Chemistry Acid and Base

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Aqueous Acid/Base Chemistry
Resources:Harris ‘Quantitative Chemical Analysis’

Review:
Pure water has a pH = 7
Autodissociation: H2O (( H3O+ + OH-

K = [H3O+][OH-]/[H2O]
-log[H3O+] = 7
[H3O+] = 10-7 M = [OH-]
[H2O] = 55.56 M
K = 1.8 x 10-16 ; pKa = 15.74

pKa is the acid dissociation constant; low pKa (strong acid, high pKa (weak acid

we can also write Kw = [H3O+][OH-]
Kw = 10-14
In water, pH + pOH = 14

pH scale

Strong Acids:
Complete dissociation of the acid in water:
HA ( H+ + A-note: H+ in H2O is hydrated (H3O+)
K (Ka since it is an acid) is large (For complete dissociation, K = infinity; however there is some very tiny amount that is not dissociated. For practical intents and purposes, dissociation is complete.), and pKa is a negative value. Ex: K = 108, pKa = -8

Example: hydrochloric acid (HCl)
What is the pH of a 0.13 M HCl solution in water?
HCl ( H+ + Cl-
0.13 M ( 0.13 M + 0.13 M
[H+] = 0.13 M
pH = -log[H+] = -log(0.13) = 0.89 (very acidic)

Strong base:
Complete dissociation of strong base results in aqueous hydroxide ion (OH-). CatOH ( Cat+ + OH-
K (Kb since it is a base) is large, but since the equilibrium is for a base, we use the value pKb to describe the degree of dissociation. Ka x Kb = Kw

Ex: sodium hydroxide, NaOH
What is the pH of a 0.003 M solution of NaOH?
NaOH ( Na+ + OH-
[OH-] = 0.003 M
pOH = 2.52
pH = 14 – pOH = 11.48 (basic)

Weak acid:
If the acid does not dissociate completely in H2O, then it is a weak acid. (note: H2O does not dissociate completely, so it is a weak acid, and a weak base).

HA (( H+ + A-
Ex: acetic acid, CH3COOH
CH3COOH (( CH3COO- + H+
It is possible to calculate pH of a weak acid solution, knowing its concentration and its pKa (acid dissociation constant). pKa acetic acid = 4.76; Ka = 1.75 x 10-5 = [CH3COO-][H+]/[CH3COOH]

Calculate the pH of 0.15 M acetic acid in H2O.
CH3COOH (( CH3COO- + H+
Concentration0.15 – x x x

1.75 x 10-5 = x2/(0.15 – x)The quadratic equation can be solved… Or, more typically, we use that [HA] >> [A-],[H+], such that 0.15 – x ( 0.15 1.75 x 10-5 ( x2/0.15
X = 1.62 x 10-3 M = [H+]
pH = 2.79 (acidic)

Weak base:
Weak bases deprotonate water, but only a small fraction of the stoichiometric equivalent.
WB + H2O (( WBH+ + OH-
Ex: ammonia, NH3Kb = 1.8 x 10-5, pKb = 4.75

What is the pH of 1.3 M NH3 ?

NH3 + H2O (( NH4+ + OH-
Concentration1.3 – x xx

1.8 x 10-5 = x2/1.3 – x again, [NH3] >> x
1.8 x 10-5 = x2/1.3
X = [OH-] = 4.84 x 10-3
pOH = 2.32
pH = 14 – pOH = 11.68 (basic)

Conjugate acid/base:

When a weak acid in water dissociates:
HA (( H+ + A-
The anion is the conjugate base.

When a weak base induces water dissociation:
WB + H2O (( WBH+ + OH-
The cation is the conjugate acid.

Conjugate bases can be isolated as salts with cations; they are weak bases. Ex: CH3COO- + Na+ ( CH3COONa (sodium acetate, colorless solid)

Conjugate acids can be isolated as salts with anions; they are weak acids. Ex: NH4+ + Cl- ( NH4Cl (ammonium chloride, colorless solid) More weak acid/base pH calculations:

Calculate the pH of 0.66 M NH4Cl
What is the chemistry?
NH4Cl ( NH4+ + Cl-complete dissociation in water
NH4+ (( NH3 + H+weak acid, pKa = 9.24, Ka NH4+ = 5.75 x 10-10

0.66 – xx x
5.75 x 10-10 = x2/0.66 – x ( x2/0.66
x = 1.95 x 10-5 = [H+]
pH = 4.71 (acidic)

Calculate the pH of 0.15 M sodium acetate in water.
What is the chemistry?
CH3COONa ( CH3COO- + Na+ complete dissociation in water
CH3COO- + H2O (( CH3COOH + OH-weak base, pKb = 9.24, Kb = 5.57 x 10-10 0.15 – x xx
5.75 x 10-10 = x2/0.15 – x ( x2/0.15
X = 9.29 x 10-6 = [OH-]
pOH = 5.03, pH = 8.97 (basic)

Multiple dissociation:

Ex: phosphoric acid, H3PO4

[pic]

K1 = 7.04 x 10-3pKa = 2.15
K2 = 6.29 x 10-8pKa = 7.20
K3 = 7.14 x 10-13pKa =...
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