1. Acid-base theory pH calculations
Joško Ivica

2. REVIEW QUESTIONS
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

3. REVIEW HCl H+ + Cl- or HCl + H2O H3O+ + Cl- KOH K+ + OH-
CH3COOH + H2O CH3COO- + H3O+
CH3COOH CH3COO- + H+
NH3 + H2O NH4+ + OH-
3) [CH3COO-] [H+]
CH3COONa CH3COO- + Na+
pH = -log[H+]
KW = [H+][OH-] = 1,008·10-14 at 25°C
pOH = -log[OH-] pH + pOH = 14 = pKW! KA KA KB
KA =
[CH3COOH]

4. 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+ + NO3-
Bases are compounds which dissociate in water producing a hydroxide ion and a cation
NaOH  Na+ + OH-
Brønsted - 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

5. pH of strong acids and bases
HA H+ + A- complete dissociation of an acid
pH = -log a(H+)
a – activity
a(H+) = γ±·c(HA)
γ± - mean activity coefficient
In very diluted solutions: γ± = 1!
c(HA) = [H+] = [A-]
pH = -log[H+]

6. pH of strong acids and bases
BOH B+ + OH-
pH = 14 - pOH = 14 + [log a(OH-)]
complete dissociation of a base
pOH = -log[OH-]
c(BOH) = [OH-] = [B+]
a(OH-) = γ±·c(BOH)
In very diluted solutions: γ± = 1!
pH = 14 - pOH = 14 + log [OH-]

7. pH of weak acids and bases
Dissociation of weak acids (Ka < 10-4)
HA + H2O A- + H3O+ Ka = = =
c-x x x
[A-][H3O+] x2 x2
[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
c >> x for diluted weak acids
[H3O+] = x = (Ka c)1/2 / log
pH = -log[H3O+]
pH = -log [H3O+] = ½ [pKa – log(c)]

8. pH of weak acids and bases
Dissociation of weak bases
c-x x x
[BH+][OH-] x2 x2
B + H2O BH+ + OH-
Kb = = =
[B]
c-x c
c-x = concentration of a base at equilibrium
x = concentration of products at equilibrium
c >> x for diluted weak bases
c = concentration of a base at the beginning
pKb = -logKb
[OH-] = x = (Kb c)1/2 / log
pOH = -log[OH-]
pH = 14 - pOH
pH = 14 – pOH = 14 – ½ [pKb – log(c)]

9. 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

10. Salts of weak acids and strong bases
[CH3COO-] [H+]
CH3COONa  CH3COO- + Na+
KA =
[CH3COOH]
[CH3COOH] [
OH- ]
CH3COO- + H2O CH3COOH + OH-
KH =
[CH3COO-]
[H+][OH-] = Kv
KH·KA = KW  KH = KW/KA
c-x x x
CH3COO- + H2O 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
[OH-]2
KW =
10-14 =
c-x = c c KA

11. [OH-]2 = KW · c (salt)
pOH = 7 – 1/2[pKA – log(c)] KA
pH = 14 - pOH
pH = 7 + ½
[pKA + log(c)]

12. Salts of weak bases and strong acids
NH4Cl  NH4+ + Cl KB =
NH4+ + H2O NH3 + H3O+ KH =
[H+][OH-] = Kv
[NH4+]
[OH-]
[NH3]
[NH3]
[H3O+]
[NH4+]
KH·KB = KW  KH = KW/KB
NH4+ + H2O NH3 + H3O [H3O+] = [NH3]
c-x x x
c = concentration of salt at the beginning KW
[H3O+]2 =
c-x = c c KB
c-x = concentration of a cation of a weak base at equilibrium
x = concentration of products at equilibrium

13. [H3O+]2 = KW· c(salt) KB
pH = 7 - ½[pKB + log(c)]

14. 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 + OH-
NH4+ + H2O = NH3 + H3O+
NH4+ + CN HCN + NH3
c-x c-x x x c-x = c
KH = [HCN][NH3]/[CN-][NH4+]
= [HCN]2/[CN-]2
KH · KA ·KB = KW  KH = KW/KA KB
KA = [H3O+][CN-]/[HCN]  (1/KH)1/2
[H3O+]2 = KA2 KH = KW · KA/KB
[H3O+]2 = KW · KA KB
pH = 7 + ½[pKA - pKB]

15. 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)

16. 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 + H2O BH+ + OH-
pOH = pKb + log[BH+]/[B] B – weak base
BH+ - conjugated acid

17. pH calculations 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 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!

18. 1.
c(KOH) = 0,001 M = [K+] = [OH-]
KOH  K+ + OH-
pOH = -log [OH-] = 3
pH = 14 – pOH = 11

19. 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

20. 3.
c(NH3) = M, pKb = 4.8
H2O
NH NH4+ + OH-
0.001-x x x x = conc. of products at equilibrium ↓
conc. of NH3 at equilibrium 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

21. 4.
c(NaCN) = 0.1 M, pKa = 9.21
NaCN  Na+ + CN HCN H+ + CN-
Ka=[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 ( log 0.1) = 11.1

22. 5.
c(NH4Cl) = 0.7 M, pKb = 4.8
NH4Cl  NH4+ + Cl NH NH4+ + OH-
Kb = [NH4+][OH-]/[NH3]
NH4+ + H2O NH3 + H3O+ KH = [NH3][H3O+]/[NH4+]
c-x = c 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
H2O

23. 6.
c(CH3CH(OH)COONH4) = M, pKa = 3.86, pKb = 4.8
CH3CH(OH)COO- + H2O CH3CH(OH)COOH + OH-
NH4+ + H2O NH3 + H3O+
CH3CH(OH)COO- + NH 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
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
(1/KH)1/2

24. 7.
0.1 M CH3COONa, 0.1 M CH3COOH, pKa = 4.8
CH3COOH + H2O CH3COO- + H3O+ Ka
pH = pKa + log [CH3COO-]/[CH3COOH] = = 4.8

25. 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