K A K B = K W pk A + pk B = 14

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Transcription:

Relationship between the ionization constants of an acid and its conjugate base HCN (aq) H 2 O(l) CN (aq) H O (aq) Conjugate couple The product between of an acid and of its conjugate base is : p p 14 1

H (aq) H 2 O (l) - (aq) H O (aq) [H O [ [H - (aq) H 2 O (l) H (aq) OH - (aq) [OH [H [ [H O [ [H 2

Interactions between the ionization constants of an acid and its conjugate base with H 2 O at 25 C HCN (aq) H 2 O (l) CN (aq) H O (aq) CN (aq) H 2 O (l) HCN (aq) OH (aq) 4.0x 5 Μ 2.5x Μ 2 H 2 O (l) H O (aq) OH (aq) 1.0x 14 Μ 2 [H O [CN [HCN [HCN[OH [CN [HO [OH p p p Therefore can be calculated: /

p of some acids and of their conjugate bases at 25 C: 14 p p Name of acid acido p base p Name of base forza dell acido crescente hydronium H O 0 H 2 O 14 water phosphoric H PO 4 2.12 H 2 PO 4 11.88 Dihydrogen phosphate fluoridirc HF.14 F.86 fluoride acetic CH COOH 4.74 CH COO 9.25 acetate carbonic H 2 CO 6.8 HCO 7.62 Hydrogen carbonate sulphidric H 2 S 7 HS 7 Hydrogen sulphide Dihydrogen phosphate H 2 PO 4 7.21 HPO 2 4 6.79 Hydrogen phosphate ammonium NH 4 9.25 NH 4.74 ammonia cianidric HCN 9.4 CN 4.6 cianide forza della base crescente Hydrogen carbonate HCO.2 CO 2.68 carbonate Hydrogen phosphate HPO 4 2 12.44 PO 4 1.56 phosphate water H 2 O 14 OH 0 hydroxide 4

Types of acid-base reactions e studied so far the reaction between acids and bases (strong and weak) with amphiprotic water has been studied. cids and bases can react in water to form a salt. acid base salt H 2 O type strong acid strong base strong acid weak base weak acid strong base weak acid weak base example HCl NaOH NaCl H 2 O HCl NH NH 4 Cl H 2 O CH COOH NaOH CH COONa H 2 O CH COOH NH CH COONH 4 H 2 O 5

Reaction between strong acid and strong base: neutralization Strong acids and bases are 0 % ionized in solution: HCl (aq) H 2 O NaOH (aq) H O (aq) Cl (aq) Na (aq) OH (aq) The net equation between HCl (strong acid) and NaOH (strong base) is : H O (aq) Cl (aq) Na (aq) OH (aq) 2 H 2 O (l) Na (aq) Cl (aq) H O (aq) OH (aq) 2 H 2 O (l) 1 1 1.0 [H O [OH The net ionic equation for the reaction of a strong acid with a strong base is always the union of a hydronium ion with a hydroxide ion to give water. Since 1 /, reagents are fully consumed to 1 /, reagents are fully consumed to yield products. If an equal amount of moles of NaOH and HCl, are mixed, the result is a neutral NaCl solution (ph 7.0 at 25 C) 14 M 2 6

hen the concentrations of strong acids and bases are much larger that [H O water self-ionization: if c c [H O 1.0x 7 M and ph 7 If c > c [H O c c and ph log (c c ) < 7 if c < c [OH c c and poh log (c c ) ph 14 poh > 7 7

Reaction between strong acid and weak base The weak base is not 0 % ionized in solution: HCl (aq) H 2 O (l) H O (aq) Cl (aq) H O (aq) NH (aq) Cl (aq) H 2 O (l) NH 4 (aq) Cl (aq) The net reaction of HCl (strong acid) e NH (weak base) is : H O (aq) NH (aq) H 2 O (l) NH 4 (aq) [NH [H O [NH 4 9 1.8 1 M 1 In this reaction H O is an acid stronger than NH 4 and NH is a base stronger than H 2 O. The reaction is strongly shifted toward the products. Mixing equimolar amounts of a strong acid with a weak base produces a salt whose cation is the conjugate acid of the weak base. The solution becomes acid, with a ph depending on the of the cation. The salt ammonium choride is formed: NH 4 Cl. 8

Reaction between weak acid and strong base In the reaction between a weak acid HCOOH, with a strong base NaOHnet ionic reaction is: HCOOH (aq) OH (aq) HCOO (aq) Η 2 Ο (l) [HCOO [HCOOH[OH 1.8 1.0 4 14 [HCOO [HO [HCOOH[OH [H O 1.8 M 1 [HCOO [HO [HCOOH In this reaction the hydroxide OH - is a base stronger than formiate HCOO - and formic acid HCOOH is an acid stronger than water: the reaction is strongly shifted toward the product. Mixing equimolar amounts of a strong acid with a weak base produces a salt whose cation is the conjugate acid of the weak base. The solution is basic with a ph that depends on the of the anion. The salt HCO 2 Na (sodium formiate) is formed. 9

Reaction between weak acid and weak base CH COOH (aq) NH (aq) CH COO (aq) NH 4 (aq) In this reaction product formation is favored since acetic acid is a stronger acid than the ammonium ion and ammonia is a stronger base than acetate: the reaction is strongly shifted to the right. Mixing of equimolar amounts of a weak base with a weak acid produces a salt whose anion is the conjugated base of the weak acid and cation is the conjugate acid of the weak base. The ph depends on the values of of the weak acid and of of the weak base. The saltammonium acetate is formed CH CO 2 NH 4

Chemistry of acids and bases Salt hydrolysis uffer solutions cid-base titrations 11

Hydrolytic equilibria The aqueous solution of a salt obtained from an acid and a strong base contains the free ionic species that form the salt. ase and acid (strong) Salt solution ph NaOH HCl NaCl H 2 O NaCl (s) Na (aq) Cl (aq) 7 NaOH HNO NaNO H 2 O NaNO (s) Na (aq) NaNO (aq) 7 OH HClO 4 ClO 4 H 2 O ClO 4 (s) (aq) ClO 4 (aq) 7 OH Hr r H 2 O r (s) (aq) r (aq) 7 Ca(OH) 2 2 HCl CaCl 2 H 2 O CaCl 2 (s) Ca 2 (aq) 2 Cl (aq) 7 In these solutions there is no proton exchange because both cation (conjugated acid of the strong base) and anion (conjugated base of the strong acid) do not tend to yield or subtract protons from water. The solution is therefore neutral. 12

e call hydrolysis those proton exchange reactions (acid-base) that are established in the aqueous solutions of salts formed by: weak acids and strong bases (which contain moderately strong conjugated bases and very weak conjugated acids): basic hydrolysis, ph > 7 weak bases and strong acids (which contain moderately strong conjugated acids and very weak conjugated bases) : acid hydrolysis, ph > 7 acids and bases both weak (which contain moderately strong conjugated acids and bases): basic, acid or neutral hydrolysis s a result of these exchanges of protons with the aqueous amphiprotic solvent there is a variation in the concentration of [H O and [OH - (ph) NH 4 Cl (s) NH 4 (aq) Cl (aq) NH 4 (aq) H 2 O (l) NH (aq) H O (aq) 1

Salt formed by weak acids and strong bases : basic hydrolysis The dissociation of the salt C is followed by the protonation equilibrium of the anion -, which is a moderately strong rønsted base. C (aq) C (aq) (aq) (aq) H 2 O (l) H (aq) OH (aq) CH COONa (aq) Na (aq) CH COO (aq) CH COO (aq) H 2 O (l) CH COOH (aq) OH (aq) CN (aq) (aq) CN (aq) CΝ (aq) H 2 O (l) HCN (aq) OH (aq) NO 2 (aq) (aq) NO 2 (aq) NO 2 (aq) H 2 O (l) HNO 2 (aq) OH (aq) Due to the hydrolytic equilibrium, the solution is enriched with OH - ions. 14

The equilibrium constant of this equilibrium is called hydrolysis constant i : (aq) H 2 O (l) H (aq) OH (aq) [H[OH [ i i is directly related to of the weak acid that is formed: [H[OH [H[OH [H O i [ [ [HO i [ OH c i S c S The weaker the acid and the higher the salt concentration, the more basic is the resulting solution. 15

lkaline hydrolisis ph determination formula [H[OH [H[OH [H O i [ [ [HO i The OH - arising from water selfprotolysis are negligible: [OH - [H nd if we consider that - is a weak base: [ - c s [ OH c i S c S Rearranging i [OH - 2 i c s 16

Salt formed by weak base and strong acid : acid hydrolysis The dissociation of the salt H is followed by the deprotonation equilibrium of the cation H, which is a moderately strong rønsted acid. H (aq) H (aq) (aq) H (aq) H 2 O (l) (aq) H O (aq) NH 4 Cl (s) NH 4 (aq) Cl (aq) NH 4 (aq) H 2 O (l) NH (aq) H O (aq) C 6 H 5 NH Cl (s) C 6 H 5 NH (aq) Cl (aq) C 6 H 5 NH (aq) H 2 O (l) C 6 H 5 NH 2 (aq) H O (aq) Due to the hydrolytic equilibrium, the solution is enriched with H O ions. The complete dissociation of H salt is accompanied by the deprotonation equilibrium of the H cation, which is a moderately strong rønsted acid. 17

The hydrolysis constant i : H (aq) H 2 O (l) (aq) H O (aq) i is related to of the weak base that is formed: [[H O [H [[H O [H [OH [OH i i [ H O ics cs The weaker the base and the higher the salt concentration, the more acidic is the resulting solution. 18

cid hydrolisis ph determination formula [[H O [H [[H O [H [OH [OH i i The H O arising from water selfprotolysis are negligible: [H O [H nd if we consider that H is a weak acid: [H c s Rearranging i [ H O ics cs [H O 2 i c s 19

Salt formed from a weak acid and base: hydrolysis The full dissociation of the salt H is accompanied by the dissociation of H and by protonation of - H (aq) H (aq) (aq) H (aq) H 2 O (l) (aq) H O (aq) (aq) H 2 O (l) H (aq) OH (aq) H O (aq) OH (aq) 2 H 2 O (l) H (aq) (aq) (aq) H (aq) 20

s usual we neglect hydrogen and hydroxide ions coming from water dissociation if the salt concentration is not too small. H (aq) (aq) (aq) H (aq) i H [ H O H H ph 7 if: H > ph < 7 H < ph > 7 21

Exercise 1. Calculate the ph of a 0.2 M of Na cetate solution ( 1.8 5 M a 25 C) CH COONa (aq) Na (aq) CH COO (aq) CH COO (aq) H 2 O (l) CH COOH (aq) OH (aq) e will have an alkaline hydrolisis [OH i # c S c S 1# 1.8# 14 5 # 0.2 1.05# 5 M poh 4.98 ph 14 4.98 9.02 22

Exercise 14. Calculate the ph of 0.1 M ammonium chloride solution ( 1.8 5 M a 25 C) NH 4 Cl (s) NH 4 (aq) Cl (aq) NH 4 (aq) H 2 O (l) NH (aq) H O (aq) e have an acid hydrolysis [H O ph 5.12 i c S c S 1 1.8 14 5 0.1 7.45 6 M 2

Exercise 15. Calculate the ph of a 0.05 M ammonium fluoride solution (ammonia 1.8 5 M; HF 7.2 4 M at 25 C) NH F (aq) NH 4 (aq) F (aq) NH 4 (aq) H 2 O (l) NH (aq) H O (aq) F (aq) H 2 O (l) HF (aq) OH (aq) H O (aq) OH (aq) 2 H 2 O (l) NH 4 (aq) F (aq) NH (aq) HF (aq) [H O ph 6.20 HF NH 7.2 1.8 4 5 1.0 14 6.2 7 M 24

Controlling ph: uffer solutions HCl to both solutions uffered solution ph 5.04 Not buffred solution ph 5.04 fter the addition of HCl, the ph of the buffered solution does not change (or changes slightly), while the nonbuffered solution drops to ph 2.. The buffered solution hampers ph variations. 25

How do we make a buffer solution? ph 0.87 ph 8.9 acetic acid 0.1 M Na acetate 0.1 M buffer 26

How do we make a buffer solution? buffer solution is an aqueous solution in which the ph does not vary appreciably upon small additions of strong acids or bases. To make a buffer solution we have to satisfy two conditions: we need the simultaneous presence of two species: an acid capable of reacting with OH - ions and a base capable reacting with H O ion. The acid and the base must not react with each other. buffer solution is usually prepared from approximately equal amounts of a conjugated acid-base pair: a weak acid and its conjugate base (e.g. acetic acid and acetate ion) a weak base and its conjugate acid (e.g. ammonia and ammonium ion) 27

How does a buffer solution work? In the acetic acid/acetate buffer, acetic acid (weak acid) is required to react and consume the added hydroxide ions : CH COOH (aq) OH (l) CH COO (aq) H 2 O (aq) [CH COO [CH COOH 1.8 5 9 1 1.8 M 14 [OH 1.0 The equilibrium constant for the reaction is very large because the OH - ion is a much stronger basis than acetate (CHCOO -). This means that all OH - ions, coming from an external source, are completely consumed. Similarly, each hydronium ion added to the solution will react with the acetate ion present in the buffer. CH COO (aq) H O (l) CH COOH (aq) H 2 O (aq) [CH COOH [CH COO 1 4 1 5.6 M 5 [HO 1.8 1 lso the equilibrium constant for this reaction is very large, since the H O ion is a much stronger acid than acetic acid (CH COOH). 28

eak acid Conjugate base (M) uffering range ph cetic acid CH COOH Common buffers in laboratory practice dhydrogen phosphate H 2 PO 4 hydrogen carbonate HCO hydrogen phosphate HPO 4 2 acetate 1.8 5.7 5.8 CH COO hydrogen phosphate HPO 4 2 carbonate CO 2 phosphate PO 4 uffer system in the blood bicarbonate hemoglobin phosphate 6.2 8 6.2 8.2 4.8 11 9. 11..6 1 11. 1. 29

The Henderson-Hasselbalch equation ph p log [coniugate base [acid The ph of a buffer solution is controlled by two factors: the strength of the acid (indicated by o p ) the relative amounts of the acid and of its conjugate base L.J. Henderson (1878 1942) The ph of a buffer solution is mainly determined by the value of p and the fine ph control is achieved by modifying the relative amounts of conjugate base and acid. [conjugate base if >1 ph > p [acid [conjugate base if 1 ph p [acid [conjugate base if <1 ph < p [acid 0

The Henderson-Hasselbalch Equation In a buffer solution containing a weak acid H and its conjugate base : H (aq) H 2 O (l) (aq) H O (aq) [H O [H [ [acid [conjugate base Let s use the co-logarithmic expression: log [H O log log [ [acid [conjugate base log log [conjugate base [acid [H O [H ph p log [conjugate base [acid 1