
HNic + H2O ⇌ Nic−1 + H3O+1
0.012
0
0
x = [H3O+] = 10−pH = 10−3.39 = 4.1 × 10−4 M
−x
+x
+x
0.012 − x
x
x
Ka =
x << 0.012, so 0.012 – x ≈ 0.012
Ka =
[H3 O+1 ][Nic−1 ]
[HNic]
(4.1 × 10−4 )2
0.012
=
x2
0.012−x
= 1.4 × 10−5
Generally, variable in denominator can be neglected for weak acid/base if [HA]0/Ka >100
Was it OK to neglect x? [HA]/Ka = (1.2 × 10−2) / (1.4 × 10−5) = 850 and
850 > 100
−1
Degree of Ionization = [A ]eqm/ [HA]0
[Nic−1] / [HNic] = x / 0.012 = (4.1 × 10−4) / (0.012) = 0.034 (or 3.4%)
Ex 16.2 For 0.10 M HNic, find pH and degree of ionization using Ka = 1.4 × 10−5
- HNic + H2O ⇌ Nic−1 + H3O+1
0.10
0
0
−x
+x
+x
0.10 − x
x
x
−5
[HA]/Ka = (0.1)/(1.4 × 10 ) = 7 × 104 > 100 We can neglect x.
[H3 O+1 ][Nic−1 ]
x2
x2
-
Ka = 1.4 × 10−5 =
-
x = 1.4 × 10 and [H3O ] = x = 1.2 × 10 M and pH = − log(1.2 × 10−3) = 2.92
Degree of Ionization = x / 0.10 = 1.2 × 10−2 = 0.012 (or 1.2%)
Note that Degree of Ionization depends on both Ka and [HA]
2
−6
[HNic]
+1
=
(0.10−x)
−3
=
0.10
Ex 16.3 For 0.325 g HAcs in 0.500 L water, find pH if Ka = 3.3 × 10−4
- 0.325g / (180.2 g/mol) = 0.00180 mol 0.00180 mol / 0.500 L = 0.00360 M
- HAcs + H2O ⇌ Acs−1 + H3O+1
0.0036
0
0
−x
+x +x
0.0036 − x
x
x
[H3 O+1 ][Acs−1 ]
x2
-
Ka = 3.3 × 10−4 =
-
Cannot neglect x b/c [HA]/Ka ≈ 11
x2 = (3.3 × 10−4)(0.0036 − x)
Rearranges to x2 + (3.3 × 10−4)(x) − (1.2 × 10−6) = 0
Use the quadratic formula to get x = (½)[− (3.3 × 10−4) ± (2.2 × 10−3)]
The result is x = 9.4 × 10−4 M = [H3O+]
pH = − log(9.4 × 10−4) = 3.03
[HAcs]
=
0.0036−x
Polyprotic Acids
- H2SO4 + H2O  HSO4−1 + H3O+1
KA1 ≈ ∞ (strong acid)
−1
−2
+1
HSO4 + H2O  SO4 + H3O
KA2 = 1.1 × 10−2
- H2CO3 + H2O  HCO3−1 + H3O+1
KA1 = 4.3 × 10−7
HCO3−1 + H2O  CO3−2 + H3O+1
KA2 = 4.8 × 10−11
- KA1 > KA2 > KA3 (for example H3PO4 > H2PO4−1 > HPO4−2)
The increasing negative charge decreases acidity.
Ex 16.4 Ascorbic acid or vitamin C is diprotic (H2Asc). Find pH of a 0.10 M solution.
- KA1 = 7.9 × 10−5 and KA2 = 1.6 × 10−12
- H2Asc + H2O ⇌ HAsc−1 + H3O+1
0.10
0
0
−x
+x
+x
0.10 − x
x
x
[H3 O+1 ][HAsc−1 ]
x2
x2
-
KA1 = 7.9 × 10−5 =
-
x2 = KA1[H2A] = 7.9 × 10−6
and
x = √K A1 [H2 A] = 2.8 × 10−3 M
Ignore change in [H3O+1] due to 2nd equation because 2nd acid is much weaker than 1st.
So:
[H3O+1] = [HAsc−1] = 2.8 × 10−3 M
And:
pH = − log(2.8 × 10−3) = 2.55
Next, use x for starting concentrations of both H3O+1 and HAsc−1 in the 2nd table (next page).
[H2 Asc]
=
(0.10−x)
=
0.10
-
HAsc−1 + H2O ⇌ Asc−2 + H3O+1
2.8×10−3
0
2.8×10−3
−y
+y
+y
−3
(2.8 × 10 ) − y
y (2.8 × 10−3) + y
-
KA2 = 1.6 × 10−12 =
-
y << 2.8 × 10−3 so [(2.8 × 10−3) + y] = [(2.8 × 10−3) − y] = 2.8 × 10−3 M
This makes both of those terms cancel and KA2 = 1.6 × 10−12 M = y = [Asc−2]
In general [A−2] = KA2 for a diprotic (or triprotic) acid
[H3 O+1 ][Asc−2 ]
[HAsc−1 ]
=
[(2.8 × 10−3 ) + y][y]
[ (2.8 × 10−3 ) − y]
=?
Triprotic Acid (has a third equilibrium table)
-
[H3O+1] = [H2A−1] = x = √K A1 [H3 A]
[HA−2] = y = KA2
HA−2 + H2O ⇌ A−3 + H3O+1
KA2
0
x
−z
+z
+z
(KA2 − z)
z
x+z
-
KA3 =
z(x + z)
(KA2 −z)
=
zx
and [A−3] = z =
KA2
𝐊 𝐀𝟐 𝐊 𝐀𝟑
𝐱
=
𝐊 𝐀𝟐 𝐊 𝐀𝟑
√𝐊 𝐀𝟏 [𝐇𝟑 𝐀]
Base Ionization Equilibria
NH3 + H2O ⇌ NH4+1 + OH−1
Kb =
[NH4 +1 ][OH−1 ]
[NH3 ]
= 1.8 × 10−5
pKb = − log(Kb) = 4.74
Ex 16.5 What is pH for 0.0075 M Morphine with Kb = 1.6 × 10−6 ?
- Mor
+ H2O ⇌ HMor+1 + OH−1
0.0075
0
0
−x
+x
+x
(0.0075 – x)
x
x
[HMor+1 ][OH−1 ]
-
Kb =
-
x2 = 1.2 × 10−8
pOH = − log(x) = 3.96
[Mor]
=
x2
(0.0075 − x)
and
and
=
x2
(0.0075)
= 1.6 × 10−6
x = 1.1 × 10−4 M = [OH−1]
pH = 14.00 – 3.96 = 10.04
Acid-Base Properties of salt sol’ns - Salts can be acidic, basic, or neutral
- NaCN(s) ⇌ Na+1 + CN−1
Na+1 is not reactive with H2O, but CN−1 is a base and has a Kb expression.
CN−1 + H2O ⇌ HCN + OH−1
Kb =
[HCN][OH−1 ]
[CN−1 ]
= 2.0 × 10−5
-
Reaction of an acid or a base with water is called a hydrolysis (acid/base can be ion as well)
NH4+1 + H2O ⇌ NH3 + H3O+1 Also is a hydrolysis, so NH4+1 has a Ka expression.
-
Ka =
[NH3 ][H3 O+1 ]
[NH4 +1 ]
= 5.6 × 10−10
Acidity and Basicity of Salts (Ex 16.6)
- Conjugate ions of strong acids/bases do not hydrolyze.
So, salts containing both conjugates together, such as KCl, are neutral.
- Salts with the conjugates of a strong base and a weak acid, such as NaF, are basic,
because the conjugate of the weak acid will hydrolyze and is a base.
- Salts with the conjugates of a weak base and a strong acid, such as Zn(NO3)2, are acidic,
because the conjugate of the weak base will hydrolyze and is an acid.
- Salts with the conjugates of a weak base and a weak acid,
such as NH4CN, have two ions that hydrolyze.
So, they can be basic or acidic depending on the components.
Ka for NH4+1 is 5.6 × 10−10, and Kb for CN−1 is 2.0 × 10−5.
The Kb is larger than the Ka, so NH4CN is basic.
Ionization Constants (Ka and Kb) for a Pair of Conjugates
- HCN + H2O ⇌ CN−1 + H3O+1
Ka
−1
−1
CN + H2O ⇌ HCN + OH
Kb
+1
−1
2H2O
⇌ H3O + OH
Kw
- For a pair of conjugates Ka × Kb = Kw = 1.00 × 10−14
- Also, pKa + pKb = pKw = 14.00
Ex 16.7 a. Ka for HCN is 4.9×10−10, and Kb for CN−1 is 2.0 × 10−5
(4.9 × 10−10)(2.0 × 10−5) = 1.00 × 10−14 = Kw
b. Ka for NH4+1 is 5.6 × 10−10, and Kb for NH3 = 1.8 × 10−5
(5.6 × 10−10)(1.8 × 10−5) = 1.00 × 10−14 = Kw
Ex 16.8 pH of a Sodium Nicotinate (NaNic) Salt Solution
Nic−1 + H2O ⇌ HNic + OH−1 Ka of HNic = 1.4 × 10−5
Kb of Nic−1 = Kw / Ka
0.10
0
0
Kb = (1.00 × 10−14) / (1.4 × 10−5) = 7.1 × 10−10
−x
+x
+x
Kb = (x)(x) / (0.10 – x) = x2 / 0.10
0.10 − x
x
x
[OH−1] = x = √0.10Kb = 8.4 × 10−6
pOH = −log[OH−1] = 5.1
pH = 14.0 – 5.1 = 8.9
Common Ion Effect
- The common ion effect is a shift in equilibrium (usually to left) that is caused
by adding a 2nd solute which possesses an ion in the reaction.
- For acetic acid, below, we can add HCl, which provides H3O+1.
HOAc + H2O ⇌ OAc−1 + H3O+1
- H3O+1 is on the product side, so the equilibrium shifts to the left.
- Less HOAc dissociates, so this decreases the degree of ionization.
- Adding OH−1 would remove H3O+1, which would create the opposite effect.
So, the equilibrium shifts to the right.
Common Ion Effect is similar for weak base Ca(OH)2(s) ⇌ Ca+2(aq) +2OH−1(aq)
- Solubility Product: Ksp = [Ca+2][OH−1]2
- Solid is not included in Ksp! Pure solids are excluded from equilibrium constants.
- Molar solubility (molsol) is the effective mol/L of the dissolved solid.
- Without a common ion, molsol = [Ca+2] = (½)[OH−1].
- If another source of Ca+2 (such as CaCl2) is added, then eqm shifts to left as product is added.
- Also, molsol no longer equals [Ca+2]. Molsol will be decreased as eqm shifts to left.
That is, less of the solid reactant dissolves because of the additional Ca+2.
- But molsol still equals (½)[OH−1], and can be calculated by titrating with an acid.
At the stoichiometric point:
mol H+ added = mol OH−1 in sample
- For any slightly soluble salt, addition of a 2nd salt containing one of the ions involved
will shift the equilibrium to the left and decrease the molsol.
- Note that Ksp is a constant and does not change.
Ex 16.9 0.10 M HOAc (Ka = 1.7 × 10−5) and 0.010 M HCl. Find Degree of Ionization
- HOAc + H2O ⇌ OAc−1 + H3O+1
0.10
0
0.010
−x
+x
+x
0.10 − x
+x
0.010 + x
(0.010+x)(x)
(0.010)(x)
-
Ka = 1.7 × 10−5 =
-
x = 1.7 × 10−4 M = [OAc−1]
Degree of Ionization = (1.7 × 10−4) / ( 0.10) = 1.7 × 10−3 = 0.17%
The degree of ionization for 0.10 M HOAc without HCl is 1.3%,
and can be determined using the same method as example 16.2.
The degree of ionization is much smaller with HCl present
because the added H3O+1 shifts the equilibrium back to the left.
-
0.10−x
=
(0.10)
= 0.10x
Buffers – A solution which resists changes in pH when small amounts of acid or base are added.
- A buffer contains either a weak acid or base, along with the conjugate ion.
So, the buffer solution possesses both an acid component and a base component.
- Two important characteristics: capacity and pH.
- Capacity is the amount of acid or base needed to create a large pH change,
which depends on how much of each conjugate is in solution.
- The pH is determined by Ka and by the ratio of the two conjugates’ concentrations.
Ex 16.10
-
Find pH for 0.10 M HOAc (Ka = 1.7 × 10−5) and 0.20 M NaOAc.
OAc−1 +
HOAc + H2O ⇌
0.10
−x
0.10 − x
0.20
+x
0.20 + x
(0.20 + x)(x)
0.20x
H3O+1
Ka = 1.7 × 10−5 =
0
+x
+x
Ka = 2.0x
[H3O+] = x = (½)Ka = 8.5 × 10−6 M
pH = − log(8.5 × 10−6) = 5.07
(0.10−x)
=
0.10
Ex 16.11
- [NH3] = (0.100 M)(0.060 L) / (0.100 L) = 0.060 M
- [NH4+] = (0.100 M)(0.040 L) / (0.100 L) = 0.040 M
- NH3 + H2O ⇌ NH4+1 + OH−1
0.060
0.040
0
−x
+x
+x
0.060 − x
0.040 + x + x
(0.040+x)(x)
0.040x
x
-
Kb = 1.8 × 10−5 =
-
x = 1.5(Kb) = 2.7 × 10−5 M = [OH−1]
pOH = − log(2.7 × 10−5) = 4.57 and pH = 14.00 − 4.57 = 9.43
(0.060−x)
=
0.060
=
1.5
Ex 16.12 75.0 ml 0.10 M HOAc (Ka = 1.7×10−5) and 0.20 M NaOAc, with 9.5 ml 0.10 M HCl
- mol H3O+ = (0.0095L)(0.10M) = 0.00095 mol H3O+1, and H3O+ will convert OAc−1 to HOAc
- mol HOAc = (0.075L)(0.10M) + 0.00095 = 0.0075 + 0.00095 = 0.00845 mol HOAc
[HOAc] = (0.00845 mol) / (0.0845 L) = 0.100 M
- mol OAc−1 = (0.075L)(0.20M) − 0.00095 mol = 0.0150 − 0.00095 mol = 0.01405 mol OAc−1
[OAc−1] = (0.01405 mol) / (0.0845 L) = 0.166 M
-
HOAc + H2O ⇌
0.100
−x
0.100 − x
OAc−1 +
0.166
+x
0.166 + x
(0.166+x)(x)
0.166x
H3O+1
Ka = 1.7 × 10−5 =
0
+x
+x
Ka = 1.66x
[H3O+] = x = Ka / 1.66 = 1.0 × 10−5 M
pH = − log(1.0 × 10−5) = 5.0
0.10−x
=
0.100
Henderson-Hasselbalch Equation
(Ka )[HA]
Derive from the log of the Ka equation:
[H3O+1] =
For weak acid buffer:
pH = pKa + 𝐥𝐨𝐠 (
For weak base buffer:
pH = (pKw − pKb) + 𝐥𝐨𝐠 (
[A−1 ]
[𝐀−1 ]
[𝐇𝐀]
)
[𝐁]
)
[𝐇𝐁 +𝟏 ]
Use the Henderson-Hasselbalch equation with Ex 16.10
pKa = − log(1.7 × 10−5) = 4.77
[A−1 ]
0.20
pH = pKa + log ( [HA] ) = 4.77 + log(0.10) = 4.77 + log(2) = 4.77 + 0.301 = 5.07
Use the Henderson-Hasselbalch equation with Ex 16.11
pKb = − log(1.8 × 10−5) = 4.74
pKa = pKw − pKb = 14.00 – 4.74 = 9.26
[B]
0.060
pH = pKa + log ([HB+1 ]) = 9.26 + log(0.040) = 9.26 + 0.18 = 9.44
Acid-Base Titration Curves – plot of pH vs. Volume of titrant
- Equivalence Point is the point in a titration where stoichiometric amount of titrant is added.
At this point, the solution is neutralized precisely so that: mol OH−1 = mol H3O+1
- pH = 7 at the equivalence point if both strong acid and strong base are used.
- Weak acids/bases have conjugates that hydrolyze, so the pH is not 7 at their equiv point.
- Titration of a weak acid yields a basic conjugate, so pH > 7 at its equivalence point.
- Conversely, a weak base has pH < 7 at its equivalence point.
- If the equivalence point is unknown, then the indicator selection is difficult.
Ex 16.14 0.025 L of 0.10 M HNic is titrated to its equivalence point with 0.10 M NaOH
- Find the equivalence point pH if Ka = 1.4 × 10−5
- mol OH−1 added = mol HNic in sample = (0.025L)(0.10M) = 0.0025 mol
- All of the 0.0025 moles of HNic are converted to Nic−1 at the equivalence point.
- Total V = 0.025 L + 0.025 L = 0.050 L
- At the equivalence point, we have [Nic−1] = moles / V = 0.0025 mol / 0.050 L = 0.050 M
- Use this concentration as the starting point on an equilibrium table for the
hydrolysis of Nic−1. Solve the table to find the equivalence point pH.
- Nic−1 + H2O ⇌ HNic + OH−1
0.050
0
0
−x
x
x
0.050 − x
x
x
Kw
Ka = 1.4 × 10−5
-
x = √36 × 10−12 = 6.0 × 10−6 = [OH−1] pOH = 5.22
Kb =
Ka
= 7.1 × 10−10 =
[OH−1 ][HNic]
-
[Nic−1 ]
=
x2
0.050
pH = 14.00 – 5.22 = 8.78