1. Chemical calculations used in medicine part 2 Pavla Balínová

2. Acid – base theories Arrhenius theory acid = substance that is able to lose H + (CH 3 COOH ↔ CH 3 COO - + 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 + H 2 O ↔ H 3 O + + Cl - acid 1 base 2 acid 2 base 1 conjugated pair 1

3. Water H 2 O The purest water is not all H 2 O → about 1 molecule in 500 million transfers a proton H + to another H 2 O molecule, giving a hydronium ion H 3 O + and a hydroxide ion OH - : H 2 O + H 2 O ↔ H 3 O + + OH - = dissociation of water The concentration of H 3 O + 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 H 3 O + and OH - ions. Equilibrium constant of water: K eq = [H 3 O + ]. [OH - ] [H 2 O] 2 Ion product of water K w : K w = [H 3 O + ]. [OH - ] K w = 1.10 -7. 1.10 -7 = 1.10 -14

4. Ion product of water K w = [H 3 O + ]. [OH - ] 10 –14 = [H 3 O + ]. [OH - ] / log log(a. b) = log a + log b log 10 –14 = log ([H 3 O + ]. [OH - ]) log 10 –14 = log [H 3 O + ] + log [OH - ] -14 = log [H 3 O + ] + log [OH - ] / · ( -1 ) ‏ 14 = - log [H 3 O + ] - log [OH - ] -log K W pH pOH - log K W = pK W pK W = pH + pOH = 14

5. Addition of an acid to pure water → increasing H 3 O + and OH - will fall until the product equals 1.10 -14. Addition of a base to pure water → increasing OH - and H 3 O + will fall until the product equals 1.10 -14. Example: Lemon juice has a [H 3 O + ] of 0.01 M. What is the [OH - ] ? [H 3 O + ]. [OH - ] = 1.10 -14 1.10 -2. [OH - ] = 1.10 -14 / : 1.10 -2 [OH - ] = 1.10 -12

6. pH scale Exponential numbers express the often minute actual concentration of H 3 O + 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/[H 3 O + ] = - log [H 3 O + ] e. g. The pH of a solution whose [H 3 O + ] is 1.10 -4 equals 4.

7. Representative pH values 9.0 – 10.0hand soap 7.35 – 7.45blood 7.0pure water 6.0urine 5.0coffee 2.9vinegar 1.5 – 2.0gastric juice 0.5lead-acid battery pH Substance

8. pH of strong acids Strong acids are those that react completely with water to form H 3 O + and anion (HCl, H 2 SO 4, HClO 4, HNO 3,…) ‏ Generally: HA → H + + A - e. g. HCl + H 2 O ↔ H 3 O + + Cl - pH = - log c H + = - log c HA 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 H 2 SO 4, pH = ? 4) Dilution of a strong acid: c 1 = 0,1 M → c 2 = 0,01 M, ∆ pH = ?

9. pH of weak acids Weak acids react only to a slight extent with water to form relatively few H 3 O + ions. Most of the molecules of the weak acids remain in the molecular form (uncharged) ‏ HA → H + + A - K dis = [H + ].[A - ] [H + ] = [A - ] [HA] [HA] = c HA K dis = [H + ] 2 c HA K dis = K HA K HA · c HA = [H + ] 2 /log log (K HA · c HA ) = 2 · log [H + ] log K HA + log c HA = 2 · log [H + ] / ½ ½ log K HA + ½ log c HA = log [H + ] / · (-1) ‏ -½ log K HA - ½ log c HA = - log [H + ] - log K HA = pK HA ½ pK HA - ½ log c HA = pH → pH = ½ pK HA - ½ log c HA pH = ½ (pK HA – log c HA ) ‏

10. pH of weak acids - calculations 1) 0.01 M acetic acid, K dis = 1.8 x 10 -5, pH = ? 2) 0.1 M lactic acid, pH = 2.4, K dis = ? 3) dilution of a weak acid: c 1 = 0.1 M → c 2 = 0.01 M, ∆ pH = ?

11. pH of strong bases Strong bases are those that are ionized completely: Generally: BOH → B + + OH - i.e. NaOH ↔ Na + + OH - Other examples: KOH, LiOH, Ba(OH) 2, Ca(OH) 2 pOH = - log c BOH 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 = ?

12. 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 + + OH - K dis = [B + ] x [OH - ] pK BOH = - log K dis [BOH] e. g. NH 3 + H 2 O ↔ NH 4 + + OH - pOH = ½ (pK BOH – log c BOH ) pH = 14 – pOH Calculations: 1) 0.2 M NH 4 OH, pK = 4.74, pH = ? 2) 0.06 M dimethylamine, pK = 3.27, pH = ?

13. 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. CH 3 COOH + CH 3 COONa) ‏ weak base + its salt (i.e. NH 4 OH + NH 4 Cl) ‏ two salts of an oxoacid (i.e. HPO 4 2- + H 2 PO 4 1- ) ‏ amphoteric compounds (i.e. aminoacids and proteins) Henderson – Hasselbalch equation: pH = pK A + log (c S x V S / c A x V A ) → for acidic buffer pOH = pK B + log (c S x V S / c B x V B ) pH = 14 – pOH → for basic buffer

14. pH of buffers - calculations: 1) 200 mL 0.5 M acetic acid + 100 mL 0.5 M sodium acetate → buffer, pK A = 4.76, pH = ? 2) 20 mL 0.05 M NH 4 Cl + 27 mL 0.2 M NH 4 OH → buffer, K = 1.85 x 10 -5, pH = ? 3) The principal buffer system of blood is a bicarbonate buffer (HCO 3 - / H 2 CO 3 ). Calculate a ratio of HCO 3 - / H 2 CO 3 components if the pH is 7.38 and pK(H 2 CO 3 ) = 6.1.