ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor

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1 ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate acid Conjugate base H 2 O(l) + NH 3 (aq) NH 4+ (aq) + OH - (aq) Acid Base Conjugate acid Conjugate base 1

2 Which of the following are conjugate acid-base pairs? A) HCl, NaOH B) H 2 O, OH - C) H 2 SO 4, SO 4 2- D) H 2 SO 3, HSO 3 - E) HClO 4, ClO 3 - F) H 3 C-NH 2, H 3 C-NH 3 + Autoionization of water Water is amphoteric as it can behave both as acid and base 2 H 2 O(l) H 3 O + (aq) + OH - (aq) Ion-product constant for water: K w = [H 3 O + ][OH - ] = [H + ][OH - ] In pure water at 25 ºC: [H + ] = [OH - ] = mol/l K w = ( mol/l) ( mol/l) = mol 2 /L 2 Constant! 2

3 ph ph= -log 10 (activity of H + ) poh= -log 10 (activity of OH - ) Ion-product of water (constant!): ph + poh = 14 Activity = f. c f is activity coefficient, f<1, c is molar concentration E.g.: ph=7 (neutral): [H + ] = 10-7 M = mol/l ph=1 (acidic): [H + ] = 10-1 M = 0.1 mol/l ph=13 (alkaline): [H + ] = M = mol/l Strong acid E.g. HCl, HNO 3, H 2 SO 4 In aqueous solution fully dissociates to H + and A ph of strong acid can be calculated as ph = log (f [H + ]) For HCl: [H + ]= [HCl] For H 2 SO 4 : [H + ]= 2 [H 2 SO 4 ] 3

4 Calculating the ph of strong acid solutions Example: Calculate the ph of 0.06 mol/l HCl. ph = log0.06 = 1.22 If activity coefficient is known/given: e.g. f = 0.879, then ph = log( ) = 1.28 Calculating the ph of strong acid solutions Example 2: Calculate the ph of 0.02 mol/l H 2 SO 4. ph = log0.04 = = ~ 1.4 4

5 Calculating the ph of strong acid solutions Example 3: Calculate the ph of M HCl. ph = log(10-10 ) = 10. alkaline? Water contributes more protons than HCl in this case (10-7 M), ph will be the same as in pure water, i.e. 7 Strong base LiOH, NaOH, KOH, Ca(OH) 2, Ba(OH) 2 In aqueous solution fully dissociate to metal ion and OH ph of strong base can be calculated as poh = log (f [OH - ]) ph = 14 poh = 14 ( log (f [OH - ]) For NaOH: [OH - ]= [NaOH] For Ba(OH) 2 : [OH - ]= 2 [Ba(OH) 2 ] 5

6 Calculating the ph of strong base solutions Example: a) Calculate the ph of NaOH 0.5 mol/l. ph = 14 ( log0.5) = ~13.7 b) If this solution is diluted 10-fold, what will be the resulting ph? ph = ~12.7 Weak acid E.g. H 2 CO 3, CH 3 COOH Only some small fraction of molecules in solution dissociates to anion and proton: CH 3 COOH CH 3 COO + H + K d [CH 3 COO ] [H + ] = [CH 3 COOH] ph = ½ pk ½ log [AH] pk = log K d 6

7 If we know pk (K d ) and concentration of a weak acid solution, we can calculate (predict) ph of the solution: ph = ½ pk ½ log [AH] If we measure ph of a weak acid solution of a known concentration, we can determine its pk (K d ): pk = 2 ph + log [AH] Reading pk of weak acid from titration curve Inflection point: If weak acid is just half-titrated, then ph = pk [CH 3 COO ]. [H + ] K d = [CH 3 COOH] 7

8 Calculating the ph of weak acid solutions Example: Calculate the ph of 0.01 mol/l acetic acid. K a = ph = ½ pk ½ log [AH] pk = log( ) = ph = ½ ½ log0.01 = = ( 1) = Calculating the ph of weak acid solutions Example 2: Calculate the ph of 0.1 mol/l hypochlorous acid. K a = ph = ½ pk ½ log [AH] pk = log( ) = ph = ½ ½ log0.1 = = ( 0.5) =

9 Titration Reaction: A + B C Substance A: unknown concentration, amount (solution volume) known Substance B: known concentration, is used to determine concentration of A added gradually to A until the reaction is just complete, and the consumed amount is recorded an indicator is needed to show that the reaction has reached completion Types of titration Neutralisation reaction (acid-base titration) HCl + NaOH NaCl + H 2 O ionic: H + + OH - H 2 O indicator e.g. phenolphthalein (in acid colourless, but violet in alkali Precipitation reaction Redox reaction 9

10 Titration calculations Example: An unknown sample of sulfuric acid H 2 SO 4 was titrated with the known KOH solution. It was found that 12 ml of the KOH c=0.1 mol/l was needed for just complete neutralisation of 10 ml H 2 SO 4 unknown sample. What is concentration of sulfuric acid in the sample? Equation: H 2 SO KOH K 2 SO H 2 O Calculation: H 2 SO 4 KOH c 1. v 1 = c 2. v 2 c 1 = c 2. v 2 / v 1 c 1 = / 10 = 0.12 Including stoichiometry : c(h 2 SO 4 ) = 0.12/2 = 0.06 mol/l Weak base E.g. NH 3 (aq), organic amines A fraction of molecules in aqueous solution accepts proton from water: NH 3 (aq) + H 2 O NH 4+ + OH - K d [NH 4+ ] [OH - ] = [NH 3 ] poh = ½ pk ½ log [B] ph = 14 poh = 14 ½ pk + ½ log [B] 10

11 Calculating the ph of weak base solutions Example: Calculate the ph of 5 mol/l aqueous ammonia. K b = ph = 14 ½ pk b + ½ log [B] pk b = log( ) = ph = 14 ½ ½ log5 = = = Hydrolysis of salts Reaction of dissolved salts with water, e.g.: A) Anion from a strong acid, cation from a weak base, e.g. NH 4 Cl: NH 4+ + H 2 O NH 3 + H 3 O + ph is acidic Cl - + H 2 O no reaction B) Anion from a weak acid, cation from a strong base, e.g. NaHCO 3 : Na + + H 2 O no reaction ph is alkaline HCO H 2 O H 2 CO 3 + OH - 11

12 ph of salt will be: strong acid e.g. HCl Anion is from weak acid e.g. H 2 CO 3 Cation is from strong base e.g. NaOH weak base e.g. NH 4 OH NEUTRAL e.g. NaCl ACIDIC e.g. NH 4 Cl BASIC e.g. NaHCO 3 ACIDIC/BASIC (depends on pk ) e.g. NH 4 HCO 3 Calculate the ph of 0.5 mol/l sodium hydrogen carbonate, NaHCO 3. The K a1 of carbonic acid is ph = 14 ½ pk b + ½ log [B] pk a = log( ) = pk b = = ph = 14 ½ ½ log0.5 = = ( ) =

13 Buffers Mixture of weak acid and its conjugate base (salt), e.g.: CH 3 COOH + CH 3 COONa NaH 2 PO 4 + Na 2 HPO 4 Mixture of weak base and its conjugate acid (salt), e.g.: Tris + TrisHCl Tris stands for: Tris(hydroxymethyl)aminomethane Henderson-Hassebalch equation ph = pk a [A ] + log [AH] pk a negative log of dissociation constant of the weak acid [A ] substance concentration of the salt/conjugate base [AH] substance concentration of the weak acid 13

14 10 Titration curve of sodium phosphate buffer 9 ph of buffer ph = pka = :1 8:2 7:3 6:4 1:1 4:6 3:7 2:8 1:9 Acid : base ratio How buffer works E.g. phosphate buffer: NaH 2 PO 4 + Na 2 HPO 4 ph = pk a + log Addition of strong acid: Na 2 HPO 4 + HCl NaH 2 PO 4 + NaCl Ionic: HPO H + H 2 PO 4 - ph = pk a + log 14

15 How buffer works E.g. phosphate buffer: NaH 2 PO 4 + Na 2 HPO 4 ph = pk a + log Addition of strong base: NaH 2 PO 4 + NaOH Na 2 HPO 4 + H 2 O Ionic: H 2 PO OH - HPO H 2 O ph = pk a + log Calculating the ph of buffer Example 1: Calculate the ph of acetate buffer 0.1 mol/l, in which the acid:salt ratio is 2:3. K a = ph = pk a + log [A - ]/[AH] pk = log( ) = ph = log 3/2 = = = =

16 Calculating the ph of buffer Example 2: Calculate the ph of sodium phosphate buffer that originated from mixing 150 ml of 0.1 M NaH 2 PO 4 and 250 ml of 0.05 M Na 2 HPO 4. pk a = ph = pk a + log [A - ]/[AH] NaH 2 PO 4 : 0.1 mol/l 0.15 = mol Na 2 HPO 4 : 0.05 mol/l 0.25 = mol ph = log /0.015 = = ( ) = = ~7.13 Calculating the ph of buffer Example 3: 500 ml of Tris free base 0.1 M was combined with 20 ml of 1 M HCl. What is predicted ph of the resulting Tris/HCl buffer? Tris pk b = ph = pk a + log [A - ]/[AH] pk a = 14 pk b Tris free base: 0.05 mol HCl: 0.02 mol ph = ( ) + log ( )/0.02 = = = = = ~8.2 16

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