Acid-base theory
pH calculations
Joško Ivica
REVIEW QUESTIONS
1)
2)
3)
4)
5)
6)
Write the formulas for hydrochloric acid, potassium
hydroxide and their dissociation reactions in water
Write the formulas for acetic acid, ammonia and their
dissociation reactions in water
Write the equation for equilibrium dissociation
constant of acetic acid
Write the formula for sodium acetate and its
dissociation reaction in water
What is pH?
Ionic product of water
REVIEW
1)
H2 O
HCl
H+ + Cl- or HCl + H2O
H3O+ + ClKOH
K+ + OHKA
2) CH3COOH + H2O
CH3COO- + H3O+
KA
CH3COOH
CH3COO- + H+
KB
NH3 + H2O
NH4+ + OH3)
[CH3COO-] [H+]
KA =
[CH3COOH]
4)
5)
6)
CH3COONa
CH3COO- + Na+
pH = -log[H+]
KW = [H+][OH-] = 1,008·10-14 at 25°C
pOH = -log[OH-]
pH + pOH = 14 = pKW!
ACIDS AND BASES
Arrhenius theory: Acids are compounds able to dissociate in
water producing a hydrogen ion (H+) and a corresponding
anion (only in aqueous solutions)
HNO3  H+ + NO3Bases are compounds which dissociate in water producing a
hydroxide ion and a cation
NaOH  Na+ + OHBrønsted - Lowry theory: Acids are compounds that release H+,
whereas bases are compounds that are able to bind H+
(applicable also in non-aqueous solutions)
acid  H+ + base
conjugated pair
pH of strong acids and bases
HA
H+ + A- complete dissociation of an acid
pH = -log a(H+)
c(HA) = [H+] = [A-]
a – activity
a(H+) = γ±·c(HA)
γ± - mean activity coefficient
In very diluted solutions: γ± = 1!
pH = -log[H+]
pH of strong acids and bases
BOH
B+ + OH-
pOH = -log[OH-]
complete dissociation of a base
c(BOH) = [OH-] = [B+]
pH = 14 - pOH = 14 + [log a(OH-)]
a(OH-) = γ±·c(BOH)
In very diluted solutions: γ± = 1!
pH = 14 - pOH = 14 + log [OH-]
pH of weak acids and bases
Dissociation of weak acids (Ka < 10-4)
c-x
HA + H2O
x
x
A- + H3O+
[A-][H3O+]
x2
x2
Ka =
=
=
[HA]
c-x
c
c-x = concentration of an acid at equilibrium
x = concentration of products at equilibrium
c = concentration of an acid at the beginning
pKa = -logKa
pH = -log[H3O+]
c >> x
for diluted weak acids
[H3O+] = x = (Ka c)1/2 / log
pH = -log [H3O+] = ½ [pKa – log(c)]
pH of weak acids and bases
Dissociation of weak bases
c-x
x
BH+
B + H2 O
+
x
OH-
[BH+][OH-]
x2
Kb =
=
[B]
c-x
c-x = concentration of a base at equilibrium
x = concentration of products at equilibrium
c = concentration of a base at the beginning
pKb = -logKb
x2
=
c
c >> x for diluted weak bases
[OH-] = x = (Kb c)1/2 / log
pOH = -log[OH-]
pH = 14 - pOH
pH = 14 – pOH = 14 – ½ [pKb – log(c)]
Salt hydrolysis
• When salts composed of ions of a strong electrolyte (acid
or base) and ions of a weak electrolyte are dissolved,
complete salt dissociation occurs because ions of a strong
electrolyte can exist only in ionized form
• Ions originating from a weak electrolyte react with water
producing their conjugated particle
• Examples: CH3COONa, KCN, NH4Cl, NH4NO3
Salts of weak acids and strong bases
[CH3COO-]
KA = +
[H[CH
] COOH]
3
CH3COONa  CH3COO- + Na+
CH3
COO-
+ H 2O
CH3COOH +
OH-
KH =
[CH3COOH] [OH-]
[H+][OH-] = Kv
[CH3COO-]
KH·KA = KW  KH = KW/KA
c-x
x
CH3COO- + H2O
x
CH3COOH + OH-
[CH3COOH] = [OH-]
c = concentration of salt at the beginning
c-x = concentration of anion of a weak acid at equilibrium
x = concentration of products at equilibrium
c-x = c
KW =10-14
KA
=
[OH-]2
c
[OH-]2 = KW · c (salt)
KA
pOH = 7 – 1/2[pKA – log(c)]
pH = 14 - pOH
pH = 7 + ½ [pKA + log(c)]
Salts of weak bases and strong acids
NH4Cl  NH4 +
+
NH4 + H2O
+
Cl-
[NH4+] [OH-]
KB =
[NH3]
NH3 + H3
O+
[NH3] [H3O+]
KH =
[NH4+]
[H+][OH-] = Kv
KH·KB = KW  KH = KW/KB
NH4+ + H2O
c-x
NH3 + H3O+
x
x
c = concentration of salt at the beginning
c-x = c
[H3O+] = [NH3]
KW
KB
[H3O+]2
=
c
c-x = concentration of a cation of a weak base at equilibrium
x = concentration of products at equilibrium
[H3O+]2 = KW· c(salt)
KB
pH = 7 - ½[pKB + log(c)]
Salts of weak acids and weak bases
Anions and cations of weak acids and bases, that produce salt –
having concentration c, react with water e.g. NH4CN
CN- + H2O = HCN + OHNH4+ + H2O = NH3 + H3O+
NH4+ + CNHCN + NH3
c-x
c-x
x
x
KH = [HCN][NH3]/[CN-][NH4+]
= [HCN]2/[CN-]2
c-x = c
KH · KA ·KB = KW  KH = KW/KA KB
[H3O+]2 = KW · KA
KB
KA = [H3O+][CN-]/[HCN]  (1/KH)1/2
[H3O+]2 = KA2 KH = KW · KA/KB
pH = 7 + ½[pKA - pKB]
BUFFERS
• BUFFERS = conjugated pair of acid or
base, which is able to maintain pH in
particular (narrow) interval after adding strong
acid or base into solution (system)
• Buffers are typically mixtures of weak acids
and their salts with strong bases or
mixtures of weak bases and their salts
with strong acids
• Buffer systems in organism are of a great
importance (blood, intercellular space, cells)
pH calculations of buffer
solutions
Buffer consisting of a weak acid and its salt with a strong base
HA + H2O
A- + H3O+ Ka
Henderson – Hasselbalch equation
pH = pKa + log[A-]/[HA]
HA – weak acid
A- – conjugated base
Buffer consisting of a weak base and its salt with a strong acid
B + H 2O
BH+ + OH-
pOH = pKb + log[BH+]/[B] B – weak base
BH+ - conjugated acid
pH calculations
1.
2.
3.
4.
5.
6.
7.
8.
Calculate the pH of 1 mM KOH
Calculate the pH of 0.01 M formic acid (HCOOH) at
25°C, pKa = 3.8!
Calculate the pH of 0.001 M NH3 at 25°C, pKb = 4.8!
Calculate the pH of 0.1 M NaCN at 25°C, pKa = 9.21!
Calculate the pH of 0.7 M NH4Cl at 25°C, pKb = 4.8!
Calculate the pH of 5 mM ammonium lactate
CH3CH(OH)COONH4 at 25°C, pKa = 3.86, pKb = 4.8
Calculate the pH of a buffer solution that contains 0.1
M CH3COONa and 0.1 M CH3COOH, pKa = 4.8!
Calculate the pH of a buffer solution that contains 0.1
M NH4Cl and 1 M NH3, pKb = 4.8!
1.
c(KOH) = 0,001 M = [K+] = [OH-]
KOH  K+ + OHpOH = -log [OH-] = 3
pH = 14 – pOH = 11
2.
c(HCOOH) = 0.01 M, pKa = 3.8
HCOOH ↔ HCOO- + H+
0.01-x=c x
x
x = conc. of products at equilibrium
↓
conc. of HCOOH at equilibrium
Ka =[HCOO-][H+]/[HCOOH] = x2/c = [H+]2/0.01
[H+] = (Ka·0.01)1/2
pH = -log[H+] = ½ [3.8 – log(0.01)] = 2.9
3.
c(NH3) = 0.001 M, pKb = 4.8
NH3
H 2O
NH4+ + OH-
0.001-x x
x
x = conc. of products at equilibrium
↓
conc. of NH3 at equilibrium 0.001-x = c
Kb=[NH4+][OH-]/[NH3] = x2/c = [OH-]2/0.001
[OH-] = (Kb·0.001)1/2
pOH = -log[OH-] = ½ [pKb - log(0.001)]
pH = 14 - ½ [4.8 - log(0.001)] = 14 – 3.9 = 10.1
4.
c(NaCN) = 0.1 M, pKa = 9.21
NaCN  Na+ + CNHCN
H+ + CNKa=[H+][CN-]/[HCN]
CN- + H2O
HCN + OH- KH = [OH-][HCN]/[CN-]
c-x = c
x
x [HCN] = [OH-]
Kv = Ka KH  Kv/ Ka = [OH-]2/c  [OH-] = (Kvc/ Ka)1/2
pOH = ½(pKv – pKa + log c)
 pH = 14 - ½(pKW – pKa + log c) = pH = 7 + ½ [pKA + log(c)]
= 7 + ½ (9.21 + log 0.1) = 11.1
5.
c(NH4Cl) = 0.7 M, pKb = 4.8
H 2O
NH4Cl  NH4+ + ClNH3
NH4+ + OHKb = [NH4+][OH-]/[NH3]
NH4+ + H2O
c-x = c
NH3 + H3O+ KH = [NH3][H3O+]/[NH4+]
x
x [NH3] = [H3O+]
Kv = Ka KH  Kv/ Ka = [H3O+]2/c  [H3O+] = (Kvc/Kb)1/2
pH = 7 - ½[pKB + log(c)] = 7 – ½ (4.8 – 0.15) = 4.68
6.
c(CH3CH(OH)COONH4) = 0.005 M, pKa = 3.86, pKb = 4.8
CH3CH(OH)COO- + H2O
NH4+ + H2O
CH3CH(OH)COOH + OH-
NH3 + H3O+
CH3CH(OH)COO- + NH4+
CH3CH(OH)COOH + NH3
c-x
c-x
x
x
KH = [CH3CH(OH)COOH][NH3]/[CH3CH(OH)COO-][NH4+]
= [CH3CH(OH)COOH]2/[CH3CH(OH)COO-]2
KW = KH KA KB  KH = KW/KA KB
(1/KH)1/2
KA = [H3O+][CH3CH(OH)COO-]/[CH3CH(OH)COOH]
[H3O+]2 = KA2 KH = KW · KA/KB
pH = 7 + ½[pKA - pKB]= 7 + ½ [3.86 – 4.8] = 6.53
7.
0.1 M CH3COONa, 0.1 M CH3COOH, pKa = 4.8
CH3COOH + H2O
CH3COO- + H3O+ Ka
pH = pKa + log [CH3COO-]/[CH3COOH] = 4.8 + 0 = 4.8
8.
0.1 M NH4Cl a 1 M NH3, pKb = 4.8
NH3 + H2O
NH4+ + OH-
Kb
pOH = pKb + log [NH4Cl]/[NH3] = 4.8 – 1 = 3.8
pH = 14 – pOH = 10.2