Acid-base theory pH calculations

Joško Ivica

REVIEW QUESTIONS1) Write the formulas for hydrochloric acid, potassium

hydroxide and their dissociation reactions in water2) Write the formulas for acetic acid, ammonia and their

dissociation reactions in water 3) Write the equation for equilibrium dissociation

constant of acetic acid 4) Write the formula for sodium acetate and its

dissociation reaction in water5) What is pH? 6) Ionic product of water

REVIEW

1) HCl H+ + Cl- or HCl + H2O H3O+ + Cl-

KOH K+ + OH- 2) CH3COOH + H2O CH3COO- + H3O+

CH3COOH CH3COO- + H+

NH3 + H2O NH4+ + OH-

3) [CH3COO-] [H+]

4) CH3COONa CH3COO- + Na+

5) pH = -log[H+] 6) KW = [H+][OH-] = 1,008·10-14 at 25°C

pOH = -log[OH-] pH + pOH = 14 = pKW!

H2O

KA

KA

KB

[CH3COOH]KA =

ACIDS AND BASESArrhenius 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

pH of strong acids and bases

HA H+ + A- complete dissociation of an acid

pH = -log a(H+) a – activitya(H+) = γ±·c(HA)

γ± - mean activity coefficient

In very diluted solutions: γ± = 1!

c(HA) = [H+] = [A-]

pH = -log[H+]

pH of strong acids and bases

BOH B+ + OH-

pH = 14 - pOH = 14 + [log a(OH-)]

complete dissociation of a base

c(BOH) = [OH-] = [B+]

a(OH-) = γ±·c(BOH)

In very diluted solutions: γ± = 1!

pH = 14 - pOH = 14 + log [OH-]

pOH = -log[OH-]

pH of weak acids and bases

Dissociation of weak acids (Ka < 10-4)

HA + H2O A- + H3O+ Ka = = =[A-][H3O+]

[HA]

c-x x x x2

c-x = concentration of an acid at equilibrium

x = concentration of products at equilibrium

c-x

c = concentration of an acid at the beginning

c >> x for diluted weak acids

x2

c

pKa = -logKa

pH = -log[H3O+] [H3O+] = x = (Ka c)1/2 / log

pH = -log [H3O+] = ½ [pKa – log(c)]

pH of weak acids and bases

Dissociation of weak bases

B + H2O BH+ + OH-c-x xx

c-x = concentration of a base at equilibrium

x = concentration of products at equilibrium

c = concentration of a base at the beginning

Kb = [BH+][OH-]

[B]

x2

c-x= =

x2

c

c >> x for diluted weak bases

pKb = -logKb

pOH = -log[OH-]

pH = 14 - pOH

[OH-] = x = (Kb c)1/2 / log

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

CH3COONa CH3COO- + Na+KA =

[CH3COO-] [H+]

[CH3COOH]

CH3COO- + H2O CH3COOH + OH- KH =[CH3COOH] OH-[ ]

[CH3COO-][H+][OH-] = Kv

KH·KA = KW KH = KW/KA

CH3COO- + H2O CH3COOH + OH-

c-x

c = concentration of salt at the beginning

c-x = concentration of anion of a weak acid at equilibrium

x x

x = concentration of products at equilibrium

[CH3COOH] = [OH-]

10-14

KA

= KW = [OH-]2

cc-x = 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

+ + Cl- KB =

NH4+ + H2O NH3 + H3O+ KH =

[H+][OH-] = KvKH·KB = KW KH = KW/KB

[NH4+] [OH-]

[NH3]

[NH3] [H3O+]

[NH4+]

NH4+ + H2O NH3 + H3O+ [H3O+] = [NH3]

c-x xx

KW

KB

=[H3O+]2

c

c = concentration of salt at the beginning

c-x = concentration of a cation of a weak base at equilibrium

x = concentration of products at equilibrium

c-x = c

[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 + OH-

NH4+ + H2O = NH3 + H3O+

NH4+ + CN- HCN + NH3

c-x c-x x x c-x = cKH = [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]

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 acidA- – 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

pH calculations1. Calculate the pH of 1 mM KOH2. Calculate the pH of 0.01 M formic acid (HCOOH) at

25°C, pKa = 3.8!

3. Calculate the pH of 0.001 M NH3 at 25°C, pKb = 4.8!

4. Calculate the pH of 0.1 M NaCN at 25°C, pKa = 9.21!

5. Calculate the pH of 0.7 M NH4Cl at 25°C, pKb = 4.8!

6. Calculate the pH of 5 mM ammonium lactate CH3CH(OH)COONH4 at 25°C, pKa = 3.86, pKb = 4.8

7. Calculate the pH of a buffer solution that contains 0.1 M CH3COONa and 0.1 M CH3COOH, pKa = 4.8!

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+ + OH-

pOH = -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

H2ONH3 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+ + 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 (9.21 + log 0.1) = 11.1

5.c(NH4Cl) = 0.7 M, pKb = 4.8

NH4Cl NH4+ + Cl- NH3 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

6.c(CH3CH(OH)COONH4) = 0.005 M, pKa = 3.86, pKb = 4.8

CH3CH(OH)COO- + H2O CH3CH(OH)COOH + OH-

NH4+ + H2O NH3 + H3O+

CH3CH(OH)COO- + NH4+ CH3CH(OH)COOH + NH3

c-x c-x x xKH = [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

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