1. Chemical calculations II
Vladimíra Kvasnicová

2. 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

3. 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-]

4. Kw = [H3O+] x [OH-] = 10-14 pKW = pH + pOH = 14
pK = - log K pH = - log [H3O+] pOH = - log [OH-]
= [H3O+] x [OH-] / log
log = log ([H3O+] x [OH-] )
log = log [H3O+] + log [OH-]
= log [H3O+] + log [OH-] / x (-1)
14 = - log [H3O+] - log [OH-]
↓ ↓ ↓
pKW = pH pOH
14 = in pure water

5. 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
acidic neutral basic
If [H+] decreases, [OH-] increases KW is 10-14
If [OH-] decreases, [H+] increases (= constant !)

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

7. weak acids (HA) [HA] ≠ [H+] Kdis ≤ 10–2 HA ↔ H+ + A-
Kdis = [H+] [A-] [H+] = [A-] [HA] = cHA Kdis = Ka [HA]
Ka = [H+] 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

8. weak acids (HA) [HA] ≠ [H+] Kdis ≤ 10–2
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 = pH = 14 - pOH

9. 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

10. Exercises 1) 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]

11. 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 ]

12. 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)

13. „bicarbonate buffer“ HCO3- NaHCO3 ↔ Na+ + HCO3-
H2CO H2CO3 ↔ H+ + HCO3-
NaHCO3
mixed → Na+ + HCO3-
H2CO3 H HCO3-
+ H2CO3
+ HCl NaOH (H+ + Cl-) (Na+ + OH- )
Na+ + HCO Na+ + HCO H+ + H2CO H2O + HCO Cl- + H2CO Na+ + H2CO3
HCO3- + H+↔ H2CO H+ + OH- ↔ H2O

14. 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)

15. Exercises
10) 200ml of 0,5M acetic acid ml 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]