Transcript Seminar 5 - Univerzita Karlova v Praze

Chemical calculations used in medicine
part 2
Pavla Balínová
Acid – base theories
• Arrhenius theory
acid = substance that is able to lose H+ (CH3COOH ↔ CH3COO- + H+)
base = substance that is able to lose OH• Brönsted – Lowry theory
acid = chemical species (molecule or ion) that is able to lose H+
base = chemical species (molecule or ion) with the ability to accept H+
HCl + H2O ↔ H3O+ + Clacid 1
base 2
acid 2
base 1
conjugated pair 1
Water H2O
The purest water is not all H2O → about 1 molecule in 500
million transfers a proton H+ to another H2O molecule, giving
a hydronium ion H3O+ and a hydroxide ion OH-:
H2O + H2O ↔ H3O+ + OH- = dissociation of water
The concentration of H3O+ in pure water is 0.000 0001 M
or 1.10-7 M.
The concentration of OH- is also 1.10-7 M.
Pure water is a neutral solution, without an excess of either
H3O+ and OH- ions.
Equilibrium constant of water: Keq = [H3O+] . [OH-]
[H2O]2
Ion product of water Kw: Kw = [H3O+] . [OH-]
Kw = 1.10-7. 1.10-7 = 1.10-14
Ion product of water
Kw = [H3O+] . [OH-]
10–14 = [H3O+] . [OH-] / log
log(a . b) = log a + log b
log 10–14 = log ([H3O+] . [OH-])
log 10–14 = log [H3O+] + log [OH-]
-14 = log [H3O+] + log [OH-] / · ( -1 )
14 = - log [H3O+] - log [OH-]
- log KW
pH
pOH
pKW = pH + pOH = 14
- log KW = pKW
Addition of an acid to pure water → increasing H3O+
and OH- will fall until the product equals 1.10-14.
Addition of a base to pure water → increasing OH- and
H3O+ will fall until the product equals 1.10-14.
Example:
Lemon juice has a [H3O+] of 0.01 M. What is the [OH-] ?
[H3O+ ] . [OH-] = 1.10-14
1.10-2 . [OH-] = 1.10-14 / : 1.10-2
[OH-] = 1.10-12
pH scale
Exponential numbers express the often minute actual
concentration of H3O+ and OH- ions.
In 1909, S. P. L. Sørensen proposed that only the number
in the exponent be used to express acidity. Sørensen´s
scale came to be known as the pH scale („power of
hydrogen“).
pH = log 1/[H3O+] = - log [H3O+]
e. g. The pH of a solution whose [H3O+ ] is 1.10-4 equals 4.
Representative pH values
Substance
pH
lead-acid battery
0.5
gastric juice
1.5 – 2.0
vinegar
2.9
coffee
5.0
urine
6.0
pure water
7.0
blood
7.35 – 7.45
hand soap
9.0 – 10.0
pH of strong acids
Strong acids are those that react completely with water
to form H3O+ and anion (HCl, H2SO4, HClO4, HNO3,…)
Generally: HA → H+ + Ae. g. HCl + H2O ↔ H3O+ + ClpH = - log cH+ = - log cHA
Calculations:
1) 0.1 M HCl, pH = ?
2) Strong acids pH: a) 1.6 c = mol/L?
b) 3.0 c = mol/L ?
3) 0.08 M H2SO4, pH = ?
4) Dilution of a strong acid: c1 = 0,1 M → c2 = 0,01 M, ∆ pH = ?
pH of weak acids
Weak acids react only to a slight extent with water to
form relatively few H3O+ ions. Most of the molecules of
the weak acids remain in the molecular form (uncharged)
HA → H+ + AKdis = [H+].[A-]
[H+] = [A-]
[HA]
[HA] = cHA
Kdis = [H+]2
cHA
Kdis = KHA
KHA · cHA = [H+]2 /log
log (KHA · cHA ) = 2 · log [H+]
log KHA + log cHA = 2 · log [H+] / ½
½ log KHA + ½ log cHA = log [H+] / · (-1)
-½ log KHA - ½ log cHA = - log [H+]
- log KHA = pKHA
½ pKHA - ½ log cHA = pH → pH = ½ pKHA - ½ log cHA
pH = ½ (pKHA – log cHA)
pH of weak acids - calculations
1) 0.01 M acetic acid, Kdis = 1.8 x 10-5, pH = ?
2) 0.1 M lactic acid, pH = 2.4, Kdis = ?
3) dilution of a weak acid: c1 = 0.1 M → c2 = 0.01 M,
∆ pH = ?
pH of strong bases
Strong bases are those that are ionized completely:
Generally: BOH → B+ + OHi.e. NaOH ↔ Na+ + OHOther examples: KOH, LiOH, Ba(OH)2, Ca(OH)2
pOH = - log cBOH
pH = 14 - pOH
Calculations:
1) 0.01 M KOH, pH = ?
2) Strong bases pH: a) 11 c = ?
b) 10.3 c = ?
3) 0,1 M Ba(OH)2, pH = ?
4) 50 mL of a solution contains 4 mg of NaOH.
Mr (NaOH) = 40, pH = ?
pH of weak bases
Weak bases are those that react with water only to a
slight extent, producing relatively few OH- ions. The
weak base remains in the molecular form.
Generally: BOH ↔ B+ + OHKdis = [B+] x [OH-]
pKBOH = - log Kdis
[BOH]
e. g. NH3 + H2O ↔ NH4+ + OHpOH = ½ (pKBOH – log cBOH)
pH = 14 – pOH
Calculations:
1) 0.2 M NH4OH, pK = 4.74, pH = ?
2) 0.06 M dimethylamine, pK = 3.27, pH = ?
pH of buffers
Buffers are solutions that keep the pH value nearly
constant when acid or base is added.
Buffer solution contains:
• weak acid + its salt (i.e. CH3COOH + CH3COONa)
• weak base + its salt (i.e. NH4OH + NH4Cl)
• two salts of an oxoacid (i.e. HPO42- + H2PO41-)
• amphoteric compounds (i.e. aminoacids and proteins)
Henderson – Hasselbalch equation:
pH = pKA + log (cS x VS / cA x VA) → for acidic buffer
pOH = pKB + log (cS x VS / cB x VB)
pH = 14 – pOH → for basic buffer
pH of buffers - calculations:
1) 200 mL 0.5 M acetic acid + 100 mL 0.5 M sodium acetate → buffer,
pKA = 4.76, pH = ?
2) 20 mL 0.05 M NH4Cl + 27 mL 0.2 M NH4OH → buffer,
K = 1.85 x 10-5, pH = ?
3) The principal buffer system of blood is a bicarbonate buffer
(HCO3- / H2CO3). Calculate a ratio of HCO3- / H2CO3 components if
the pH is 7.38 and pK(H2CO3) = 6.1.