Download Equilibria of Polyprotic Acids and Bases: H3PO4 and Ammonia and more Lecture notes Chemistry in PDF only on Docsity!
POLYPROTIC ACIDS AND BASES: Very important!
-- acids that can lose, and bases that can pick up, more than one H +^ (e.g. diprotic H 2 A and triprotic H 3 A acids).
Ionization occurs stepwise, and each step has its own Ka.
H 3 PO 4 (aq) ⇌ H+^ (aq) + H 2 PO 4 -^ (aq) Ka1 = 7.2 x 10 -3^ increasing H 2 PO 4 -^ (aq) ⇌ H+^ (aq) + HPO 4 2-^ (aq) Ka2 = 6.3 x 10 -8^ acid HPO 4 2-^ (aq) ⇌ H +^ (aq) + PO 4 3-^ (aq) Ka3 = 4.2 x 10-13^ strength
Note: (i) K a1 > Ka2 > Ka3 Always true for polyprotic acids, i.e., each ionization
step is more difficult because it is more difficult to remove H+^ from a molecule as its negative charge increases.
BUT this does not apply for ionic salts of these acids, e.g., Na 3 PO 4 , Na 2 SO 4 , etc.
They dissociate 100% in one step.
Na 3 PO 4 (aq) → 3 Na+^ (aq) + PO 4 3-^ (aq)
(ii) The conjugate bases are polyprotic bases, i.e., PO 4 3-^ is a triprotic base – it can pick up 3 H+^.
Table 18.5 Successive K a values for Some Polyprotic Acids at
25°C
(b) Second dissociation : H 2 PO 4 -^ (aq) ⇌ H+^ (aq) + HPO 4 2-^ (aq)
Ka2 = 6.3 x 10 -8^ [init] 0.19 M 0.19 0 [change] -y +y +y [equil] (0.19-y) (0.19+y) y
K (^) a2 = = = 6.3 x 10-8^ = y (assume y << 0.19)
y = 6.3 x 10-8^ (very small - assumption definitely okay !!)
Note: (1) [H+] from second dissociation = Ka2 = [HPO 4 2-] (2) [H+] from second dissociation is negligible!
(c) Third dissociation : HPO 4 2-^ (aq) ⇌ H+^ (aq) + PO 4 3-^ (aq)
Ka3 = 4.2 x 10 -13^ [init] 6.3 x 10-8^ M 0.19 0 [change] -z +z +z [equil] (6.3 x 10-8)-z (0.19+z) z
Ka3 = = ≈ = 4.2 x 10 -
∴ z = = 1.4 x 10 -19^ M (assumption okay !!) [H+^ ] from third dissociation completely negligible – almost nothing.
−
2 4
2 4 [HPO
[H ][HPO ] ( 0. 19 y
y( 0. 19 y) −
[HPO ]
[H ][PO ]
2 4
3 4 −
( 6. 3 x 10 ) z
( 0. 19 z)z (^8) −
−
(d) Calculate [OH-], pH and pOH pH = -log[H+^ ] = -log (0.19) = 0.72 pOH = -log[OH-^ ] and pOH = 14.00 – pH
via [OH-^ ] [H+^ ][OH-^ ] = Kw = 1.0 x 10 -14^ (at 25 °C) ∴ [OH-^ ] = = = 5.3 x 10 -
pOH = -log[OH-^ ] = -log(5.3 x 10-14) = 13.
(via pH pH + pOH = 14.00 ∴ pOH = 14.00 – pH = 14.00 = 0.72 = 13.
Summary: In a 5.0 M H 3 PO 4 solution: [H 3 PO 4 ] = 4.8 M [H +^ ] = [H 3 O+^ ] = 0.19 M [H 2 PO 4 -^ ] = 0.19 M [OH -^ ] = 5.3 x 10-14^ M [HPO 4 2-^ ] = 6.3 x 10-8^ M pH = 0. [PO 4 3-^ ] = 1.4 x 10-19^ M pOH = 13.
**** Remember:** (1) The [H 3 O+^ ] is determined by K (^) a1. [H 3 O+^ ] from other dissociations (Ka2, Ka3) is negligible, and is ignored. (2) [HPO 4 2-^ ] from second dissociation = Ka2 **
[H ]
1.0 x 10 -
1.0 x 10 -
K (^) w = [H +][OH
pH pOH
pH = -log[H^3 O^
] (^) pOH = -log[OH
pH + pOH = 14.
H 3 O +^ OH
Figure 18.9 Abstraction of a proton from water by the base
methylamine.
Lone pair of N
pair binds H +
Example 1. What is Kb for quinine (anti-malarial drug) if the pH of a 1.5 x
10 -3^ M solution is 9.84? ( NOTE: pH > 7.00 ∴ quinine is a base) Qui (aq) + H 2 O (aq) ⇌ HQui+^ (aq) + OH-^ (aq) [init] 1.5 x 10 -3^ M 0 0 [change] -x +x +x [equil] (1.5 x 10 -3)-x x x
Kb = = ≈
pH = 9.84 ∴ pOH = 4.16 ∴ [OH-^ ] = x = 6.9 x 10-5^ M (check: x = [OH-^ ] << 1.5 x10-3^ (4.6%) ∴ assumption okay) ∴ Kb = = 3.1 x 10-
Example 2. Calculate the pH of a 15.0 M solution of NH 3 (Kb = 1.8x10-5)
NH 3 (aq) + H 2 O (aq) ⇌ NH 4 +^ (aq) + OH-^ (aq) [init] 15.0 M 0 0 [change] -x +x +x [equil] 15.0-x x x
Kb = = ≈ = 1.8 x 10-
∴ x 2 = 2.7 x 10-4^ ∴ x = 1.6 x 10-2^ (<< 15.0 ∴ assumption okay)
∴ [OH-^ ] = 1.6 x 10-2^ M ∴ pOH = 1.80 ∴ pH = 14.00 - 1.80 = 12.
[Qui]
[HQui +][OH-] ( 1. 5 x 10 ) x
x 3
2 − (^) − 3
2
- 5 x 10
x −
-5 2 1.5x 10
(6.9x 10 )
[NH ]
[NH ][OH ] 3
4
x 2 − (^15). 0
x 2
Table 18.6 K b Values for Some Molecular (Amine) Bases at 25°C
Example 1: Ka for HF is 6.8 x 10-4. What is Kb for F -^? F -^ is the conjugate base of HF, therefore its Kb is related to the Ka of HF by
Ka Kb = Kw ∴ Kb = = = 1.5 x 10-
Note: Kb very small ∴ do not get much [OH -^ ] in a solution of e.g. NaF.
Example 2: What is the pH of a 2.5 M solution of sodium acetate (NaAc)? Ka for acetic acid (HAc) is 1.8 x 10-5.
Acetate (Ac-^ = CH 3 COO-^ ) is the conjugate base of acetic acid (HAc = CH 3 COOH).
Ac-^ (aq) + H 2 O(l) ⇌ HAc(aq) + OH-^ (aq) [init] 2.5 M 0 0 [change] -x +x +x [equil] (2.5-x) x x
Kb = = ≈ (assume x << 2.5)
We need Kb. We were given Ka. ** be careful not to use the wrong K! **
∴ Kb = = = 5.55 x 10-
∴ = 5.55 x 10-10^ ∴ x = 3.72 x 10-5^ = 3.7 x 10-5^ M (small! - assumption okay)
∴[OH-^ ] = 3.7 x 10-5^ ∴ pOH = -log [OH-^ ] = 4.43 ∴ pH = 14.00 – pOH = 9.
a
w K
K
6.8x 10
1.0x 10
[Ac ]
[HAc ][OH ] −
− (2.5- x)
x 2
x 2
a
w K
K
1.8x 10
1.0x 10
x 2
B. Acidic solutions : (Three ways to get this.)
(i) The anion of a strong acid (Cl-^ , Br -^ , I -^ , NO 3 -^ , ClO 4 -^ ) and the cation
(conjugate acid) of a weak base (e.g., NH 4 +^ ).
The anion does not react with water, but the cation is a weak acid and will form H 3 O+^. NH 4 +^ (aq) + H 2 O (l) ⇌ NH 3 (aq) + H 3 O+^ (aq) As with any weak acid, this makes the solution acidic (pH <7.00). Same for other cations such as NH 3 Me +^ , NH 2 Et 2 +^ , etc.
(ii) [Fe(H 2 O) 6 ]3+^ (aq) + H 2 O (l) ⇌ [Fe(H 2 O) 5 (OH)]2+^ (aq) + H 3 O+^ (aq)
This is an example of a “complex ion” – the Fe atom is attached to six H 2 O molecules to give [Fe(H 2 O) 6 ]3+^. H 2 O molecules attached to Mn+ ions become more acidic than a free H 2 O molecule—therefore, they behave as weak acids.
(iii) Certain salts such as NaHSO 4 (containing Na+^ and HSO 4 -^ ions) also give acidic solutions because the HSO 4 -^ is a weak acid. HSO 4 -^ (aq) + H 2 O(l) ⇌ H 3 O+^ (aq) + SO 4 2-^ (aq) This is the second dissociation of the diprotic acid H 2 SO 4. We do not see the first dissociation because we are not starting with H 2 SO 4 , but NaHSO 4.
C. Basic Solutions (Only one way.)
The anion (i.e., the conjugate base) of a weak acid (F -^ , NO 2 -^ , ClO -^ ,
acetate -^ , PO 4 3-^ , etc.) and the cation of a strong base (Na +, K +, Ca 2+,
etc.). The cation does not react with water, but the anion is a weak
base and thus gives OH -^ , e.g., NaCN, KF, sodium acetate
(CH 3 COONa), Na 2 CO 3 , Ca(HCO 3 ) 2 , Na 2 SO 3 , etc., etc., etc. (lots !!).
Example : NaCN dissolves in H 2 O to give Na +^ (aq) and CN -^ (aq) ions.
Then:
CN -^ (aq) + H 2 O (l) ⇌ HCN (aq) + OH-^ (aq)
This is a typical base-ionization reaction ∴ pH > 7.
Table 18.7 The Acid-Base Behavior of Salts in Water
Sample Problem 18.13 Predicting Relative Acidity of Salt Solutions from Reactions of the Ions with Water
PROBLEM: Predict whether aqueous solutions of the following are
acidic, basic, or neutral.
(a) Potassium perchlorate, KClO 4
(b) Sodium benzoate, C 6 H 5 COONa
(c) Chromium(III) nitrate, Cr(NO 3 ) 3
(a) This solution will be neutral.
(b) This solution will be basic.
(c) This solution will be acidic.
Molecules as Lewis Acids Many neutral molecules function as Lewis acids, if they contain an atom that can accept an electron pair (for example, an atom that is electron deficient with less than 8 valence electrons). e.g. BF 3 + :NH 3 ⇌ BF 3 -NH 3 acid base adduct also AlCl 3 + :Cl-^ ⇌ AlCl 4 - (See the textbook for the structures of these adducts)
Lewis Acids with Polar Multiple Bonds Molecules that contain polar multiple bonds can function as Lewis acids e.g. CO 2 , SO 2 , etc. In such cases, the central atom is electron deficient with a vacant orbital that can accept an electron pair. e.g. when CO 2 dissolves in water, it forms the weak diprotic acid H 2 CO 3 (carbonic acid) from a Lewis acid/Lewis base reaction: O=C=O + H 2 O: ⇌ H 2 CO 3 acid base adduct
Metal Cations as Lewis Acids We saw earlier that a “complex ion” such as Fe(H 2 O) 6 3+^ is a weak acid. Such a complex ion is the adduct from a Lewis acid/base reaction. Fe 3+^ + 6 H 2 O: ⇌ Fe(H 2 O) 6 3+ acid base adduct Cr 3+^ + 6 :NH 3 ⇌ Cr(NH 3 ) 6 3+ Fe2+^ + 6 :CN-^ ⇌ Fe(CN) 6 4-
Electron-Deficient Molecules as Lewis Acids
B and Al often form electron-deficient molecules, and
these atoms have an unoccupied p orbital that can accept
a pair of electrons:
BF 3 accepts an electron pair from ammonia to form a covalent bond.
