Chemical calculations IIVladimíra Kvasnicová

Calculation of pH

pH = - log a(H3O+)

a = γ x c

a = activity

γ = activity coefficient

c = concentration (mol /L)

in diluted (mM) solutions: γ = 1 a = c

pH = - log c(H3O+)

c(H3O+) = [H3O+] = molar concentration

Dissociation of water:

H2O ↔ H+ + OH-

H2O + H+ + OH- ↔ H3O+ + OH-

H2O + H2O ↔ H3O+ + OH-

Kdis = [H3O+] x [OH-]

[H2O]2

Kdis x [H2O]2 = [H3O+] x [OH-]

Kdis x [H2O]2 = constant, because [H2O] is manifold

higher than [H3O+] or [OH-]

Kw = constant = ionic product of water

Kw = [H3O+] x [OH-]

Kw = [H3O+] x [OH-] = 10-14

pKW = pH + pOH = 14

pK = - log K pH = - log [H3O+] pOH = - log [OH-]

10-14 = [H3O+] x [OH-] / log

log 10-14 = log ([H3O+] x [OH-] )

log 10-14 = log [H3O+] + log [OH-]

-14 = log [H3O+] + log [OH-] / x (-1)

14 = - log [H3O+] - log [OH-]

↓ ↓ ↓ pKW = pH + pOH

14 = 7 + 7 in pure water

pKW = pH + pOH = 14

=> water: [H3O+] = 10–7 (pH = 7)

[OH-] = 10–7 (pOH = 7)

simplification: [H3O+] = [H+] = c(H+)

=> pH = – log c(H+) pH = 0 – 14

pH 0 -------------- 7 --------------14 acidic neutral basic If [H+] decreases, [OH-] increases KW is 10-14

If [OH-] decreases, [H+] increases (= constant !)

strong acids (HA) [HA] = [H+]HA → H+ + A-

pH = - log c(H+) = - log cHA

strong bases (BOH) [BOH] = [OH-]BOH → B+ + OH-

pOH = - log cBOH

pH = 14 - pOH

weak acids (HA) [HA] ≠ [H+] Kdis ≤ 10–2

HA ↔ H+ + A-

Kdis = [H+] [A-] [H+] = [A-] [HA] = cHA Kdis = Ka

[HA] Ka = [H+]2

cHA

Ka x cHA = [H+]2 / log

log (Ka x cHA ) = 2 x log [H+]

log Ka + log cHA = 2 x log [H+] / ½

½ log Ka + ½ log cHA = log [H+] / x (-1)

-½ log Ka - ½ log cHA = - log [H+] - log Ka = pKa

½ pKa - ½ log cHA = pH

=> pH = ½ pKa - ½ log cHA

weak acids (HA) [HA] ≠ [H+] Kdis ≤ 10–2

HA ↔ H+ + A-

pH = ½ pKa - ½ log cHA

weak bases (BOH) [BOH] ≠ [OH-] Kdis = [B+] [OH-]

BOH ↔ B+ + OH- [BOH]

pOH = ½ pKb - ½ log cBOH

=> pH of basic solutions: pH + pOH = 14 pH = 14 - pOH

Important equations

pH = - log c(H+) pK = - log K

pH + pOH = 14

ACIDS: pH = - log cHA

pH = ½ pKa - ½ log cHA

BASES: pOH = - log cBOH

pOH = ½ pKb - ½ log cBOH

pH = 14 – pOH

Exercises1) 0,1M HCl, pH = ?, [H+] = ?

[10-1 M, pH =1]2) 0,01M KOH, pH = ?, [H+] = ?

[10-12 M, pH = 12]

3) 0,01M acetic acid, K = 1,8 x 10–5 , pH = ?[pK = 4,74; pH = 3,4]

4) 0,2M NH4OH; pK = 4,74; pH = ?

[pOH = 2,72; pH = 11,3]

5) 0,1M lactic acid; pH = 2,4; Ka = ?

[pK=3,8; Ka = 1,58 x 10-4]

6) strong acid: pH = 3 c = ? [10–3 M ]

7) strong base: pH = 11 c = ?

[pOH = 3; c = 10–3 M ]

8) dilution of a weak acid: c1 = 0,1 c2 = 0,01 ? ∆ pH

[∆ pH = 0,5 ]

9) dilution of a strong acid:c1 = 0,1 c2 = 0,01 ? ∆ pH

[∆ pH = 1 ]

BUFFERS

= solutions which have the ability to absorb small additions of either a strong acid or strong base with a very little change of pH.

• buffers are used to maintain stable pH• composition of buffers:

„conjugated pair: acid /base“* weak acid + it`s salt

* weak base + it`s salt* 2 salts of a polyprotic acid* amphoteric compound (e.g.

protein)

„bicarbonate buffer“ HCO3- NaHCO3 ↔ Na+ + HCO3

-

H2CO3 H2CO3 ↔ H+ + HCO3-

NaHCO3

mixed → Na+ + HCO3-

H2CO3 H+ + HCO3-

+ H2CO3

+ HCl + NaOH (H+ + Cl-) (Na+ + OH- )

Na+ + HCO3- Na+ + HCO3

-

H+ + H2CO3 H2O + HCO3-

Cl- + H2CO3 Na+ + H2CO3

HCO3- + H+↔ H2CO3 H+ + OH- ↔ H2O

Henderson-Hasselbalch equation

pH = pKa + log (cs / ca) (for acidic buffer )

pOH = pKb + log (cs / cb ) (for basic buffer)

pH = 14 - pOH

pK = dissociation constant of the weak acid (pKa) or base (pKb)

cs = actual concentration of salt

ca = actual concentration of weak acid

cb = actual concentration of weak base

c = c´ x V c´ = concentration before mixing the components V = volume of a component (acid or base or salt)

Exercises

10) 200ml of 0,5M acetic acid + 100ml of 0,5M sodium acetate => buffer; pKa = 4,76 pH = ?

[pH = 4,46 ]

11) 20ml of 0,05M NH4Cl + ? ml 0,2M NH4OH

=> buffer of pH = 10; Kb = 1,85 x 10–5 pK = ?

[pK = 4,73; 27 ml]